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3 years ago

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a street light is mounted at the top of a 15 ft-tall pole. A man 6 ft tall walks away from the pole with a speed of 5 ft/s along

a straight path. How fast is the tip of his shadow moving when he is 40 ft from the pole

1 answer:

Virty [35]3 years ago

8 0

See attached. The trick is to use similar triangles.

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In quadrilateral DBCA, name the angles which are consecutive with angle D.

lisov135 [29]

Answer: The consecutive angles with angle D are B and A.

Explanation:

It is given that DBCA is a quadrilateral. Since it is a quadrilateral it means the have 4 vertices and 4 angles D, B,C,A.

The angles are formed in the order of the figure name. If the quadrilateral name is DBCA, So the order of the angles are D,B,C,A. It means the angle immediate after D is B, the angle immediate after B is C, the angle immediate after C is A and the angle immediate after A is D.

If the quadrilateral name is DBCA it means the sides are DB, BC, CA and AD as shown in the figure.

Consecutive angles of D means the angle immediate before and after the angle D.

In the figure there are some types of quadrilateral and from the figure we can easily noticed that the consecutive angles with angle D are B and A.

3 0

3 years ago

Read 2 more answers

Find the first five terms of the sequence. an=-n2+2n

Julli [10]

If your sequence is this

$an = - {n}^{2} + 2n \\ then \\ a1 = - 1 + 2 = 1 \\ a2 = - 4 + 4 = 0 \\ a3 = - 9 + 6 = - 3 \\ a4 = - 16 + 8 = - 8 \\ a5 = - 25 + 10 = - 15$

3 0

3 years ago

Read 2 more answers

Show with work please.

kolbaska11 [484]

Answer:

$\csc \left(\theta-\frac{\pi }{2}\right)=0.73$

Step-by-step explanation:

The identity you will use is:

$\csc \left(x\right)=\frac{1}{\sin \left(x\right)}$

So,

$\csc \left(\theta-\frac{\pi }{2}\right)$

$\csc \left(\theta-\frac{\pi }{2}\right)=\frac{1}{\sin \left(-\frac{\pi }{2}+\theta\right)}$

Now, using the difference of sin

Note: state that $\text{sin}(\alpha\pm \beta)=\text{sin}(\alpha) \text{cos}(\beta) \pm \text{cos}(\alpha) \text{sin}(\beta)$

$\csc \left(\theta-\frac{\pi }{2}\right)=\frac{1}{-\cos \left(\theta\right)\sin \left(\frac{\pi }{2}\right)+\cos \left(\frac{\pi }{2}\right)\sin \left(\theta\right)}$

Solving the difference of sin:

$-\cos \left(\theta\right)\sin \left(\frac{\pi }{2}\right)+\cos \left(\frac{\pi }{2}\right)\sin \left(\theta\right)$

$-\cos \left(\theta\right) \cdot 1+0\cdot \sin \left(\theta\right)$

$-\text{cos} \left(\theta\right)$

Then,

$\csc \left(\theta-\frac{\pi }{2}\right)=-\frac{1}{\cos \left(\theta\right)}$

Once

$\text{sec}(-\theta)=\text{sec}(\theta)$

And, $\text{sec}(\theta)=-0.73$

$-\frac{1}{\cos \left(\theta\right)}=-\text{sec}(\theta)$

$-\frac{1}{\cos \left(\theta\right)}=-(-0.73)$

$-\frac{1}{\cos \left(\theta\right)}=0.73$

Therefore,

$\csc \left(\theta-\frac{\pi }{2}\right)=0.73$

3 0

3 years ago

Simplify the expression below 6m^5+15m^2/3m^2

densk [106]

Answer:6m^5+5

Step-by-step explanation:

To simplify 6m^5+15m^2/3m^2

We get

6m^5+5

8 0

3 years ago

Help with my geometry question

Vaselesa [24]

For an inscribed angle like angle AHB, the measure will be half of the intercepted arc which we can see is 90 degrees.. therefore the measure of angle AHB is 45 degrees.

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3 years ago