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Oduvanchick [21]

3 years ago

9

What is the answer to 3/4-3/8?

1 answer:

Norma-Jean [14]3 years ago

8 0

6/8-3/8
3/8
the answer to 3/4-3/8 is 3/8

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Which ordered pair is the solution to the system of equations? {y=3x−102x+3y=3

uysha [10]

(3,-1) for sure i took the test 100%
hope i helped out

8 0

3 years ago

The line integral of xydx×x^2y^3dy where c is the triangle with points (0,0) (1,0) (1,2) using green's theorem

lyudmila [28]

If $\mathcal C$ is the boundary of the triangle $D$ , then by Green's theorem

$\displaystyle\int_{\mathcal C}xy\,\mathrm dx+x^2y^3\,\mathrm dy=\iint_D\left(\frac{\partial(x^2y^3)}{\partial x}-\frac{\partial(xy)}{\partial y}\right)\,\mathrm dA$
$=\displaystyle\int_{x=0}^{x=1}\int_{y=0}^{y=2x}(2xy^3-x)\,\mathrm dy\,\mathrm dx$
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$=\displaystyle\int_{x=0}^{x=1}x(8x^4-2x)\,\mathrm dx=\frac23$

4 0

3 years ago

The lines below are parallel. If the slope of the green line is -4, what is the slope of the red line? 5 m = O A. 4 B. -4 O c. 1

Kobotan [32]

Answer:

-4

Step-by-step explanation:

Parallel lines have the same slope

-4

6 0

3 years ago

Read 2 more answers

Help pls! will give brainliest!<br><br> find the percent markdown of 175.80-141.27

polet [3.4K]

Answer:

141.2/175.8=.803185

1-.803185=.1968=20% markdown

ples give brainliest! :D

8 0

3 years ago

Evaluate cube root of 5 multiplied by square root of 5 over cube root of 5 to the power of 5.

Andre45 [30]

Answer:

Option d) 5 to the power of negative 5 over 6 is correct.

$\dfrac{\sqrt[3]{\bf 5} \times \sqrt{\bf 5}}{\sqrt[3]{\bf 5^{\bf 5}}}= 5^{\frac{\bf -5}{\bf 6}}$

Above equation can be written as 5 to the power of negative 5 over 6.

ie, $5^\frac{\bf -5}{\bf 6}$

Step-by-step explanation:

Given that cube root of 5 multiplied by square root of 5 over cube root of 5 to the power of 5.

It can be written as below

$\dfrac{\sqrt[3]{5} \times \sqrt{5}}{\sqrt[3]{5^5}}$

$\dfrac{\sqrt[3]{5} \times \sqrt{5}}{\sqrt[3]{5^5}}= \dfrac{5^{\frac{1}{3}} \times 5^{\frac{1}{2}}}{5^{\frac{5}{3}}}$

$\dfrac{\sqrt[3]{5} \times \sqrt{5}}{\sqrt[3]{5^5}}= \dfrac{5^{\frac{1}{3}+\frac{1}{2}}}{5^{\frac{5}{3}}}$

$\dfrac{\sqrt[3]{5} \times \sqrt{5}}{\sqrt[3]{5^5}}= \dfrac{5^{\frac{2+3}{6}}}{5^{\frac{5}{3}}}$

$\dfrac{\sqrt[3]{5} \times \sqrt{5}}{\sqrt[3]{5^5}}= 5^{\frac{5}{6}} \times 5^{\frac{-5}{3}}$

$\dfrac{\sqrt[3]{5} \times \sqrt{5}}{\sqrt[3]{5^5}}= 5^{\frac{5-10}{6}}$

$\dfrac{\sqrt[3]{5} \times \sqrt{5}}{5^5}= 5^{\frac{-5}{6}}$

Above equation can be written as 5 to the power of negative 5 over 6.

7 0

4 years ago