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3 years ago

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Simplify 3 square root of 125

1 answer:

RSB [31]3 years ago

6 0

We have:

$3 \sqrt{125}$
$3 \sqrt{25*5}$
$15 \sqrt{5}$

As 5 has no perfect square factors, we have completely simplified this expression.

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PLEASE HELP!! the measures of two complementary angles are in the ratio 2:3. What is the measure of the smaller angle?

ivolga24 [154]

The answer is 36°.

Complementary angles are those angles whose sum is 90°.

Let angles be x and y.
x + y = 90°
x : y = 2 : 3
_____
3 * x = 2 * y
3x = 2y
x + y = 90°
_____
x = 2/3*y
x + y = 90°
2/3y + y = 90°
2/3y + 3/3y = 90°
5/3y = 90°
5y = 90° * 3
5y = 270°
y = 54°

x + 54° = 90°
x = 90° - 54°
x = 36°

The smaller angle is 36°

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balu736 [363]

14 is the answer to this.

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Caroline is making a triangular flag with the base of 10 inches and a height of 8 inches what is the area of the flag

bearhunter [10]

Answer:

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Step-by-step explanation:

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3 years ago

8x -5(x - 1) = 4(x + 2)<br><br> solve for x, and show work please.

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Find Mx, My, and (x, y) for the lamina of uniform density rho bounded by the graphs of the equations. y = x2/3, y = 0, x = 1

erik [133]

Answer:

$\mathbf{(\overline x , \overline y ) = (\dfrac{5}{8}, \dfrac{5}{14})}$

Step-by-step explanation:

Given that:

$y = x^{2/3}$ at y = 0 , x = 1

Then:

Area = $\int^{1}_{0} x^{2/3} \ dx$

Area = $\begin {bmatrix} \dfrac{3}{5}x^{5/3} \end {bmatrix} ^1_0$

Area = $\dfrac{3}{5}$

Then:

$\overline x = \dfrac{1}{A} \int^b_a x (f(x) -g(x) ) \ dx$

$\overline x = \dfrac{5}{3} \int^1_0 x (x^{2/3} -0 ) \ dx$

$\overline x = \dfrac{5}{3} \int^1_0 x^{5/3} \ dx$

$\overline x = \dfrac{5}{3} \ [\dfrac{3}{8}x^{8/3}]^1_0$

$\overline x = \dfrac{5}{3} \times \dfrac{3}{8}$

$\overline x = \dfrac{5}{8}$

Similarly;

$\overline y = \dfrac{1}{A} \int^b_a \dfrac{1}{2} \begin{bmatrix} (f(x)^)2 - (g(x))^2 \end {bmatrix} \ dx$

$\overline y = \dfrac{5}{3} \int^1_0 \dfrac{1}{2} \begin{bmatrix} (f(x^{2/3})^2 -0 \end {bmatrix} \ dx$

$\overline y = \dfrac{5}{3} \int^1_0 \dfrac{1}{2} \begin{bmatrix} (x^{4/3} ) \end {bmatrix} \ dx$

$\overline y = \dfrac{5}{3} \begin{bmatrix} \dfrac{1}{2} (x^{7/3} ) \times \dfrac{3}{7} \end {bmatrix} ^1_0$

$\overline y = \dfrac{5}{3} \begin{bmatrix} \dfrac{3}{14} (x^{7/3} ) \end {bmatrix} ^1_0$

$\overline y = (\dfrac{5}{3} \times \dfrac{3}{14} )$

$\overline y = \dfrac{5}{14}$

Thus; $\mathbf{(\overline x , \overline y ) = (\dfrac{5}{8}, \dfrac{5}{14})}$

4 0

2 years ago