I need help with this question plz

Mathematics

1 answer:

Rudiy272 years ago

8 0

Answer:

y = 10/3x + 20

Step-by-step explanation:

First, we need to find the slope of the function. We can do that by using two points and inserting them into the slope formula. It can be any two but I'll just pick (3, 30) and (6, 40).

m = y₁ - y₂ / x₁ - x₂

m = 30 - 40 / 3 - 6

m = -10/-3

m = 10/3

Now we know the slope is 10/3.

In slope-intercept form, the y-intercept is the constant at the end of the equation. The table of values tells us that the y-intercept is located at (0, 20). So the equation of the linear function is y = 10/3x + 20.

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A square has a side length of 5 1/4 inches. What is the area of the square?

ElenaW [278]

The area of the square is : A = 25

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2 years ago

If 1/√a-√b=1/3 and 1/√a+√b=1/2, then find the difference of a and b.

kow [346]

ANSWER:

If $\frac{1}{\sqrt{a}-\sqrt{b}}=\frac{1}{3}$ and $\frac{1}{\sqrt{a}+\sqrt{b}}=\frac{1}{2}$ then the difference of a and b is 6

SOLUTION:

Given, $\frac{1}{\sqrt{a}-\sqrt{b}}=\frac{1}{3}$ → $\sqrt{a}-\sqrt{b}=3$ ----- (1)

And $\frac{1}{\sqrt{a}+\sqrt{b}}=\frac{1}{2}$ → $\sqrt{a}+\sqrt{b}=2$ --- (2)

We have to find difference of a and b.

Now, add (1) and (2)

$\sqrt{a}-\sqrt{b}=3$

$\sqrt{a}+\sqrt{b}=2$

Adding above two equations, we get,

$2 \sqrt{a}+0=2+3$

$\begin{array}{l}{2 \sqrt{a}=5} \\\\ {\sqrt{a}=\frac{5}{2}} \\\\ {a=\frac{25}{4}}\end{array}$

substitute $\sqrt{a}$ value in (2)

$\begin{array}{l}{\frac{5}{2}+\sqrt{b}=2} \\\\ {\sqrt{b}=\frac{2}{\sin \frac{5}{2}}} \\\\ {\sqrt{b}=\frac{4-5}{2}} \\\\ {\sqrt{b}=\frac{-1}{2}} \\\\ {b=\frac{1}{4}}\end{array}$