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3 years ago

14

A flagpole that is 30 feet tall casts a shadow that extends 42 feet from the base of the flagpole. What is the distance from the

top of the flagpole to the end of the shadow?

1 answer:

uranmaximum [27]3 years ago

6 0

It’s the hypotenuse, so you would do a^2 + b^2 = c^2, let a equal 30 feet and let b equal 42 feet. hope this helps

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Mrs . Porcelli’s classroom bulletin board is 2/1/4 feet long. Ms .Smith bulletin board is 3 times as long Mrs . Porcelli’s.How l

snow_lady [41]

Answer:

$6\frac{3}{4}\ ft$

Step-by-step explanation:

we know that

To find out how long is Ms.Smith’s bulletin board, multiply Mrs . Porcelli’s bulletin board by 3

so

$2\frac{1}{4}(3)$

but first convert mixed number to an improper fraction

so

$2\frac{1}{4}\ ft=2+\frac{1}{4}=\frac{9}{4}\ ft$

substitute

$\frac{9}{4}(3)=\frac{27}{4}\ ft$

convert to mixed number

$\frac{27}{4}\ ft=\frac{24}{4}+\frac{3}{4}=6\frac{3}{4}\ ft$

3 0

2 years ago

Suppose given △ABD and △CBD.

patriot [66]

Answer:

The required result is proved with the help of angle bisector theorem.

Step-by-step explanation:

Given △ABD and △CBD, AE and CE are the angle bisectors. we have to prove that $\frac{AD}{AB}=\frac{DC}{CB}$

Angle bisector theorem states that an angle bisector of an angle of a Δ divides the opposite side in two segments that are proportional to the other two sides of triangle.

In ΔADB, AE is the angle bisector

∴ the ratio of the length of side DE to length BE is equal to the ratio of the line segment AD to the line segment AB.

$\frac{DE}{EB}=\frac{AD}{AB}$ → (1)

In ΔDCB, CE is the angle bisector

∴ the ratio of the length of side DE to length BE is equal to the ratio of the line segment CD to the line segment CB.

$\frac{DE}{EB}=\frac{CD}{CB}$ → (2)

From equation (1) and (2), we get

$\frac{AD}{AB}=\frac{CD}{CB}$

Hence Proved.

5 0

2 years ago

Which equation is equivalent to 3[x+3(4x-5)]=15x-24

Klio2033 [76]

3[x+3(4x-5)]=15x-24; x+12x-15=5x-8 (divide both sides by 3); 8x=7 (move all the x's to one side), x=7/8

7 0

2 years ago

Read 2 more answers

Suppose a bag contains many marbles, 54% of which are purple. You draw 5 marbles from the bag with replacement.

dolphi86 [110]

Part A

Use the binomial probability distribution formula.

p = 0.54 = probability of getting a purple marble

n = 5 = sample size

x = 2 = number of purple we want to get

$P(x) = \frac{n!}{x!(n-x)!}*p^x*(1-p)^{n-x}\\\\P(2) = \frac{5!}{2!(5-2)!}*0.54^2*(1-0.54)^{5-2}\\\\P(2) = 10*0.54^2*0.46^3\\\\P(2) = 0.283831776\\\\$

The $\frac{n!}{x!(n-x)!}$ portion is from the nCr combination formula. The exclamation marks indicate a factorial.

Alternatively, you could use Pascal's Triangle for that portion.

<h3>Answer: 0.283831776</h3>

This decimal value is exact. Round it however you need to.

============================================================

Part B

To find the expected value, aka the mean, we multiply the sample size and probability of getting a purple marble on any single selection.

n*p = 5*0.54 = 2.7

<h3>Answer: 2.7</h3>

6 0

1 year ago

3х - 5 = -11 what is x

Gennadij [26K]

Answer:

-2

Step-by-step explanation:

Add 5 to both sides to get

3x-5+5=-11+5

and simplify to get

3x = -6

x = -2

7 0

2 years ago

Read 2 more answers