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3 years ago

10

Please help me on my math problems screenshots are below

1 answer:

xeze [42]3 years ago

7 0

Answer:

1st. (0,-2)

2nd. yes

Step-by-step explanation:

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Hey how to do this and what the answer

frez [133]

Answer:

5x + y - 4 = 0

Step-by-step explanation:

When we are given two points and are asked to find the equation of the line we use the two - point form.

Two - point form: $\frac{y - y_1}{y_2 - y_1} = \frac{x - x_1}{x_2 - x_1}$ .

where: $(x_1,y_1)$ and $(x_2,y_2)$ are the two points on the line.

Given the two points are: (5,21) and (-5,-29).

Substituting in the two point form:

⇒ $\frac{y - 21}{-29 - 21} = \frac{x - 5}{-5 -5}$

$\implies \frac{y - 21}{-50} = \frac{x - 5}{-10}$

$\implies y - 21 = 5x - 25$

⇒ 5x + y - 4 = 0

8 0

3 years ago

Solve for c.<br> 55c+13 < 750 +39

iren [92.7K]

Answer:

c = anything less than 14.1

Step-by-step explanation:

55c+13 < 750 + 39

55c+13 < 789

55c < 776

c < 14.1

Hope It Helps

5 0

3 years ago

Read 2 more answers

etta bought a calculator for $15. Glenn found the same model for $9.79. How much more did Etta pay than Glenn

ololo11 [35]

5.21.

To find this, give both numbers decimal points before setting up a subtraction equation as so:

15.00 - 9.79

Then proceed to solve to get the final answer of 5.21.

Hope this helps!

6 0

4 years ago

Read 2 more answers

Need help with my homework

Volgvan

Answer:

$\displaystyle y=\frac{16-9x^3}{2x^3 - 3}$

$\displaystyle y=-\frac{56}{13}$

Step-by-step explanation:

<u>Equation Solving</u>

We are given the equation:

$\displaystyle x=\sqrt[3]{\frac{3y+16}{2y+9}}$

i)

To make y as a subject, we need to isolate y, that is, leaving it alone in the left side of the equation, and an expression with no y's to the right side.

We have to make it in steps like follows.

Cube both sides:

$\displaystyle x^3=\left(\sqrt[3]{\frac{3y+16}{2y+9}}\right)^3$

Simplify the radical with the cube:

$\displaystyle x^3=\frac{3y+16}{2y+9}$

Multiply by 2y+9

$\displaystyle x^3(2y+9)=\frac{3y+16}{2y+9}(2y+9)$

Simplify:

$\displaystyle x^3(2y+9)=3y+16$

Operate the parentheses:

$\displaystyle x^3(2y)+x^3(9)=3y+16$

$\displaystyle 2x^3y+9x^3=3y+16$

Subtract 3y and $9x^3$ :

$\displaystyle 2x^3y - 3y=16-9x^3$

Factor y out of the left side:

$\displaystyle y(2x^3 - 3)=16-9x^3$

Divide by $2x^3 - 3$ :

$\mathbf{\displaystyle y=\frac{16-9x^3}{2x^3 - 3}}$

ii) To find y when x=2, substitute:

$\displaystyle y=\frac{16-9\cdot 2^3}{2\cdot 2^3 - 3}$

$\displaystyle y=\frac{16-9\cdot 8}{2\cdot 8 - 3}$

$\displaystyle y=\frac{16-72}{16- 3}$

$\displaystyle y=\frac{-56}{13}$

$\mathbf{\displaystyle y=-\frac{56}{13}}$

8 0

3 years ago

I WILL GIVE BRIANLEST TO THE PERSON WHO ANSWER IT WRITE.

kykrilka [37]

The answer to the first part is D

4 0

3 years ago

Read 2 more answers