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2 years ago

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69. The O'Brien's had $400.0 in a special

1 answer:

charle [14.2K]2 years ago

5 0

Hello :D

$40.00 x 3 = $120.00

$20.00 x 2 = $40.00

$120.00 + $40.00 = $160.00

$400.00 - $160.00 = $240.00

There is $240.00 left in the account.

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A 20 foot ladder leaning against a wall is used to reach a window that is 17 feet above the ground. How far from the wall is the

Liono4ka [1.6K]

Answer:

The wall is 10.5 foot far from the bottom of the ladder.

Step-by-step explanation:

Given:

Height of the ladder leaning against wall (Hypotenuse) = 20 foot.

Height from where the window is above from the ground (Leg1) = 17 feet.

To find the distance of the bottom of the ladder far from the wall (Leg2) = ?

Now, by using the pythagorean theorem:

$Hypotenuse^{2} = (Leg1)^{2} + (Leg2)^{2}$

$20^{2} =17^{2}+(Leg2) ^{2}$

$400=289+(Leg2)^{2}$

$400-289=(Leg2)^{2}$

$111=(Leg2)^{2}$

by squaring both sides

$Leg2=10.535$

10.535 foot rounded nearest tenth is 10.5 foot.

Therefore, the wall is 10.5 foot far from the bottom of the ladder.

6 0

2 years ago

55% of 19.6 is what number?

tiny-mole [99]

10.78 is the answer.
19.6 x .55

3 0

2 years ago

Use implicit differentiation to find an equation of the tangent line to the curve at the given point. x2/3 + y2/3 = 4 (−3 3 , 1)

vovikov84 [41]

Answer with Step-by-step explanation:

We are given that an equation of curve

$x^{\frac{2}{3}}+y^{\frac{2}{3}}=4$

We have to find the equation of tangent line to the given curve at point $(-3\sqrt3,1)$

By using implicit differentiation, differentiate w.r.t x

$\frac{2}{3}x^{-\frac{1}{3}}+\frac{2}{3}y^{-\frac{1}{3}}\frac{dy}{dx}=0$

Using formula : $\frac{dx^n}{dx}=nx^{n-1}$

$\frac{2}{3}y^{-\frac{1}{3}}\frac{dy}{dx}=-\frac{2}{3}x^{-\frac{1}{3}}$

$\frac{dy}{dx}=\frac{-\frac{2}{3}x^{-\frac{1}{3}}}{\frac{2}{3}y^{-\frac{1}{3}}}$

$\frac{dy}{dx}=-\frac{x^{-\frac{1}{3}}}{y^{-\frac{1}{3}}}$

Substitute the value x= $-3\sqrt3,y=1$

Then, we get

$\frac{dy}{dx}=-\frac{(-3\sqrt3)^{-\frac{1}{3}}}{1}$

$\frac{dy}{dx}=-(-3^{\frac{3}{2}})^{-\frac{1}{3}}=-\frac{1}{-(3)^{\frac{3}{2}\times \frac{1}{3}}}=\frac{1}{\sqrt3}$

Slope of tangent=m= $\frac{1}{\sqrt3}$

Equation of tangent line with slope m and passing through the point $(x_1,y_1)$ is given by

$y-y_1=m(x-x_1)$

Substitute the values then we get

The equation of tangent line is given by

$y-1=\frac{1}{\sqrt3}(x+3\sqrt3)$

$y-1=\frac{x}{\sqrt3}+3$

$y=\frac{x}{\sqrt3}+3+1$

$y=\frac{x}{\sqrt3}+4$

This is required equation of tangent line to the given curve at given point.

8 0

3 years ago

State which theorem you can use to show that the quadrilateral is a parallelogram.

tekilochka [14]

Answer:

If one pair of opposite sides in a four sided figure are both opposite and parallel, then the figure is a parallelogram.

Step-by-step explanation:

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victus00 [196]

8x2 20-8 3x4 Hope it helped

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3 years ago

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