Which line has a slope of frac{1}{3} and goes through (3, -1)?

Answer:

Last choice

Step-by-step explanation:

V means or.

So we have x<0 or y<0.

This means we want the quadrant that have negative x's and we want the quadrants that have negative y's

x is negative to the left of the y-axis so it's negative in quadrants 2 and 3

y is negative below the x-axis so it's negative in quadrants 3 and 4.

So we are looking at all quadrants except quadrant 1.

Answer:

see explanation

Step-by-step explanation:

The equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

Here slope m = [tex]\frac{1}{2}[/tex], hence

y = [tex]\frac{1}{2}[/tex] x + c ← is the partial equation

To find c substitute (- 2, - 4) into the partial equation

- 4 = - 1 + c ⇒ c = - 4 + 1 = - 3

y = [tex]\frac{1}{2}[/tex] x - 3 ← equation of line

First find the decreased value,

[tex]777-777\cdot0.12=777-93.24=683.76[/tex]

Then from the value found, find increased value,

[tex]683.76+683.76\cdot0.04\approx\boxed{711.11}[/tex]

Hope this helps.

r3t40

Answer:

3.162 units

Step-by-step explanation:

First identify the points that undergo transformation

You have;

A = (0,0)   and B =(1,3)

The transformation is T: (x,y)= (x+2, y+1), this means to get the image you add the x coordinate of the object to 2, and the y coordinate to 1.

Finding coordinates of the image points A' and B'

A'= (0+2,0+1) = (2,1)

B'=(1+2, 3+1)=(3,4)

Finding the distance A'B'

The formula for distance d is

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

where

x₁=2

x₂=3

y₁=1

y₂=4

d=distance between two points

Applying the formula

[tex]d=\sqrt{(3-2)^2+(4-1)^2} \\\\\\d=\sqrt{1^2+3^2} \\\\\\d=\sqrt{1+9} \\\\\\d=\sqrt{10} \\\\\\d=3.162[/tex]

The distance A'B' is √10 =3.162 units

Answer:

what bout the AA distance???

Step-by-step explanation:

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

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.

To know more about an area follow

https://brainly.com/question/25292087

#SPJ1

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)

[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:

22cm,24cm,24cm

Step-by-step explanation:

Let us call one of the other sides x

the shortest side = 2x-26

in an isosceles, 2 sides are equal (x in this case)

so we now have sides of x,x and 2x-26

form an eqution from this.

4x-26=70

4x=96

x=24

24 x 2 = 48 - 26 = 22

thus, the shortest side is 22cm and the other sides are both 24cm

Answer:

The lengths of the three sides in ascending order is.

_22__cm, __24__cm, __24__cm

Step-by-step explanation:

The perimeter of a triangle is equal to the sum of the length of its three sides.

By definition, an isosceles triangle has two equal sides.

We know that the short side measures  26 cm less than twice as long as the other sides, and that the other two sides are of equal length.

We also know that the perimeter of the triangle is 70 cm

Then we propose the following equation

[tex]P = b + 2s[/tex]

Where P is the perimeter, b is the shortest side of the triangle and s is the length of the equal sides.

Then:

[tex]b= 2s -26[/tex]

We substitute this equation in the first equation and solve for s

[tex]P = 2s -26 + 2s[/tex]

[tex]P = 4s -26=70[/tex]

[tex]4s -26=70[/tex]

[tex]4s=70 +26[/tex]

[tex]4s=96[/tex]

[tex]s=\frac{96}{4}[/tex]

[tex]s=24[/tex]

Then

[tex]b= 2(24) -26[/tex]

[tex]b= 22[/tex]

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:

[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:

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.

[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:

5(-9+1)

= -45+5

= -40 (answer)

✌Pls mark my answer as brainliest ✌

❤and follow me ❤

Answer is provided in the image attached.

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

Answer:

14000 lbs.

Step-by-step explanation:

7 tons = 14000 lbs

1 ton = 2000 lbs.

7 x 2000 = 14000 Lbs.

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:

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]

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.

Answer:

[tex]\large\boxed{\dfrac{5x\sqrt{x}}{8y^4}}[/tex]

Step-by-step explanation:

[tex]\sqrt{\dfrac{25x^9y^3}{64x^6y^{11}}}\qquad\text{use}\ \sqrt{ab}=\sqrt{a}\cdot\sqrt{b},\ \sqrt{\dfrac{a}{b}}=\dfrac{\sqrt{a}}{\sqrt{b}};\ \dfrac{a^m}{a^n}=a^{m-n}\\\\=\dfrac{\sqrt{25}}{\sqrt{64}}\cdot \sqrt{x^{9-6}y^{3-11}}=\dfrac{5}{8}\sqrt{x^3y^{-8}}=\dfrac{5}{8}\sqrt{x^3}\cdot\sqrt{y^{-8}}\\\\\text{use}\ a^na^m=a^{n+m},\ (a^n)^m=a^{nm}\\\\=\dfrac{5}{8}\sqrt{x^{2+1}}\cdot\sqrt{y^{(-4)(2)}}=\dfrac{5}{8}\sqrt{x^2x}\cdot\sqrt{(y^{-4})^2}[/tex]

[tex]=\dfrac{5}{8}\sqrt{x^2}\cdot\sqrt{x}\cdot\sqrt{(y^{-4})^2}\qquad\text{use}\ \sqrt{a^2}=a\ \text{for}\ a\geq0\\\\=\dfrac{5}{8}x\sqrt{x}\cdot y^{-4}\qquad\text{use}\ a^{-n}=\dfrac{1}{a^n}\\\\=\dfrac{5x\sqrt{x}}{8y^4}[/tex]

Answer:

D is the correct answer on edge2021!!!

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:

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:

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:

D describes the concept of natural price accurately

Step-by-step explanation:

Adam Smith said that the natural price is the price which covers all the costs to produce it for instance, rent of land, wages of labour, cost of machinery.

A. The price of any item that is less than costs of producing it

This is incorrect since natural price covers all costs of production,

B. The price of any item that is equal to its market price.

C. The price of any item that is less than its market price

Both B and C are incorrect because market price is determined by the forces of demand and supply and changes accordingly whereas the natural price is the everyday normal price.

D. The price of any item that is equal to the costs of producing it.

This is correct because the natural price covers all costs involved in producing a product and is neither low nor high.

!!

Answer:

The answer is D.

Step-by-step explanation:

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

Answer:

x + 10 = x² + 2x + 1

Step-by-step explanation:

[tex]\sqrt{x + 10}[/tex] - 1 = x

x + 1 = [tex]\sqrt{x + 10}[/tex]

Squaring both sides gives;

(x + 1)² = ([tex]\sqrt{x + 10}[/tex])²

x + 10 = x² + 2x + 1

Answer:

option B

Step-by-step explanation:

the given equation is,

[tex]\sqrt{x+10}-1 =x[/tex].................................(1)

adding  both side by 1 in equation (1)

[tex]\sqrt{x+10}= x + 1[/tex]

squaring both side              

x + 10 = ( x + 1 )²                

we know,

( a + b )²  = a² + b² + 2 a b

x + 10 = x² + 2 x + 1

hence, the correct answer is option B.

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.

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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:

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:

10

---------

9 t^2

Step-by-step explanation:

10t             20t-40

----------- * ------------

6t-12         30 t^3

Factor

10t             20(t-2)

----------- * ------------

6(t-2)         30 t^3

We can cancel 10t

1            20(t-2)

----------- * ------------

6(t-2)         3 t^2

We can cancel t-2

1            20

----------- * ------------

6               3 t^2

We can cancel a 2 from the 20 and from the 6

1               10

----------- * ------------

3               3 t^2

     10

-----------

9 t^2