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2 years ago

What is the equation of the line through (-2,5) and (0,3)?

Mathematics

2 answers:

statuscvo [17]2 years ago

7 0

I’m pretty sure the answer is D

chubhunter [2.5K]2 years ago

6 0

Answer:

A. y= -x + 3

Hope this helps!

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In right triangle AABC,<br> MZA= 37º. Find mzB.<br> 1

Nadya [2.5K]

53 degrees
Because 180 - 90 ( right angle ) - 37 = 53

4 0

2 years ago

Please help me on this one would be great help.

dusya [7]

T = 60

please mark me brianliest

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2 years ago

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The radius of a circle is 8 kilometers . What is the diameter?

dimaraw [331]

Answer:

16 km

Step-by-step explanation:

the radius is half of the diameter

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3 years ago

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Jose’s school has 426 students. His principal has promised the Student Council that their idea will be carried out if they can g

Kryger [21]

Answer: Step 4

<u>Step-by-step explanation:</u>

1. Define variable x as student signatures still needed (Correct)

$2)\ \dfrac{x+82}{426}$ (Correct)

3. 25% = 0.25 (Correct)

$4)\ \dfrac{x+82}{426}\leq 0.25$ (Error)!

The inequality symbol should be GREATER THAN or equal to (≥) because Jose needs 25% or more (not less).

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3 years ago

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Use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function.

seropon [69]

Answer:

$h'(x)=\frac{3r^{2}}{2\sqrt{r^3+5}}$

Step-by-step explanation:

1) The Fundamental Theorem of Calculus in its first part, shows us a reciprocal relationship between Derivatives and Integration

$g(x)=\int_{a}^{x}f(t)dt \:\:a\leqslant x\leqslant b$

2) In this case, we'll need to find the derivative applying the chain rule. As it follows:

$h(x)=\int_{a}^{x^{2}}\sqrt{5+r^{3}}\therefore h'(x)=\frac{\mathrm{d} }{\mathrm{d} x}\left (\int_{a}^{x^{2}}\sqrt{5+r^{3}}\right )\\h'(x)=\sqrt{5+r^{3}}\\Chain\:Rule:\\F'(x)=f'(g(x))*g'(x)\\h'=\sqrt{5+r^{3}}\Rightarrow h'(x)=\frac{1}{2}*(r^{3}+5)^{-\frac{1}{2}}*(3r^{2}+0)\Rightarrow h'(x)=\frac{3r^{2}}{2\sqrt{r^3+5}}$

3) To test it, just integrate:

$\int \frac{3r^{2}}{2\sqrt{r^3+5}}dr=\sqrt{r^{3}+5}+C$

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2 years ago