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Nata [24]
2 years ago
12

What is the first term in this expansion? a 2x15 b 215x15 c 10y15 d 1015y15

Mathematics
2 answers:
Oduvanchick [21]2 years ago
8 0

Answer:

2^15x^15 is the correct answer.

Step-by-step explanation:

maw [93]2 years ago
4 0

Answer:

The answer is 215x15

Step-by-step explanation:

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Number 5 please help me​
Vesna [10]

Answer:

I think C or D

Step-by-step explanation:

3 0
3 years ago
I need help please!!
shepuryov [24]
The first one the height of the candle remains the same after many minutes the second one the water in a storage tank decreas. Find slope by using slope formula
3 0
3 years ago
Can some one give me this answer
iris [78.8K]
Check picture below, solve for "x" and "y"

unless you've already covered the 30-60-90 triangle, then check your book on them, to see the ratios, using the 30-60-90 ratios is simpler

3 0
3 years ago
Old faithful in Yellowstone National Park shoots water 60 feet into the air that casts a shadow of 42 feet. What is the height o
Serhud [2]

Answer:

The height of near by tree will be 90 feet.

Step-by-step explanation:

Considering the water triangle as shown in figure a.

As the old faithful shoots water 60 feet into the air that casts a shadow of 42 feet.

Considering the tree triangle as shown in figure a. The tree triangle has a shadow 63 feet long.

Let x be the height of near by tree that cost a shadow 63 feet long.

Assuming that the triangles are similar

So,

the ratios of similar triangles will be related as:

\frac{63}{42}=\frac{x}{60}

\mathrm{Switch\:sides}

\frac{x}{60}=\frac{63}{42}

{x}=60.\frac{63}{42}

x=90 feet

Therefore, the height of near by tree will be 90 feet.

Keywords: ratio, similar triangle

Learn more about similar triangle from brainly.com/question/10161974

#learnwithBrainly

7 0
3 years ago
If a pool is 4 m wide and 16 m diagonally across, how long is the pool? (Pyth. Theorem)
alukav5142 [94]

Answer:

L=4\sqrt{15}\ m

or

L=15.49\ m

Step-by-step explanation:

we know that

Applying the Pythagorean Theorem in the rectangular pool

d^2=L^2+W^2

where

d is the diagonal of the pool

L is the length of the pool

W is the wide of the pool

we have

d=16\ m\\W=4\ m

substitute the given values

16^2=L^2+4^2

solve for L

L^2=16^2-4^2\\L^2=240\\L=\sqrt{240}\ m

simplify

L=4\sqrt{15}\ m

or

L=15.49\ m

4 0
2 years ago
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