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2 years ago

What is the answer to the problem: Solve (The square root of 2x-7) = 7?

1 answer:

4 0

The equation is
$\sqrt{2x-7}=7\\\text{Elevating to 2 the two sides of the equation we get:}\\2x-7=49\\\text{Therefore:}\\2x=49+7=56\\x=\frac{56}{2}=28\\\text{Solution : }x=28$

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You have 5 pounds and 10 ounces of laundry. Your washer has a capacity of 7 pounds. How many more ounces of laundry would you ne

irga5000 [103]

7 pounds = 16 pounds and 1 pound = 6 pounds and 16 ounces.

(6 pounds + 16 ounces) minus (5 pounds + 10 ounces) = 1 pound + 6 ounces

7 0

3 years ago

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What are the quotient and remainder of (5x^4-3x^2-4x+6) / (x-7)?

Phantasy [73]

Use long or synthetic division

The correct choice is B

7 0

3 years ago

Suppose f(x, y) is a differentiable function of x and y and let g(r, s) = f (2rs, 8s − 2r). Use

worty [1.4K]

By the chain rule,

$\dfrac{\partial g}{\partial r} = \dfrac{\partial f}{\partial r} \\\\ ~~~~~ = \dfrac{\partial f}{\partial x} \dfrac{\partial x}{\partial r} + \dfrac{\partial f}{\partial y} \dfrac{\partial y}{\partial r} \\\\ ~~~~~ = 2 s \dfrac{\partial f}{\partial x} - 2 \dfrac{\partial f}{\partial y}$

where $x(r,s)=2rs$ and $y(r,s)=8s-2r$ .

If $r=2$ and $s=1$ , then

$x = 2\cdot2\cdot1 = 4$

$y=8\cdot1-2\cdot2 = 4$

so that

$g_r(2,1) = 2\cdot1 f_x(4,4) - 2 f_y(4,4) = 2\cdot2-2\cdot3 = \boxed{-2}$

5 0

9 months ago

Solve the following quadratic equation for all values of x in simplest form.

NNADVOKAT [17]

Answer:

x = ± 1

Step-by-step explanation:

Given

3(x² - 4) + 7 = - 2 ( subtract 7 from both sides )

3(x² - 4) = - 9 ( divide both sides by 3 )

x² - 4 = - 3 ( add 4 to both sides )

x² = 1 ( take the square root of both sides )

x² = ± $\sqrt{1}$ = ± 1

Thus

x = - 1, x = 1

8 0

2 years ago

How to solve this??<br><br> (r^-7b^-8) ^0<br> -------------<br> (t^-4w)

professor190 [17]

keeping in mind that anything raised at the 0 power, is 1, with the sole exception of 0 itself.

$\bf ~~~~~~~~~~~~\textit{negative exponents}
\\\\
a^{-n} \implies \cfrac{1}{a^n}
\qquad \qquad
\cfrac{1}{a^n}\implies a^{-n}
\qquad \qquad a^n\implies \cfrac{1}{a^{-n}}
\\\\[-0.35em]
\rule{34em}{0.25pt}\\\\
\cfrac{(r^{-7}b^{-8})^0}{t^{-4}w}\implies \cfrac{1}{t^{-4}w}\implies \cfrac{1}{t^{-4}}\cdot \cfrac{1}{w}\implies t^4\cdot \cfrac{1}{w}\implies \cfrac{t^4}{w}$

6 0

2 years ago

Read 2 more answers