When the factors of a trinomial are (x+p) and (x-q) then the coefficient of the x-term in the trinomial is:

The area of the slice of the pizza will be 13.5π square inches. So the correct answer is option B.

The space occupied by any two-dimensional figure in a plane is called the area. The space occupied by the circle in a two-dimensional plane is called the area of the circle.

Given that:-

slice of pizza from a large pizza with a radius of 9 inches cut into 6 slices

The area of the slice will be 1 / 6 th of the total area of the pizza since the pizza is cut into six parts.

Area of slice = ( 1 / 6 ) x π ( 9² )

Area of slice = 13.5π square inches.

Therefore the area of the slice of the pizza will be 13.5π square inches.

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Answer:

D. (x, y, z) = (1, 1, 2). That is:

[tex]\left\{\begin{aligned}&x = 1\\ & y = 1 \\&z = 2\end{aligned}\right.[/tex].

Step-by-step explanation:

Step One: Make sure that the first coefficient of the first row is 1. In this case, the coefficient of [tex]x[/tex] in the first row is already 1.

Step Two: Using row 1, eliminate the first unknown of row 1 [tex]x[/tex] in the rest of the rows. For example, to eliminate [tex]x[/tex] from row 2, multiply row 1 by the opposite of the coefficient of [tex]x[/tex] in row 2 and add that multiple to row 2. The coefficient of [tex]x[/tex] in row 2 is [tex]3[/tex]. Thus, multiply row 1 by [tex]-3[/tex] to get its multiple:

[tex]-3x - 3y + 3z = 0[/tex].

Add this multiple to row 2 to eliminate [tex]x[/tex] in that row:

[tex]\begin{array}{lrrrcr}&-3x &-3y&+3z& =&0 \\ + & 3x& -y& +z& =& 4\\\cline{1-6}\\[-1.0em]\implies&&-4y &+ 4z&=&4\end{array}[/tex].

Similarly, for the third row, multiply row 1 by [tex]-5[/tex] to get:

[tex]-5x - 5y + 5z = 0[/tex].

Do not replace the initial row 1 with this multiple.

Add that multiple to row 3 to get:

[tex]\begin{array}{lrrrcr}&-5x &-5y&+5z& =&0 \\ + & 5x& & +z& =& 7\\\cline{1-6}\\[-1.0em]\implies&&-5y &+6z&=&7\end{array}[/tex].

After applying step one and two to all three rows, the system now resembles the following:

[tex]\left\{\begin{array}{rrrcr}x& + y & -z& = &0\\&-4y &+4z&=&4\\ &-5y &+6z&=&7\end{array}\right.[/tex].

Ignore the first row and apply step one and two to the second and third row of this new system.

[tex]\left\{\begin{array}{rrcr}-4y &+4z&=&4\\ -5y &+6z&=&7\end{array}\right.[/tex].

Step One: Make sure that the first coefficient of the first row is 1.

Multiply the first row by the opposite reciprocal of its first coefficient.

[tex]\displaystyle (-\frac{1}{4})\cdot (-4y) + (-\frac{1}{4})\cdot 4z = (-\frac{1}{4})\times 4[/tex].

Row 1 is now [tex]y - z = -1[/tex].

Step Two: Using row 1, eliminate the first unknown of row 1 [tex]y[/tex] in the rest of the rows.

The coefficient of [tex]y[/tex] in row 2 is currently [tex]-5[/tex]. Multiply row 1 by  [tex]5[/tex] to get:

[tex]5y - 5z = -5[/tex].

Do not replace the initial row 1 with this multiple.

Add this multiple to row 2:

[tex]\begin{array}{lrrcr}&-5y&+6z& =&7 \\ + & 5y& -5z& =& -5\\\cline{1-5}\\[-1.0em]\implies&&z &= &2\end{array}[/tex].

The system is now:

[tex]\left\{\begin{array}{rrcr}y & - z&=&-1\\& z &=&2\end{array}\right.[/tex].

Include the row that was previously ignored:

[tex]\left\{\begin{array}{rrrcr}x& + y & -z& = &0\\&y &-z&=&-1 \\ & &z&=&2\end{array}\right.[/tex].

This system is now in a staircase form called Row-Echelon Form. The length of the rows decreases from the top to the bottom. The first coefficient in each row is all [tex]1[/tex]. Find the value of each unknown by solving the row on the bottom and substituting back into previous rows.

From the third row: [tex]z = 2[/tex].

Substitute back into row 2:

[tex]y -2 = -1[/tex].

[tex]y = 1[/tex].

Substitute [tex]y = 1[/tex] and [tex]z = 2[/tex] to row 1:

[tex]x + 1 - 2 = 0[/tex].

[tex]x = 1[/tex].

In other words,

[tex]\left\{\begin{aligned}&x = 1\\ & y = 1 \\&z = 2\end{aligned}\right.[/tex].

Answer:

[tex]6+\frac{i}{3}[/tex]

Step-by-step explanation:

[tex]\frac{1}{3\imath}-(-6+\frac{2}{3\imath})[/tex]

[tex]\frac{1}{3\imath}+6-\frac{2}{3\imath}[/tex]

taking like terms together

[tex]\frac{1}{3\imath}-\frac{2}{3\imath}+6[/tex]

taking LCM

[tex]\frac{1-2}{3\imath}+6[/tex]

[tex]\frac{-1}{3\imath}+6[/tex]

taking LCM

[tex]\frac{-1+18\imath}{3\imath}[/tex]

splitting the term

[tex]\frac{-1+18\imath}{3\imath}[/tex]

splitting the term

[tex]-\frac{1}{3\imath}+\frac{18\imath}{3\imath}[/tex]

[tex]-\frac{1\times3\imath}{3\imath \times \imath}+6[/tex]

[tex]-\frac{i}{3\imath^2}+6[/tex]

we know that

[tex]\imath^2=-1[/tex]

putting this value in above equation

[tex]\frac{\imath}{3}+6[/tex]

Answer:

Step-by-step explanation:

The formula of a distance between two points:

[tex]d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]

We have the points (3, 7) and (15, 16). Substitute:

[tex]d=\sqrt{(15-3)^2+(16-7)^2}=\sqrt{12^2+9^2}=\sqrt{144+81}=\sqrt{225}=15[/tex]

[tex]x^5 + x^4 - 7x^3 - 7x^2 -144x - 144 = 0 \\x^4(x+1)-7x^2(x+1)-144(x+1)=0\\(x^4-7x^2-144)(x+1)=0\\\\x+1=0\Rightarrow x=-1\\\\x^4-7x^2-144=0\\x^4-16x^2+9x^2-144=0\\x^2(x^2-16)+9(x^2-16)=0\\(x^2+9)(x^2-16)=0\\(x^2+9)(x-4)(x+4)=0\\\\x^2+9=0\vee x-4=0 \vee x+4=0\\x=4 \vee x=-4\\\\x\in\{-4,-1,4\}[/tex]

Answer:

{ -1, -3i,3i,4,-4}

Step-by-step explanation:

I'm going to try to get the Rational Root Theorem to work for us.

Since the coefficient of leading term is 1 we just need to look at the factors of the constant.

Possible rational zeros are going to be the factors of -144.

So here are some possible rational zeros: 1,2,3,4,6,8,9,12,16,18,24,36,48,72 and also the negative version of these numbers are numbers we must consider.

I'm going to see if -1 works.  

(-1)^5+(-1)^4-7(-1)^3-7(-1)^2-144(-1)-144

-1   +   1     +7      -7       +144  -144=0

So -1 is a zero so x+1 is a factor. I'm going to use synthetic division to see what multiplies to x+1 that will me the initial polynomial expression we had.

-1   |    1       1      -7        -7      -144    -144

     |           -1      0          7        0       144

     | ________ __________________

          1      0     -7         0       -144      0

So the (x+1)(x^4-7x^2-144)=0

The cool thing is that other factor is a sort of quadratic in disguise. That is it becomes a quadratic if you let u=x^2. So let's do that.

u^2-7u-144=0

(u+9)(u-16)=0

u=-9 or u=16

So x^2=-9 or x^2=16.

Square rooting both sides gives us:

[tex] x= \pm 3i \text{ or } x=\pm 4 [/tex]

So the solution set is { -1, -3i,3i,4,-4}

Answer:

Step-by-step explanation:

Givens

Tan(60) = opposite / adjacent

adjacent = 26

Cos(39) = adjacent / hypotenuse

Solution

Tan(60) = opposite (which is the height) / 26            Multiply both sides by 26

26*tan(60) =  opposite                                                  Multiply the left.

tan60 = 1.732

26 * 1.732 = opposite

45.033

====================                

Cos(39) = adjacent / hypotenuse

cos(39) = 0.7771

0.7771 = 45.0333 / x              Multiply both sides by x

0.7771*x = 45.0333                Divide by 0.7771

x = 45.0333 / 0.7771

x = 57.94

Answer:

option C

Step-by-step explanation:

We  have to find the value of x

[tex]tan\theta=\frac{perpendicular side}{base}[/tex]

[tex]tan60^{\circ}=\frac{height}{26}[/tex]

We know that [tex] tan60^{\circ}=\sqrt3[/tex]

[tex]\sqrt3=\frac{height}{26}[/tex]

[tex]height=1.732\times 26[/tex]

Where [tex]\sqrt3=1.732[/tex]

Height=45.032

Height=45.0

[tex]cos\theta=\frac{bas}{hypotenuse}[/tex]

[tex] cos 39^{\circ}=\frac{45}{x}[/tex]

[tex]0.777=\frac{45}{x}[/tex]

[tex]x=\frac{45}{0.777}[/tex]

x=57.9

Hence, option C is true.

The segment bisects \AEC because \AEB and \BEC are both 45\u00b0, proven using the Angle Addition Postulate and the Subtraction Property of equality.

To prove that the segment bisects \AEC, begin by acknowledging the given information that m\AEB = 45\u00b0 and \AEC is a right angle with a measure of 90\u00b0. According to the Angle Addition Postulate, m\AEB + m\BEC = m\AEC. Substitute the known values to get 45\u00b0 + m\BEC = 90\u00b0. Utilizing the Subtraction Property of equality allows us to solve for m\BEC, finding it to also be 45\u00b0. This means that \BEC ≅ \AEB which leads us to conclude that since they have equal measures, the segment indeed bisects \AEC.

Answer:

210

Step-by-step explanation:

Here comes the problem from Combination.

We are being asked to find the number of ways out in which 3 students may sit on 7 seats in a row. Please see that in this case the even can not be repeated.

Let us start with the student one. For him all the 7 seats are available to sit. Hence number of ways for him to sit = 7

Let us see the student second. For him there are only 6 seats available to sit as one seat has already been occupied. Hence number of ways for him to sit = 6

Let us see the student third. For him there are only 5 seats available to sit as two seat has already been occupied. Hence number of ways for him to sit = 5

Hence the total number of ways for three students to be seated will be

7 x 6 x 5

=210

Answer:

GCF is 4ab

And the expression will be: 4ab ( 3a^2+2ab-5b^2)

Step-by-step explanation:

Factor the GCF :

12a^3b + 8a^2b^2-20 ab^3

We need to find the common terms that are common in each of the term given above

12,8 and 2 are all divisible by 4

a is common in all terms and b is also common in all terms,

So, GCF is 4ab

Taking 4ab common

12a^3b + 8a^2b^2-20 ab^3=4ab ( 3a^2+2ab-5b^2)

Answer:

Part A) The system of inequalities is

[tex]x\geq2[/tex]  and  [tex]y\geq2[/tex]

Part B) In the procedure

Part C) The schools that Natalie is allowed to attend are A,B and D

Step-by-step explanation:

Part A: Using the graph above, create a system of inequalities that only contain points C and F in the overlapping shaded regions

we have

Points C(2,2), F(3,4)

The system of inequalities could be

[tex]x\geq2[/tex] -----> inequality A

The solution of the inequality A is the shaded area at the right of the solid line x=2

[tex]y\geq2[/tex] -----> inequality B

The solution of the inequality B is the shaded area above of the solid line y=2

see the attached figure N 1

Part B: Explain how to verify that the points C and F are solutions to the system of inequalities created in Part A

we know that

If a ordered pair is a solution of the system of inequalities, then the ordered pair must satisfy both inequalities

Verify point C

C(2,2)    

Inequality A

[tex]x\geq2[/tex] -----> [tex]2\geq2[/tex] ----> is true

Inequality B

[tex]y\geq2[/tex] ------> [tex]2\geq2[/tex] ----> is true

therefore

Point C is a solution of the system of inequalities

Verify point D

F(3,4)    

Inequality A

[tex]x\geq2[/tex] -----> [tex]3\geq2[/tex] ----> is true

Inequality B

[tex]y\geq2[/tex] ------> [tex]4\geq2[/tex] ----> is true

therefore

Point D is a solution of the system of inequalities

Part C: Natalie can only attend a school in her designated zone. Natalie's zone is defined by y < −2x + 2. Explain how you can identify the schools that Natalie is allowed to attend.

we have

[tex]y < -2x+2[/tex]

The solution of the inequality is the shaded area below the dotted line [tex]y=-2x+2[/tex]

The y-intercept of the dotted line is the point (0,2)

The x-intercept of the dotted line is the point (1,0)

To graph the inequality, plot the intercepts and shade the area below the dotted line

see the attached figure N 2

therefore

The schools that Natalie is allowed to attend are A,B and D

The airplane traveled 855 miles in 9 hours.

Based on the given conditions, formulate 9x 475-:5

Reduce the fraction to the lowest term by canceling the

greatest common factor: 9x95

Calculate the first two terms: 855

The answer is 855 miles.

Problem-solving is the act of defining a problem; figuring out the purpose of the trouble; identifying, prioritizing, and selecting alternatives for an answer; and imposing an answer.

Problem-solving starts with identifying the issue. As an example, a trainer may need to parent out a way to improve a scholar's overall performance on a writing talent test. To do that, the instructor will overview the writing exams looking for regions for improvement.

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Answer:

B)

Step-by-step explanation:

First, lets try to build the second equation, it has the general shape

[tex]y=m *x+b[/tex]

Where b is a constant (a number) and m the slope, another constant.

Using the given condition (0,-7) you can find the first constant of our second equation, just put "0" where you see "x", and put -7 where there is a "y".

That gives us that b is equals to -7.

Now we only need to know that the condition for 2 linear equations to be parallels is that their slope have to be the same or multiple between each others, this means that "m" of our second equation has to be equal to 1/2

thus, our second equation is [tex]y=\frac{1}{2}*x-7[/tex]

Another way to see this is that you can compare two linear equations y1=m1*x1 + b1, and y2=m2*x2+b2, if these two intersect in somewhere, this condition should meet:

X= (b1-b2)/(m2-m1).

If the slopes are the same, the above equation gives us an error, meaning that the linear equations are, in fact parallels.

Answer:

No.

Step-by-step explanation:

The circle of that equation lies on the point (3,1) as its furthest point to the right (x-axis) and therefore could never also lie on the point (3,-1).

Your equation is in center radius form, which is as follows:

[tex](x-h)^2+(y-k)^2=r^2[/tex]

Noting that your radius is 4 (since your radius squared is 16).

To graph your circle, simply go to your origin (h, k) which in this case is (-1, 1) and then count out in any direction (up, down, left, or right -- no diagonals) 4 units (since your radius is 4). This will give you the four outermost edges of your circle. Simply fill in the gaps from there, and you'll have sketched your circle.

No. The circle of that equation lies on the point (3,1) as its furthest point to the right (x-axis) and thus could never  lie on the point (3,-1).

A circle is a shape consisting of all points in a plane that are given the same distance from a given point called the center.

The equation is in center radius form, which is as follows:

(x - h)² + (y - k)² = r²

the radius is 4 (since your radius squared is 16).

Substituting 3 for x and -1 for y:

(3 + 1)^2 + (-1 - 1)^2 = 16

4^2 + (-2)^2 = 16

16 + 4 = 16

20 = 16

Hence, we can see 20 is clearly not equal to 16, so we know the point (3, -1) can’t lie on that circle.

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The answer is:

The value of Tan(x) is:

[tex]\frac{7}{24}[/tex]

Since we want to know the value of Tan(x), we need to use the following formula:

[tex]Tan(x)=\frac{Opposite}{Adjacent(XZ)}[/tex]

We already know the opposite side, but we need to find the value of the distance "XZ",so, we can calculate it using the Pythagorean Theorem since it's a right triangle.

Let be:

[tex]a=25\\b=7[/tex]

Then, finding "XZ" we have:

[tex]a^{2}=b^{2}+c^{2}\\ \\25^{2}=7^{2}+c^{2} \\\\c^{2}=625-49\\\\c=\sqrt{576}=24[/tex]

Now, finding the value of Tan(x), we have:

[tex]Tan(x)=\frac{opposite}{adjacent}\\\\Tan(x)=\frac{7}{24}[/tex]

Hence, we have that the value of Tan(x) is:

[tex]\frac{7}{24}[/tex]

Have a nice day!

Answer:

Tan X = 7/24

Step-by-step explanation:

From the figure we can see a right angled triangle XYZ.

XY = 25 and YZ = 7

To find value of XZ

XZ² = XY² - YZ²

 = 25² - 7²

 = 625 - 49

 = 576

XZ = √576 = 24

To find the value of X

Tan X = YZ/XZ

 = 7/24

[tex]\bf \begin{cases} 2.34\\ 2.341 \end{cases}\implies \begin{cases} 2.340\\ 2.341 \end{cases}\implies \begin{cases} 2.3400\\ 2.3410 \end{cases} \\\\[-0.35em] ~\dotfill\\\\ \boxed{2.3400}\rule[0.35em]{8em}{0.25pt}2.3403\rule[0.35em]{10em}{0.25pt}\boxed{2.3410}[/tex]

Answer:

7

Step-by-step explanation:

Those are alternate interior angles. Alternate interior angles are the ones that happen at different intersections along the transversal but on opposite sides while inside the lines the transversal goes through. If these lines are parallel, then the alternate interior angles are congruent. Same thing the other way around. Alternate interior angles being congruent implies those lines are parallel.

So we are looking to solve:

3x-2=2x+5

Subtract 2d on both sides:

x-2=5

Add 2 on both sides:

x=7

The phrase "A || B" in a mathematical context usually indicates that line A is parallel to line B. To find the value of x that makes A parallel to B, we must generally consider the properties of parallel lines, specifically in relation to their slopes.
If A and B are lines in a coordinate plane, they will be parallel if and only if their slopes are equal and they are not the same line (in which case they would be coincident). If you are given the equations of the lines in the form y = mx + b, where m represents the slope and b represents the y-intercept, then A and B are parallel if the m values of both equations are the same.
Here's how you would generally find x when A and B are lines defined by equations:
1. Start with the equations of lines A and B. Both will typically be in a format that allows you to solve for their slopes.
   - Line A's equation might look like y = mx + c, where m is the slope of line A.
   - Line B's equation might look like y = nx + d, where n is the slope of line B.
2. Set their slopes equal to each other, as parallel lines have the same slope. This would give you the equation:
   m = n
3. If either slope contains a variable x that you need to solve for, create an equation for x by equating the two slopes:
   mx = nx
4. Solve for x. If the slopes contain x, they might be presented in a linear equation with x, or they could involve more complex expressions.
This is a high-level overview since the specifics would depend greatly on the actual equations of lines A and B. Without the explicit equations for lines A and B, I cannot give a numerical answer. If the question provided equations for lines A and B, please provide them, and I will assist you in finding the value of x that makes A parallel to B.

Answer:

[tex]\large\boxed{y-5=-\dfrac{2}{3}(x+3)}[/tex]

Step-by-step explanation:

The point-slope form of an equation of a line:

[tex]y-y_1=m(x-x_1)[/tex]

m - slope

(x₁, y₁) - point on a line

We have the slope [tex]m=-\dfrac{2}{3}[/tex] and the point [tex](-3,\ 5)[/tex].

Substitute:

[tex]y-5=-\dfrac{2}{3}(x-(-3))\\\\y-5=-\dfrac{2}{3}(x+3)[/tex]

Answer:

The equation in slope-intercept form is y=-2x+6.

The equation in standard form is 2x+y=6.

The equation in point-slope form is y-2=-2(x-2).

Step-by-step explanation:

The slope-intercept form of a line is y=mx+b where m is the slope and b is the y-intercept.

Parallel lines will have the same slope and different y-intercept.

Anyways the slope of y=-2x+1 is -2.

So the equation of our line we are looking for is -2.

So we know our equation is in the form y=-2x+b.

We must inf b using y=-2x+b with (x,y)=(2,2).

y=-2x  +b  with (x,y)=(2,2)

2=-2(2)+b

2=-4+b

Add 4 on both sides:

2+4=b

Simplify:

6=b

The equation is y=-2x+6.

Now it didn't say what form it wanted.

There are some forms I can give you like standard and point-slope form.

There is also general form but it is not too much different from standard form.

Standard form is ax+by=c where a,b, and c are integers if possible.

Point-slope form is y-y1=m(x-x1) where (x1,y1) is a point on the line and m is the slope.

So let's go for standard form (ax+by=c) first:

y=-2x+6

add 2x on both sides:

2x+y=6

This is standard form because it is in the form

ax+by=c.

Ok we know point (2,2) is on our line and we also know we have slope,m, is -2.

Point-slope form is

y-y1=m(x-x1)

y-2=-2(x-2)

Answer: B

Step-by-step explanation:

Answer:

[tex]x^6+x^4+4x^3-2x^2-x+3[/tex]

Step-by-step explanation:

[tex]x^3+2x+3[/tex]

[tex]\times(x^3-x+1)[/tex]

---------------------------------

First step multiply your terms in your first expression just to the 1 in the second expression like so:

[tex]x^3+2x+3[/tex]

[tex]\times(x^3-x+1)[/tex]

---------------------------------

[tex]x^3+2x+3[/tex]  Anything times 1 is that anything.

That is, [tex](x^3+2x+3) \cdot 1=x^3+2x+3[/tex].

Now we are going to take the top expression and multiply it to the -x in the second expression. [tex]-x(x^3+2x+3)=-x^4-2x^2-3x[/tex].  We are going to put this product right under our previous product.

[tex]x^3+2x+3[/tex]

[tex]\times(x^3-x+1)[/tex]

---------------------------------

[tex]x^3+2x+3[/tex]

[tex]-x^4-2x^2-3x[/tex]  

We still have one more multiplication but before we do that I'm going to put some 0 place holders in and get my like terms lined up for the later addition:

[tex]x^3+2x+3[/tex]

[tex]\times(x^3-x+1)[/tex]

---------------------------------

[tex]0x^4+x^3+0x^2+2x+3[/tex]

[tex]-x^4+0x^3-2x^2-3x+0[/tex]  

Now for the last multiplication, we are going to take the top expression and multiply it to x^3 giving us [tex]x^3(x^3+2x+3)=x^6+2x^4+3x^3[/tex]. (I'm going to put this product underneath our other 2 products):

[tex]x^3+2x+3[/tex]

[tex]\times(x^3-x+1)[/tex]

---------------------------------

[tex]0x^4+x^3+0x^2+2x+3[/tex]

[tex]-x^4+0x^3-2x^2-3x+0[/tex]  

[tex]x^6+2x^4+3x^3[/tex]

I'm going to again insert some zero placeholders to help me line up my like terms for the addition.

[tex]x^3+2x+3[/tex]

[tex]\times(x^3-x+1)[/tex]

---------------------------------

[tex]0x^6+0x^4+x^3+0x^2+2x+3[/tex]

[tex]0x^6-x^4+0x^3-2x^2-3x+0[/tex]  

[tex]x^6+2x^4+3x^3+0x^2+0x+0[/tex]

----------------------------------------------------Adding the three products!

[tex]x^6+x^4+4x^3-2x^2-x+3[/tex]

Answer:

5(-9+1)

= -45+5

= -40 (answer)

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Answer is provided in the image attached.

Answer: cross section = 25π in²

Step-by-step explanation:

Cylinder volume is the product of the cross section by height.

Then cross section = cylinder volume/height = 200 π in³/8in = 25π in²

Answer: 25π in²

[tex]\textit{\textbf{Spymore}}[/tex]

The area of the cross-section of the cylinder is 25π in² square inches.

The volume V of a cylinder is given by the formula:

[tex]\[ V = \pi r^2 h \][/tex]

Where:

- r is the radius of the cylinder's base

- h is the height of the cylinder

Given that the volume of the cylinder is [tex]\( 200\pi \)[/tex] cubic inches and the height h is 8 inches, we can solve for the radius r:

[tex]\[ 200\pi = \pi r^2 \times 8 \][/tex]

[tex]\[ 200 = r^2 \times 8 \][/tex]

[tex]\[ 25 = r^2 \][/tex]

[tex]\[ r = 5 \][/tex]

Now that we have found the radius r to be 5 inches, we can calculate the area A of the cross-section of the cylinder using the formula for the area of a circle:

[tex]\[ A = \pi r^2 \][/tex]

[tex]\[ A = \pi \times 5^2 \][/tex]

[tex]\[ A = \pi \times 25 \][/tex]

[tex]\[ A = 25\pi \][/tex]

Therefore, the area of the cross-section of the cylinder is [tex]\( 25\pi \)[/tex] square inches.

[tex]\bf \begin{cases} f(x)=&10x+25\\ g(x)=&x+8 \end{cases}~\hspace{5em} \begin{array}{llll} f(~~g(x)~~)=&10[g(x)]+25\\\\ f(~~g(x)~~)=&10[x+8]+25\\\\ f(~~g(x)~~)=&10x+80+25\\\\ f(~~g(x)~~)=&10x+105 \end{array}[/tex]

Answer:

f(g(x)) = 10x + 105

Step-by-step explanation:

Start with f(x) = 10x + 25.  Replace this x with (x + 8), which is g(x):

f(g(x)) = 10(x + 8) + 25, or

         = 10x + 80 + 25, or 10x + 105

f(g(x)) = 10x + 105

Answer:

(2,-3) and (-2,5)

Step-by-step explanation:

Let us graph the two equations one by one.

1. [tex]f(x)=-2x+1[/tex]

If we compare this equation with the slope intercept form of a line which is given as

[tex]y=mx+c[/tex]

we see that m = -1 and c =1

Hence the slope of the line is -2 and the y intercept is 1. Hence one point through which it is passing is (0,1) .

Let us find another point by putting x = 1 and solving it for y

[tex]y=-2(1)+1[/tex]

[tex]y=-2+1 = -1[/tex]

Let us find another point by putting x = 2 and solving it for y

[tex]y=-2(2)+1[/tex]

[tex]y=-4+1 = -3[/tex]

Hence the another point will be (2,-3)

Let us find another point by putting x = -2 and solving it for y

[tex]y=-2(-2)+1[/tex]

[tex]y=+1 = 5[/tex]

Hence the another point will be (-2,5)

Now we have two points (0,1) ,(1,-1) ,  (2,-3) and (-2,5) we joint them on line to obtain our line  

2.

[tex]g(x)=y=x^2-2x-3[/tex]

[tex]y=x^2-2x+1-1-3[/tex]

[tex]y=(x-1)^2-4[/tex]

[tex](y+4)=(x-1)^2[/tex]

It represents the parabola opening upward with vertices (1,-4)

Let us mark few coordinates so that we may graph the parabola.

i) x=0 ; [tex]y=y=(0)^2-2(0)-3=0-0-3=-3[/tex] ; (0,-3)

ii)x=-1 ; [tex]y=(-1)^2-2(-1)-3=1+2-3=0[/tex] ; (-1,0)

iii) x=2 ; [tex]y=(2)^2-2(2)-3 = 4-4-3 =-3[/tex] ;(2,-3)

iii) x=1 ; [tex]y=(1)^2-2(1)-3 = 1-2-3 =-4[/tex]  ;(1,-4)

iii) x=-2 ; [tex]y=(-2)^2-2(-2)-3 = 4+4-3 =5[/tex]  ;(-2,5)

Now we plot them on coordinate axis and line them to form our parabola

When we plot them we see that we have two coordinates (2,-3) and (-2,5) are common , on which our graphs are intersecting. These coordinates are solution to the two graphs.

Answer:

so you dont have to suffer lol heres a screenshot

Step-by-step explanation:

Answer:

Step-by-step explanation:

The slope-intercept form of an equation of a line:

[tex]y=mx+b[/tex]

m - slope

b - y-intercept

The formula of a slope:

[tex]m=\dfrac{y_2-y_1}{x_2-x_1}[/tex]

=======================================

We have the points (-2, 1) and (-4, -3). Substitute:

[tex]m=\dfrac{-3-1}{-4-(-2)}=\dfrac{-4}{-2}=2[/tex]

[tex]y=2x+b[/tex]

Put the coordinates of the point (-2, 1) to the equation:

[tex]1=2(-2)+b[/tex]

[tex]1=-4+b[/tex]           add 4 to both sides

[tex]5=b\to b=5[/tex]

Finally:

[tex]y=2x+5[/tex]

Answer:

Slope = 2

Step-by-step explanation:

Slope is the change in y/change in x. So (8-4)/4-2)=2.

Answer: Using the formula below you will get -4/-2 which is slope of 2

Step-by-step explanation:

So you would use the slope formula which is y_{2}- y_{1} / x_{2}-x_{1}  

Answer:

For plato users

D. Jacob made a mistake at step 5. He should have used  x = 3/2   as the new upper bound.

Step-by-step explanation:

The mistake made by Jacob is; D: Jacob made a mistake at step 5. He should have used x=32 as the new upper bound.

In Mathematics, successive approximation can be defined as a classical method that is used in Calculus for solving integral equations or initial value problems.

In this question, Jacob started the first iteration of successive approximation by using the lower and upper bounds of the graph. However, we can deduce that Jacob made a mistake instep 5 because he should have used x = 3/2 as the new upper bound.

Read more about Successive Approximations at; https://brainly.com/question/25219621

#SPJ2

Answer:

3+(6-2)*4=19

Step-by-step explanation:

Due to PEMDAS, it would first be required to do "6-2", which is 4.

Then, the 4 in the parenthesis is multiplied by the 4 on the outside, making 16.

Finally, 3 would be added to 16, making 19.

Answer:

Law of Cosines; A ≈ 61°, B ≈ 89°, C ≈ 30°

Step-by-step explanation:

In this problem the given values are the length sides of the triangle, therefore, the triangle should be solved by beginning with the Law of Cosines

step 1

Applying the law of cosines find the value of angle C

we know that

[tex]c^{2}=a^{2}+b^{2}-2(a)(b)cos(C)[/tex]

we have

[tex]a = 8, b = 7, c = 4[/tex]

substitute the values and solve for cos(C)

[tex]4^{2}=8^{2}+7^{2}-2(8)(7)cos(C)[/tex]

[tex]16=64+49-112cos(C)[/tex]

[tex]16=113-112cos(C)[/tex]

[tex]112cos(C)=113-16[/tex]

[tex]cos(C)=97/112[/tex]

[tex]C=arccos(97/112)=30\°[/tex]

step 2

Applying the law of cosines find the value of angle B

we know that

[tex]b^{2}=a^{2}+c^{2}-2(a)(c)cos(B)[/tex]

we have

[tex]a = 8, b = 7, c = 4[/tex]

substitute the values and solve for cos(B)

[tex]7^{2}=8^{2}+4^{2}-2(8)(4)cos(B)[/tex]

[tex]49=64+16-64cos(B)[/tex]

[tex]49=80-64cos(B)[/tex]

[tex]64cos(B)=80-49[/tex]

[tex]cos(B)=31/64[/tex]

[tex]B=arccos(31/64)=61\°[/tex]

step 3

Find the measure of angle A

we know that

The sum of the interior angles of a triangle must be equal to 180 degrees

so

[tex]A+B+C=180\°[/tex]

we have

[tex]C=30\°[/tex]

[tex]B=61\°[/tex]

substitute and solve for A

[tex]A+61\°+30\°=180\°[/tex]

[tex]A+91\°=180\°[/tex]

[tex]A=180\°-91\°=89\°[/tex]