What is the slope of a line that passes through the points (-2,3) and (4,-12)

Answer:

The correct answer is option D. (3h + 7)(h – 6)

Step-by-step explanation:

It is given a quadratic function ,

3h² - 11h - 42

To find the factors of given function

3h² - 11h - 42  = 3h² - 18h  + 7h - 42

 = 3h(h -6) + 7(h - 6)

 =(h - 6)(3h + 7)

 = (3h + 7)(h – 6)

The correct answer is option D. (3h + 7)(h – 6)

Answer:

B) 46.67°

Step-by-step explanation:

Step 1 : Write the cosine formula to find the angle C

c² = a² + b² - 2ab cos C

Step 2 : Substitute the values in the formula

6² = 7² + 8² - 2(8)(7) cos C

cos C = 0.6875

C = cos^-1 (0.6875)

C = 46.567 °

Step 3 : Round off to 2 decimal places

Angle C = 46.57°

!!

Answer:

Tangent of 45º When a square is divided by a diagonal into two equal right triangles, the angles measure 90º, 45º and 45º. The diagonal (hypotenuse of the triangle) is then obtained by applying the Pythagorean theorem: Trigonometric Ratios. Trig.

Step-by-step explanation:

Answer:

Step-by-step explanation:

The point-slope of an equation of a line:

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

m - slope

The formula of a slope:

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

We have the points (-2, 0) and (2, 8).

Substitute:

[tex]m=\dfrac{8-0}{2-(-2)}=\dfrac{8}{4}=2[/tex]

for the point (-2, 0):

[tex]y-0=2(x-(-2))\\\\y-0=2(x+2)[/tex]

for the point (2, 8):

[tex]y-8=2(x-2)[/tex]

Answer:

Answer is m∠4=40

Step-by-step explanation:

According to the image which i have posted below,we first want to take note that m∠6 & m∠7 are vertical angles. Vertical angles are equal to each other, therefore m∠6 is equal to m∠7.

m∠6 = m∠7 (vertical angles)

11x + 8 = 12x – 4

12x - 11x = 8 + 4

x = 12

so

m∠6 = 11x + 8  

m∠6 = 11(12) + 8

m∠6 = 132 + 8

m∠6 = 140

m∠4 = 180 - m<6

m∠4 = 180 - 140

m∠4 = 40

Answer is m∠4=40....

Answer:

[tex]96x^4y+104x^2y^2[/tex]

is the simplified form of

your given problem:

[tex](12x^2+13y)(4x \times 2xy)[/tex].

Step-by-step explanation:

So the given problem is this:

[tex](12x^2+13y)(4x \times 2xy)[/tex]

I'm going to do the multiplication in the second ( ).

Nothing can be done in the first ( ) because the operation is addition and those two terms aren't like terms.

[tex](12x^2+13y)(8x^2y)[/tex]

I got x^2 because of x(x) part.

Now we get to distribute 8x^2y to both terms in the ( ).

[tex]12x^2 \cdot 8x^2y+13y \cdot 8x^2y[/tex]

Adding exponents on bases that have the same variable:

[tex]96x^4y+104x^2y^2[/tex]

Answer:

B. 40

Step-by-step explanation:

All triangle angles equal 180

100+40=140

180-140=40

B. 40 is the answer.

Answer:

x < 3/2.

Step-by-step explanation:

1/3( 5x + 3) < 1/4(6x + 5)

5/3 x + 1 < 3/2x + 5/4

5/3 x - 3/2 x < 5/4 - 1

1/6 x < 1/4

x < 6/4

x < 3/2.

The value of the variable x is less than 3/2.

The distribution of weights to the variables involved that establishes the equilibrium in the calculation is referred to as a result.

Let x be the number.

One-third of the sum of 5 times a number and 3 is less than one-fourth the sum of six times that number and 5. Then the equation will be

(1/3)(5x + 3) < (1/4)(6x + 5)

Simplify the inequality, then the value of x will be

20x + 12 < 18x + 15

2x < 3

x < 3/2

The value of the variable x is less than 3/2.

More about the solution of the equation link is given below.

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[tex]\left(1-\dfrac{2}{3}\right)\cdot\dfrac{6}{9}=\dfrac{1}{3}\cdot \dfrac{2}{3}=\dfrac{2}{9}[/tex]

The pharmacist has 2 more prescriptions left to review that you filled.

To find out how many prescriptions the pharmacist has left to review, we can follow these steps:

1. Calculate the total number of prescriptions filled by you:

[tex]\[ \text{Total filled prescriptions} = \frac{6}{9} \times \text{Total prescriptions} \][/tex]

2. Calculate the number of prescriptions reviewed by the pharmacist:

[tex]\[ \text{Prescriptions reviewed by pharmacist} = \frac{2}{3} \times \text{Total filled prescriptions} \][/tex]

3. Calculate the number of prescriptions left for the pharmacist to review:

[tex]\[ \text{Prescriptions left to review} = \text{Total filled prescriptions} - \text{Prescriptions reviewed by pharmacist} \][/tex]

Let's put the numbers into these equations. Assuming there were initially 9 prescriptions:

1. Total filled prescriptions:

[tex]\[ \text{Total filled prescriptions} = \frac{6}{9} \times 9 = 6 \][/tex]

2. Prescriptions reviewed by pharmacist:

[tex]\[ \text{Prescriptions reviewed by pharmacist} = \frac{2}{3} \times 6 = 4 \][/tex]

3. Prescriptions left to review:

[tex]\[ \text{Prescriptions left to review} = 6 - 4 = 2 \][/tex]

Answer:

it would be the 4th option

Step-by-step explanation:

all the above are wrong except the fourth option because a 3-D square or cube is made up of 6 equal squares. I hope this helps a-lot. : )

The 3-D figure Scarlett can make and the combinations of shapes she can use is a square prism with six squares.

A square prism is a three-dimensional object that is made up of six squares.

The volume of a square prism = length x width x height.

To learn more about right square prism, please check: https://brainly.com/question/13048128

Answer:

3

Step-by-step explanation:

g(-1) means what is g(x) when x=-1.

So find -1 under the column labeled x and then scroll directly to the right of that and you should see what g(-1).  It is 3

Here are my examples:

g(-8)=6

g(-5)=-2

g(-1)=3

g(0)=-5

Answer:

A.   y = 9(x +1/2)^2 - 13/4.

Step-by-step explanation:

y = 9x^2 + 9x - 1

y = 9(x^2 + x) - 1

y = 9 [ (x + 1/2)^2 - 1/4] - 1

y = 9 (x + 1/2)^2  - 9/4 - 1

y = 9(x +1/2)^2 - 13/4.

Answer: First Option

[tex]y = (x+\frac{1}{2}) ^ 2 -\frac{13}{4}[[/tex]

Step-by-step explanation:

For a quadratic function of the form:

[tex]y = ax ^ 2 + bx + c[/tex]

The vertex form of the equation is:

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

Where the vertex is the point (h, k) and [tex]h =-\frac{b}{2a}[/tex]

In this case the equation is: [tex]y=9x^2+9x-1[/tex]

So:

[tex]a=9\\b=9\\c=-1[/tex]

Therefore:

[tex]h =-\frac{9}{2*(9)}[/tex]

[tex]h =-\frac{1}{2}[/tex]

[tex]k=9(-\frac{1}{2})^2+9(-\frac{1}{2})-1\\\\k=-\frac{13}{4}[/tex]

Finally the equation in vertex form is:

[tex]y = (x+\frac{1}{2}) ^ 2 -\frac{13}{4}[[/tex]

Answer:

The farmers will have the same amount of feed (300 lbs) in 3 days

Step-by-step explanation:

After dividing the total amounts of feed by the daily feed amounts, it’s found that the two farmers will have the same amount of feed left after approximately 3.5 days.

To solve this problem, you need to find out how many days it takes for both farmers to run out of feed. For the first farmer, divide the total amount of feed (700 pounds) by the amount he uses each day (200 pounds). This comes to 3.5 days. For the second farmer, divide his total feed (1000 pounds) by the amount he uses each day (350 pounds), which comes to approximately 2.857 days.

Since we want to know when they will have the same amount of feed left, we look for the higher number, because the first farmer will still have feed left when the second farmer runs out. Therefore, the two farmers will have the same amount of feed left after around 3.5 days.

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Answer:

x=2, y=7 -------> y=11-2x  and 4x-3y=-13

x=5, y=2 ------> 2x+y=12 and x=9-2y

x=3, y=5  -----> 2x+y=11 and x-2y=-7

x=7, y=3 ------> x+3y=16  and  2x-y=11

Step-by-step explanation:

Part 1) we have

2x+y=12 -----> equation A

x=9-2y -----> equation B

Solve by substitution

Substitute equation B in equation A and solve for y

2(9-2y)+y=12

18-4y+y=12

4y-y=18-12

3y=6

y=2

Find the value of x

x=9-2(2)=5

therefore

The solution is

x=5, y=2

Part 2) we have

x+2y=9 -----> equation A

2x+4y=20 ---> equation B

Multiply equation A by 2 both sides

2(x+2y)=9*2

2x+4y=18 -----> equation C

Compare equation C with equation B

Both equations have the same slope with different y-intercept

therefore

The lines are parallel and the system has no solution

Part 3) we have

x+3y=16 ------> equation A

2x-y=11 -----> equation B

Solve the system by elimination

Multiply equation B by 3 both sides

3(2x-y)=11*3

6x-3y=33 -----> equation C

Adds equation A and equation C

x+3y=16

6x-3y=33

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

x+6x=16+33

7x=49

x=7

Find the value of y

x+3y=16

7+3y=16

3y=16-7

3y=9

y=3

therefore

The solution is

x=7, y=3

Part 4) we have

y=11-2x -----> equation A

4x-3y=-13 ---> equation B

Solve by substitution

Substitute equation A in equation B and solve for x

4x-3(11-2x)=-13

4x-33+6x=-13

10x=-13+33

10x=20

x=2

Find the value of y

y=11-2(2)=7

therefore

The solution is

x=2, y=7

Part 5) we have

y=10+x -----> equation A

-3x+3y=30 ---> equation B

Multiply equation A by 3 both sides

3*y=3*(10+x)

3y=30+3x

Rewrite

-3x+3y=30 ----> equation C

equation B and equation C are identical

therefore

The system has infinitely solutions

Part 6) we have

2x+y=11 -----> equation A

x-2y=-7 ----> equation B

Solve by elimination

Multiply equation A by 2 both sides

2(2x+y)=11*2

4x+2y=22 ----> equation C

Adds equation B and equation C and solve for x

x-2y=-7

4x+2y=22

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

x+4x=-7+22

5x=15

x=3

Find the value of y

x-2y=-7

3-2y=-7

2y=3+7

2y=10

y=5

therefore

The solution is

x=3, y=5  

Answer:

See below.

Step-by-step explanation:

The y-coordinate , giving the value of the function at this point. It is a part of the range of the function.

Answer:

It is the y-value

Step-by-step explanation:

Took the test on edg.

Answer:

x = -16

Step-by-step explanation:

  3          

 (— • x) -  -24  = 0  

  2        

       -24     -24 • 2

   -24 =  ———  =  ———————

           1         2  

For this case we must solve the following equation:

[tex]\frac {3} {- 2x} = 24[/tex]

Multiplying by 2x on both sides of the equation:

[tex]-3 = 24 * 2x\\-3 = 48x[/tex]

Dividing between 48 on both sides of the equation:

[tex]x = - \frac {3} {48}[/tex]

We simplify:

[tex]x = - \frac {1} {16}[/tex]

Answer:[tex]x = - \frac {1} {16}[/tex]

Answer:

The answer is 12a^3+2a²+23a+18 ....

Step-by-step explanation:

The given terms are:

(3a + 2)(4a² - 2a + 9)

Now multiply each value of second bracket with the first bracket:

=4a²(3a+2) -2a(3a+2)+9(3a+2)

=12a^3+8a²-6a²-4a+27a+18

Solve the like terms:

=12a^3+2a²+23a+18

Therefore the answer is 12a^3+2a²+23a+18 ....

Answer:

1.Intersecting segments theorem

In this scenario, two secant segments intersect each other inside the circle.The relationship is that the product of the segment pieces of one segment is equal to the product of the segment piece of the other.

Hint⇒identify the corresponding segment pieces for multiplication

2.Two secant segments that intersect outside circle

In this scenario the product of the whole secant with its external part is equal to the product of the other whole secant segment with its external part.

Hint ⇒identify the segment pieces outside the circle and the whole segments that include the external parts

3.One secant and one Tangent

In this case, the relation is that the product of whole segment with its external part is equal to square of the tangent segment.

Hint⇒ Identify the tangent segment and the whole secant segment that has an external part.

Hope this Helps.

Answer:

Part 1) The length of the diagonal of the outside square is 9.9 units

Part 2) The length of the diagonal of the inside square is 7.1 units

Step-by-step explanation:

step 1

Find the length of the outside square

Let

x -----> the length of the outside square

c ----> the length of the inside square

we know that

[tex]x=a+b=4+3=7\ units[/tex]

step 2

Find the length of the inside square

Applying the Pythagoras Theorem

[tex]c^{2}= a^{2}+b^{2}[/tex]

substitute

[tex]c^{2}= 4^{2}+3^{2}[/tex]

[tex]c^{2}=25[/tex]

[tex]c=5\ units[/tex]

step 3

Find the length of the diagonal of the outside square

To find the diagonal Apply the Pythagoras Theorem

Let

D -----> the length of the diagonal of the outside square

[tex]D^{2}= x^{2}+x^{2}[/tex]

[tex]D^{2}= 7^{2}+7^{2}[/tex]

[tex]D^{2}=98[/tex]

[tex]D=9.9\ units[/tex]

step 4

Find the length of the diagonal of the inside square

To find the diagonal Apply the Pythagoras Theorem

Let

d -----> the length of the diagonal of the inside square

[tex]d^{2}= c^{2}+c^{2}[/tex]

[tex]d^{2}= 5^{2}+5^{2}[/tex]

[tex]d^{2}=50[/tex]

[tex]d=7.1\ units[/tex]

If a = 4 units and b = 3 units, the length of the diagonal of the outside square rounded to the nearest tenth is  9.9 units

units and the length of the diagonal of the inside square rounded to the nearest tenth is 7.1 units

Let's solve its step by step

step 1

Find the length of the outside square

Let

x -----> the length of the outside square

c ----> the length of the inside square

we know that

x=a+b=4+3=7 units

step 2

Find the length of the inside square

Applying the Pythagoras Theorem

[tex]c^(2)= a^(2)+b^(2)[/tex]

substitute

[tex]c^(2)= 4^(2)+3^(2)[/tex]

[tex]c^(2)=25[/tex]

c=5 units

step 3

Find the length of the diagonal of the outside square

To find the diagonal Apply the Pythagoras Theorem

Let

D -----> the length of the diagonal of the outside square

[tex]D^(2)= x^(2)+x^(2)[/tex]

[tex]D^(2)= 7^(2)+7^(2)[/tex]

[tex]D^(2)=98[/tex]

D=9.9 units

step 4

Find the length of the diagonal of the inside square

To find the diagonal Apply the Pythagoras Theorem

Let

d -----> the length of the diagonal of the inside square

[tex]d^(2)= c^(2)+c^(2)[/tex]

[tex]d^(2)= 5^(2)+5^(2)[/tex]

[tex]d^(2)=50[/tex]

d=7.1 units

Answer:

y-4 =-1(x+1)  point slope form

y-2 = -1(x-1)

y = -x +3  slope intercept form

Step-by-step explanation:

We have 2 points, we can find the slope

m = (y2-y1)/(x2-x1)

   = (2-4)/(1--1)

   (2-4)/(1+1)

     -2/2

   =-1

The slope is -1

Then we can use point slope form to find an equation

y-y1 =m(x-x1)

y-4 = -1(x--1)

y-4 =-1(x+1)  point slope form

Using the other point

y-2 = -1(x-1)

Distribute the -1

y-2 = -1x +1

Add 2 to each side

y-2+2 = -x+1+2

y = -x +3  slope intercept form

Answer:

[tex]\boxed{\text{0.675 h}}[/tex]

Step-by-step explanation:

18 min = 0.3 h

Car 1 started 0.3 h before Car 2.

  Let t = time of Car 2. Then

t + 0.3 = time of Car 1

Distance = speed × time, and both cars travel the same distance. Then

[tex]\begin{array}{rcl}45(t + 0.3) & = & 65t\\45t + 13.5 & = & 65t\\20t & = & 13.5\\t & = & \textbf{0.675 h}\\\end{array}\\\text{Car will overtake Car 1 in } \boxed{\textbf{0.675 h}}[/tex]

Check:

[tex]\begin{array}{rcl}45(0.675 + 0.3) & = & 65 \times 0.675\\45 \times 0.975 & = & 43.875\\43.875 & = & 43.875\\\end{array}[/tex]

OK.

Answer:

Car2 overtakes Car1 after 0.675 hours

Step-by-step explanation:

To solve this question, we must know that

Speed = distance / time

Speed_car1 = 45 mph = distance_car1/ time_1

Speed_car2= 65 mph = distance_car2/ time_2

We know that

time1 - time2 = 18 minutes = 0.3 h

And, at the time of the overtake, both cars will have traveled the same distance.

So,

distance_car1 = 45 mph * time1 = distance_car2 = 65 mph * time2

time1 / time2 = 65/45

time1 = 1.444*time2

Then,

1.444*time2- time2 =  0.3 h

time2 = 0.675 h

time1 = 0.975 h

Car2 overtakes Car1 after 0.675 hours

Answer:

[tex]\large\boxed{y-4=\dfrac{1}{2}(x+2)\text{- point-slope form}}\\\boxed{y=\dfrac{1}{2}x+5\text{- slope-intercept form}}[/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

We have the slope [tex]m=\dfrac{1}{2}[/tex] and the point [tex](-2, 4)[/tex].

Substitute:

[tex]y-4=\dfrac{1}{2}(x-(-2))[/tex]

[tex]y-4=\dfrac{1}{2}(x+2)[/tex] - point-slope form

Convert to the slope-intercept form (y = mx + b):

[tex]y-4=\dfrac{1}{2}(x+2)[/tex]         use the distributive property

[tex]y-4=\dfrac{1}{2}x+1[/tex]         add 4 to both sides

[tex]y=\dfrac{1}{2}x+5[/tex] - slope-intercept form

Answer:

(5, 9)

Step-by-step explanation:

Given the 2 equations

x = 5 → (1)

y = 2x - 1 → (2)

x = 5 is the value of the x- coordinate.

Substitute x = 5 into (2) for the corresponding value of y

y = (2 × 5) - 1 = 10 - 1 = 9

Solution is (5, 9 )

Answer:

See attachment

The relation is not a function.

The domain is [-4.1,3.9]

The range is [-2.1,5.9]

Step-by-step explanation:

The given relation has ordered pairs (3.9, 5.9), (–1.1, 5.9), (–4.1, 3.9), (–4.1, –2.1)

It is implied in the ordered pairs that the relation is a continuous function.

We plot the points and connect them with straight lines as shown in the attachment.

The relation is not a function because its graph fails the vertical line test.

In other words, we have an x-coordinate that corresponds to more than one y-coordinate.

-4.1 corresponding to 3.9 and -2.1 at the same time.

The domain is the set of values for which the function is defined.

The domain is [-4.1,3.9]

The range refers to the corresponding y-values for which the function exists.

The range is [-2.1,5.9]

Answer:

kerri's speed is 50 MPH (miles per hour)

Step-by-step explanation:

You look at the one, and follow the line upward until it stops, and that shows the speed, per hour. I know thats correct so i hope this helps you and good luck on the rest of the test :)

Answer:

50 miles per hour.

Step-by-step explanation:

In this graph, distance covered by Kerri has been shown on y-axis and time on x-axis.

Speed of Kerri will be defined by the rate of change in distance which is slope of the line.

Speed = [tex]\frac{\text{change in distance}}{\text{change in time}}[/tex]

           = [tex]\frac{300-0}{6-0}[/tex] = 50 miles per hour.

The situation "The time it takes to read a book depends on the number of pages in the book" in function notation is: A. Time(pages).

In Mathematics, a function is typically used in mathematics for uniquely mapping an input variable (domain or independent value) to an output variable (range or dependent value).

This ultimately implies that, an independent value represents the value on the x-axis of a cartesian coordinate while a dependent value represents the value on the y-axis of a cartesian coordinate.

In this context, we can logically deduce that time is a function of the number of pages, so it should be written in function notation as follows:

Time(pages).

Complete Question:

Which of the following shows the situation below in function notation?

The time it takes to read a book depends on the number of

pages in the book.

A. Time(pages)

B. Book(pages)

C.Pages(time)

D. Book(time)

Answer:

The shortest distance from A to C is [tex]AC=5\sqrt{13}\ units[/tex]

Step-by-step explanation:

see the attached figure to better understand the problem

we know that

The shortest distance from A to C is the hypotenuse of the right triangle AYC

Applying the Pythagoras Theorem

[tex]AC^{2}=AY^{2} +YC^{2}[/tex]

step 1

Find the length YC (hypotenuse of the right triangle YBC)

Applying the Pythagoras Theorem

[tex]YC^{2}=YB^{2} +BC^{2}[/tex]

substitute the given values

[tex]YC^{2}=6^{2} +15^{2}[/tex]

[tex]YC^{2}=261[/tex]

[tex]YC=\sqrt{261}\ units[/tex]

step 2

Find the shortest distance from A to C

[tex]AC^{2}=AY^{2} +YC^{2}[/tex]

substitute the given values

[tex]AC^{2}=8^{2} +\sqrt{261}^{2}[/tex]

[tex]AC^{2}=325[/tex]

[tex]AC=\sqrt{325}\ units[/tex]

[tex]AC=5\sqrt{13}\ units[/tex]

Final answer:

The vertex of a parabola in the form y = ax + bx² can be found using the vertex formula x = -b/(2a). The vertex represents the highest or lowest point on the graph depending on whether a is positive or negative.

Explanation:

The vertex of a parabola in the form y = ax + bx² can be found using the vertex formula. The vertex formula is x = -b/(2a), which gives the x-coordinate of the vertex. To find the y-coordinate of the vertex, substitute the x-coordinate into the equation y = ax + bx². The vertex of the parabola represents the highest or lowest point on the graph, depending on whether the coefficient a is positive or negative.

Well,

Given that [tex]x=\sin(\theta)[/tex],

We can rewrite the equation like,

[tex]\dfrac{\sin(\theta)}{\sqrt{1-\sin(\theta)^2}}[/tex]

Now use, [tex]\cos(\theta)^2+\sin(\theta)^2=1[/tex] which implies that [tex]1-\sin(\theta)^2=\cos(\theta)^2[/tex]

That means that,

[tex]\dfrac{\sin(\theta)}{\sqrt{1-\sin(\theta)^2}}\Longleftrightarrow\dfrac{\sin(\theta)}{\sqrt{\cos(\theta)^2}}[/tex]

By def [tex]\sqrt{x^2}=x[/tex] therefore [tex]\sqrt{\cos(\theta)^2}=\cos(\theta)[/tex]

So the fraction now looks like,

[tex]\dfrac{\sin(\theta)}{\cos(\theta)}[/tex]

Which is equal to the identity,

[tex]\boxed{\tan(\theta)}=\dfrac{\sin(\theta)}{\cos(\theta)}[/tex]

Hope this helps.

r3t40