Here is a picture of a math problem.
Here is a picture of a math problem. - 1

Answers

Answer 1
Answer: the coordinates are (6,-9) (-12,-3) (0,-7)

you put 6 in the place of x to get 6+3y=-21 then you must take 6 away from each side to get 3y= -27 then divide each side by 3 so -27 divided by 3 is -9. do this for all of them! hope it helped!

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A soccer ball is kicked into the air from the ground. If the ball reaches a maximum height of 25 ft and spends a total of 2.5 s in the air, which equation models the height of the ball correctly? Assume that acceleration due to gravity is –16 ft/s2.
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Which of the following is a solution to the quadratic equation below? x^2+x-56=0

a.2
b.-55
c.-8
d.28

Answers

There are two solutions to that quadratic quation: -8 and 7, but as there is only one in answers, you need to pick C.

A chef has 1/2 a pound of flour to divide equally into two containers. How much flour will be in each container?

Answers

There is 1/4 flour will be in each container

What is Division?

A division is a process of splitting a specific amount into equal parts.

Given that A chef has 1/2 a pound of flour

The chef has to divide 1/2 a pound of flour to two containers equally.

We need to find how much flour will be in each container.

Which means we have to divide 1/2 by 2.

(1/2)/2

One by two whole divided by 2

1/2 is the numerator and 2 is the denominator

=1/2×2

=1/4

Hence, there is 1/4 flour will be in each container

To learn more on Division click:

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Answer:

1/4

Step-by-step explanation:

Sixteen is fourteen less than the product of a number and five. What is the number?

Answers

16=5x-14
+14 +14
30=5x
/5 /5
6=x
X=6
5*?=?-14=16
5*6=30
30-14=16

Find the center of the circle that can be circumscribed about ABC with A(0,0) B(6,0) C(6,4)

Answers

This can be solved by the distance formula.
If the center of the circle is at (h,k)
r² = (h - 0)² + (k - 0)²
r² = (h - 6)² + (k - 0)²
r² = (h - 6)² + (k - 4)²

Equating the equations
h² + k² = (h - 6)² + k²
h = 3

(h - 6)² + k² = (h - 6)² + (k - 4)²
k = 2

Therefore, the center of the circle is at (3,2).

Justin wants to buy a new bike that costs $115. He has $40 and plans tosave $15 each week. What is the equation the represents this situation.

Answers

15x+40=115

15x means 15 per week solve for x if you want to know how many weeks 40 is what he has only once

Which functions are symmetric with respect to the origin?y = arcsinx and y = arccosx
y = arccosx and y = arctanx
y = arctanx and y = arccotx
y = arcsinx and y = arctanx

Answers

The function are symmetric with respect to the origin is y = arcsinx and y = arctanx and this can be determined by using the trigonometric property.

The following steps can be used in order to determine which functions are symmetric with respect to the origin:

Step 1 - Check all the options given in order to determine which functions are symmetric with respect to the origin.

Step 2 - Substitute (x = 0) in option a), that is:

\rm y = sin^(-1)(0)=0

\rm y = cos^(-1)(0)=1.57

Step 3 - Substitute (x = 0) in option b), that is:

\rm y = cos^(-1)(0)=1.57

\rm y = tan^(-1)(0)=0

Step 4 - Substitute (x = 0) in option c), that is:

\rm y = tan^(-1)(0)=0

\rm y = cot^(-1)(0)=1.57

Step 5 - Substitute (x = 0) in option d), that is:

\rm y = tan^(-1)(0)=0

\rm y = sin^(-1)(0)=0

For more information, refer to the link given below:

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Answer:

The correct option is;

y = arcsinx and y = arctanx

Step-by-step explanation:

The given options are;

1) y = arcsinx and y = arccosx

Here, we have at the origin, where x = 0,  arccosx ≈ 1.57 while arcsinx = 0

Therefore arccosx does not intersect arcsinx at the origin for it to be symmetrical to arcsinx or the origin

2)  y = arccosxy and y = arctanx

Here arctanx = 0 when x = 0 and arcos x = 1.57 when x = 0 therefore, they are not symmetrical

3) y = arctanx and y = arccotx

Similarly, At x = 0, arccotx = 1.57 therefore, they are not symmetrical

4) y = arcsinx and y = arctanx

Both functions arcsinx and arctanx pass through the origin and their shapes are similar but inverted as they go from negative to positive.