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r-ruslan [8.4K]
3 years ago
5

In triangle DEF, ZE is a right angle. If cos(D) = ? and sin(D) = 24, what is tan(D)?

Mathematics
1 answer:
musickatia [10]3 years ago
6 0

Answer:

ans=24/7

Step-by-step explanation:

cosD=7/25=b/h

sinD=24/25=p/h

now....tanD=p/b=24/7

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Can i get help please ?
wariber [46]

Answer:

D -- the last one

Step-by-step explanation:

The formula tells you what to do.

t = 3

A0 = 25

A(t) = ?

A(3)  = 25 * (1/2)^3

A(t) = 25 * 1/8

A(t) = 25/8

A(t) = 3.125

4 0
3 years ago
Helpelelelepelelelelelelellel
Romashka-Z-Leto [24]

Answer:

0.83 cents

Step-by-step explanation:

$25.35 + $2.99 = $28.34

$30 - $28.34 = $1.66

1.66 ÷ 2 = 0.88

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3 0
3 years ago
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Emily studies 40 minutes after lunch for a science exam. She studies z more minutes that evening.
vladimir2022 [97]

Answer:

y = 40 + z

Step-by-step explanation:

4 0
3 years ago
Mr. Sánchez owns a square field. The side lengths are 0.9 kilometers. There are 1,980 prairie dog burrows in the fields. What is
Simora [160]
The first thing we are going to do is find the area of the field. To do this we are going to use the area of a square formula: A=s^2
Were 
A is the area in square kilometers 
s is one of the sides of the square
We know for our problem that  the side lengths of the field are 0.9 kilometers, so s=0.9. Lets replace that value in our formula to find A:
A=(0.9)^2
A=0.81 

Now, to find the population density of the filed, we are going to use the population density formula: P_{d}= \frac{I}{A}
where
P_{d} is the population density in <span>in burrows per square kilometer
</span>I is the number of burrows 
A is the are of the field
We know that I=1980 and A=0.81, so lets replace those values in our formula:
P_{d}= \frac{1980}{0.81}
P_{d}=2444.4

We can conclude that the <span>density of prairie dog burrows is approximately 2444 burrws per square kilometer.</span>
5 0
3 years ago
Simplify. <br><br> 4√6+2√6−√6<br><br> A- 5√18<br> B-5√6<br> C-6√6<br> D- 6√18
Aleksandr [31]

Answer:

B

Step-by-step explanation:

Simplifying radicals is similar to collecting like terms in algebra.

Here, the like terms are the radicals

Given

4\sqrt{6} + 2\sqrt{6} - \sqrt{6}

= \sqrt{6}(4 + 2 - 1 )

= 5\sqrt{6} → B

3 0
3 years ago
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