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3 years ago

AsKevin Selma madelyn Louis

Mathematics

1 answer:

snow_tiger [21]3 years ago

7 0

Very cool yes yes yes

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*brainliest for best answer thank you* Solve for b.

vekshin1

Hello from MrBillDoesMath!

Answer:

The fourth choice, b = +\- sqrt( sg + a^2)

Discussion:

s = (b^2 - a^2)/g => multiply both sides by "g"

sg = b^2 - a^2 => add a^2 to both sides

sg + a^2 = b^2 => take the square root of each side

b = +\- sqrt( sg + a^2)

which is the fourth choice.

Thank you,

MrB

7 0

3 years ago

What is the slope of the line passing through the points (1, 2) and (5, 4)?

bulgar [2K]

<h2>SOLVING</h2>

$\Large\maltese\underline{\textsf{A. What is Asked}}$

What is the slope of the line passing through the point (1,2) and (5,4)

$\Large\maltese\underline{\textsf{This problem has been solved!}}$

Formula used, here $\bf{\dfrac{y2-y1}{x2-x1}$

_______________________________________________________

$\bf{\dfrac{4-2}{5-1}$ | simplify

$\bf{\dfrac{2}{4}$ | reduce

$\bf{\dfrac{1}{2}$

$\rule{300}{1.7}$

$\bf{Result:}$

$\bf{=Slope:\dfrac{1}{2}$

$\boxed{\bf{aesthetic\not101}}$

3 0

2 years ago

Read 2 more answers

0.75(3.55a - 6b) simplified please

GREYUIT [131]

Answer:

Step-by-step explanation:

2.66a-6b

3 0

3 years ago

A model car is 7.5 inches long made with a scale of 2 inches = 48 inches. How long is the actual car?

slava [35]

Answer:

360

Step-by-step explanation:

3 0

2 years ago

Express p in terms of q if

Irina-Kira [14]

$p=x^2+\dfrac{1}{x^2}\\\\p=\dfrac{x^4}{x^2}+\dfrac{1}{x^2}\\\\p=\dfrac{x^4+1}{x^2}\qquad(*)$

$q=x+\dfrac{1}{x}\\\\q=\dfrac{x^2}{x}+\dfrac{1}{x}\\\\q=\dfrac{x^2+1}{x}\qquad\text{square both sides}\\\\q^2=\left(\dfrac{x^2+1}{x}\right)^2\\\\q^2=\dfrac{(x^2+1)^2}{x^2}\qquad\text{use}\ \ (a+b)^2=a^2+2ab+b^2\\\\q^2=\dfrac{(x^2)^2+2(x^2)(1)+1^2}{x^2}\\\\q^2=\dfrac{x^4+2x^2+1}{x^2}\\\\q^2=\dfrac{x^4+1}{x^2}+\dfrac{2x^2}{x^2}\\\\q^2=\dfrac{x^4+1}{x^2}+2\qquad\text{subtract 2 from both sides}\\\\q^2-2=\dfrac{x^4+1}{x^2}\\\\\text{From (*) we have}\\\\\boxed{p=q^2-2}$

8 0

3 years ago