I WILL GIVE BRAINLIEST!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Help me with this question

Colt1911 [192]

Step-by-step explanation:

just look at the graphic : what value do we have on the cost side, when we have 1 pounds on the amount side ?

all the answer options specify a price per pound. that is why we only need to focus on the price for one pound.

we see in the graphic, that for 1 pound the price is $4.

just to make sure, we can also work with the directly specified point (0.5, 2), which means that 0.5 pounds cost $2.

and the function is clearly a straight line.

so, every point on the function has the same y/x ratio (slope).

therefore,

2/0.5 = 4

now, what point with x=1 has the same ratio ?

y/1 = 4

y = $4

perfect, the results are identical. so, $4 per pound is the correct answer.

4 0

3 years ago

Assume y≠60 which expression is equivalent to (7sqrtx2)/(5sqrty3)

Drupady [299]

Answer:

The equivalent will be:

$\frac{\sqrt[7]{x^2}}{\sqrt[5]{y^3}}=\left(\:x^{\frac{2}{7}}\right)\left(y^{-\frac{3}{5}}\right)$

Therefore, option 'a' is true.

Step-by-step explanation:

Given the expression

$\frac{\sqrt[7]{x^2}}{\sqrt[5]{y^3}}$

Let us solve the expression step by step to get the equivalent

$\frac{\sqrt[7]{x^2}}{\sqrt[5]{y^3}}$

as

$\sqrt[7]{x^2}=\left(x^2\right)^{\frac{1}{7}}$ ∵ $\mathrm{Apply\:radical\:rule}:\quad \sqrt[n]{a}=a^{\frac{1}{n}}$

$\mathrm{Apply\:exponent\:rule:\:}\left(a^b\right)^c=a^{bc},\:\quad \mathrm{\:assuming\:}a\ge 0$

$=x^{2\cdot \frac{1}{7}}$

$=x^{\frac{2}{7}}$

also

$\sqrt[5]{y^3}=\left(y^3\right)^{\frac{1}{5}}$ ∵ $\mathrm{Apply\:radical\:rule}:\quad \sqrt[n]{a}=a^{\frac{1}{n}}$

$\mathrm{Apply\:exponent\:rule:\:}\left(a^b\right)^c=a^{bc},\:\quad \mathrm{\:assuming\:}a\ge 0$

$=y^{3\cdot \frac{1}{5}}$

$=y^{\frac{3}{5}}$

so the expression becomes

$\frac{x^{\frac{2}{7}}}{y^{\frac{3}{5}}}$

$\mathrm{Apply\:exponent\:rule}:\quad \:a^{-b}=\frac{1}{a^b}$

$=\left(\:x^{\frac{2}{7}}\right)\left(y^{-\frac{3}{5}}\right)$ ∵ $\:\frac{1}{y^{\frac{3}{5}}}=y^{-\frac{3}{5}}$

Thus, the equivalent will be:

$\frac{\sqrt[7]{x^2}}{\sqrt[5]{y^3}}=\left(\:x^{\frac{2}{7}}\right)\left(y^{-\frac{3}{5}}\right)$

Therefore, option 'a' is true.

5 0

3 years ago

100x+300y=1.200<br> (2/3x+10)/100=(y+20)/300

AlexFokin [52]

Answer:

x= -4.284 and y= 1.432

Step-by-step explanation:

5 0

3 years ago

Simplify 33/34 divided by 66/85

GREYUIT [131]

The answer would be 1 1/4, or 1.25.

5 0

3 years ago

Read 2 more answers

For f(x) = 4x + 2 and g(x) = x2 - 6, find (f+ g)(x).

enot [183]

Answer:

A

Step-by-step explanation:

(f + g)(x) = f(x) + g(x)

f(x) + g(x) = 4x + 2 + x² - 6 = x² + 4x - 4 → A

3 0

3 years ago