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3 years ago

12

If you place a 65-foot ladder against the top a 16-foot building ,how many feet will the bottom of ladder be from the bottom of

the building?

1 answer:

Annette [7]3 years ago

8 0

Answer:

You can use the Pythagoren Theorem:

Ladder distance ² = hypotenuse ² - height ²

Ladder distance ² = 65 ² - 16 ²

Ladder distance ² = 4,225 -256

Ladder distance ² = 3,969

Ladder distance = 63 feet

Step-by-step explanation:

You might be interested in

A local phone company charges a monthly fee of $34.99 plus $.05 for each minute of long distance calls. Parts of minutes are rou

stepladder [879]

53.24-34.99=18.25

$1 will buy 20 minutes of long distance calling

x$18.25 = 365 minutes @ $34.99

$132-$34.99=$97.01

20x97=1,940 minutes

minimum=365 minutes

maximum=1,940 minutes

Hope this helps.

3 0

4 years ago

Read 2 more answers

Help me pleeeeeeaaaseeee

Gelneren [198K]

Answer:

I'm hoping the answer should be D. 9 units.

8 0

3 years ago

You decide to enter in a rowing competition. To train, you go to the boat house and begin rolling down stream. You row for the s

fredd [130]

Answer:

Time spent rowing down stream $=100\ seconds$

Speed of boat in still water $=14\ ms^{-1}$

Step-by-step explanation:

Let speed of boat in still water be $= x\ ms^{-1}$

Speed of current $= 10\ ms^{-1}$

Speed of boat down stream = $\textrm{Speed of boat in still water +Speed of current}= (x+10)\ ms^{-1}$

Distance rowed down stream = 2400 m

Time spent rowing down stream = $\frac{Distance}{Speed}=\frac{2400}{x+10}\ s$ = $\frac{Distance}{Speed}=\frac{2400}{x+10}\ s$

Speed of boat up stream = $\textrm{Speed of boat in still water -Speed of current}= (x-10)\ ms^{-1}$

Distance rowed up stream = $\frac{1}{6} \textrm{ of distance rowed downstream}=\frac{1}{6}\times 2400 = 400\ m$

Time spent rowing up stream = $\frac{Distance}{Speed}=\frac{400}{x-10}\ s$

We know that,

$\textrm{Time spent rowing down stream =Time spent rowing up stream}$

So,

$\frac{2400}{x+10}=\frac{400}{x-10}$

Cross multiplying

$2400(x-10)=400(x+10)$

Dividing both sides by $400$

$\frac{2400(x-10)}{400}=\frac{400(x+10)}{400}$

$6(x-10)=x+10$

$6x-60=x+10$

Adding 60 to both sides.

$6x-60+60=x+10+60$

$6x=x+70$

Subtracting both sides by $x$

$6x-x=x+70-x$

$5x=70$

Dividing both sides by 5.

$\frac{5x}{5}=\frac{70}{5}$

∴ $x=14$

Speed of boat in still water $=14\ ms^-1$

Time spent rowing down stream = $\frac{2400}{14+10}=\frac{2400}{24}=100\ s$

3 0

3 years ago

How do you do this question?

Mekhanik [1.2K]

Answer:

(-∞, -2), (-2, -0.618), and (1.618, 3)

Step-by-step explanation:

The red arrows indicate the regions in which the function is decreasing.

The corresponding intervals are (-∞, -0.618) and (1.618, 3),

Let's examine the intervals given.

(-∞, -2): Yes, decreasing.

(-2, -0.618): Yes, decreasing.

(-2, 0.689): No. Decreasing in (-2, -0.618) but increasing in (-0.618, 0.689).

(-1.17, 0.689): No. Decreasing in (-1.17, -0.618) but increasing

in (-0.618, 0.689).

(0.698, 71.457): No. Increasing from (0.698, 1.618), decreasing in (1.618, 3) and increasing in (3, 71.457).

(1.618, 3): Yes, decreasing.

(2.481, ∞): No. Decreasing in (2.481, 3) but increasing in (3, ∞).

The three choices that indicate a decreasing interval are (-∞, -2),

(-2, -0.618), and (1.618, 3).

3 0

3 years ago

Two towers face each other separated by a distance = 15 m. As seen from the top of the first tower, the angle of depression of t

Crazy boy [7]

Answer:

Height of second tower = 17.32m

Step-by-step explanation:

I have attached a diagram depicting the question.

From the diagram, The first tower is depicted by side AEB and the second tower CD.

While d is the distance that separates the two towers and it's given as 15m.

Now, since the angle of depression of the second tower’s base is 60°, then for triangle BAC. Angle C = 60°.

Thus; using trigonometric ratios;

tan 60° = AB/AC.

This gives; AB = d*tan 60°

Similarly, for the triangle BED, BE = d*tan 30°

Since, AE = CD, thus ;

CD = AB − BE

CD = d (tan 60° − tan 30°)

CD = 15(1.7321 − 0.5774)

CD = 15 × 1.1547

CD ≈ 17.32 m.

So, height of second tower = 17.32 m

7 0

3 years ago