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iogann1982 [59]

3 years ago

12

One end of a ladder 32 feet long is placed 10 feet from the outer wall of a building that stands on the ground level. How far up

the building to the nearest foot will the ladder reach?

1 answer:

ladessa [460]3 years ago

4 0

Answer:

30 ft

Step-by-step explanation:

This is a classic right triangle problem, where the length of the ladder represents the hypotenuse, where the ladder is lengthwise from the building is the base of the triangle, and what we are looking for is the height of the triangle. Pythagorean's Theorem will help us find this length.

$32^2-10^2=y^2$ so

1024 - 100 = y^2 and

y = 30.4 so 30 feet

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Use the Perfect Square Trinomial and provide the first step to calculating 382 without a calculator.

dusya [7]

Answer:

$(40-2)^2$

Step-by-step explanation:

Given: $(38)^2$

To find: the correct option

Solution:

A binomial polynomial is a polynomial consisting of two terms.

A trinomial polynomial is a polynomial consisting of three terms.

On multiplying a binomial $(x-y)$ to itself, a perfect square trinomial $(x-y)^2$ is obtained.

Here, $38=40-2$

So, $(38)^2=(38)(38)=(40-2)(40-2)=(40-2)^2$

Here, $(40-2)$ is a binomial and it is multiplied to $(40-2)$ to get a perfect square trinomial $(40-2)^2$

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3 years ago

Beth can run 3 miles in 25.8 minutes. How long does it take her to run1 one mile you

Dmitry_Shevchenko [17]

Answer: 8.6

Step-by-step explanation:

25.8 divided by 3

3 0

3 years ago

Read 2 more answers

Find an equation of the tangent plane to the given parametric surface at the specified point.

Neko [114]

Answer:

Equation of tangent plane to given parametric equation is:

$\frac{\sqrt{3}}{2}x-\frac{1}{2}y+z=\frac{\pi}{3}$

Step-by-step explanation:

Given equation

$r(u, v)=u cos (v)\hat{i}+u sin (v)\hat{j}+v\hat{k}$ ---(1)

Normal vector tangent to plane is:

$\hat{n} = \hat{r_{u}} \times \hat{r_{v}}\\r_{u}=\frac{\partial r}{\partial u}\\r_{v}=\frac{\partial r}{\partial v}$

$\frac{\partial r}{\partial u} =cos(v)\hat{i}+sin(v)\hat{j}\\\frac{\partial r}{\partial v}=-usin(v)\hat{i}+u cos(v)\hat{j}+\hat{k}$

Normal vector tangent to plane is given by:

$r_{u} \times r_{v} =det\left[\begin{array}{ccc}\hat{i}&\hat{j}&\hat{k}\\cos(v)&sin(v)&0\\-usin(v)&ucos(v)&1\end{array}\right]$

Expanding with first row

$\hat{n} = \hat{i} \begin{vmatrix} sin(v)&0\\ucos(v) &1\end{vmatrix}- \hat{j} \begin{vmatrix} cos(v)&0\\-usin(v) &1\end{vmatrix}+\hat{k} \begin{vmatrix} cos(v)&sin(v)\\-usin(v) &ucos(v)\end{vmatrix}\\\hat{n}=sin(v)\hat{i}-cos(v)\hat{j}+u(cos^{2}v+sin^{2}v)\hat{k}\\\hat{n}=sin(v)\hat{i}-cos(v)\hat{j}+u\hat{k}\\$

at u=5, v =π/3

$=\frac{\sqrt{3} }{2}\hat{i}-\frac{1}{2}\hat{j}+\hat{k}$ ---(2)

at u=5, v =π/3 (1) becomes,

$r(5, \frac{\pi}{3})=5 cos (\frac{\pi}{3})\hat{i}+5sin (\frac{\pi}{3})\hat{j}+\frac{\pi}{3}\hat{k}$

$r(5, \frac{\pi}{3})=5(\frac{1}{2})\hat{i}+5 (\frac{\sqrt{3}}{2})\hat{j}+\frac{\pi}{3}\hat{k}$

$r(5, \frac{\pi}{3})=\frac{5}{2}\hat{i}+(\frac{5\sqrt{3}}{2})\hat{j}+\frac{\pi}{3}\hat{k}$

From above eq coordinates of r₀ can be found as:

$r_{o}=(\frac{5}{2},\frac{5\sqrt{3}}{2},\frac{\pi}{3})$

From (2) coordinates of normal vector can be found as

$n=(\frac{\sqrt{3} }{2},-\frac{1}{2},1)$

Equation of tangent line can be found as:

$(\hat{r}-\hat{r_{o}}).\hat{n}=0\\((x-\frac{5}{2})\hat{i}+(y-\frac{5\sqrt{3}}{2})\hat{j}+(z-\frac{\pi}{3})\hat{k})(\frac{\sqrt{3} }{2}\hat{i}-\frac{1}{2}\hat{j}+\hat{k})=0\\\frac{\sqrt{3}}{2}x-\frac{5\sqrt{3}}{4}-\frac{1}{2}y+\frac{5\sqrt{3}}{4}+z-\frac{\pi}{3}=0\\\frac{\sqrt{3}}{2}x-\frac{1}{2}y+z=\frac{\pi}{3}$

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3 years ago

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Anastaziya [24]

Answer:

8 is the GCF so the expression would be: 8(3+2)

Step-by-step explanation:

(The answer would be 40 =)) Good Luck and Have a Great Day!

5 0

3 years ago

Let g(x) = 2x and h(x) = x2 + 4. Evaluate (h ∘ g)(−5).

Zolol [24]

Hi! the answer should be 104

----

Formula: h(g(x))

1) First you want to take the -5 and plug that into the given equation for g(x)

• 2(-5)= -10

2) Next you want to take the -10 and plug it into the equation for h(x)

• (-10)²+4= 104

6 0

2 years ago