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RideAnS [48]
3 years ago
9

Which expression is equivalent to sqr 400i8j4k6

Mathematics
1 answer:
Gelneren [198K]3 years ago
4 0

I have attached the correct answer. Brainliest will be much appreciated!

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What is the ratio of cosine of C?<br><br> A 24/7<br><br> B 24/25<br><br> C 7/25<br><br> D 25/24
Alecsey [184]

Answer:

Cos C = adjecent/hypotenuse

Cos C = 14/50

divide the numerator and the denominator by 2

Cos C = 7/25

The answer is C

8 0
2 years ago
In na hour , a jogger burns about 4 calories pero pound of body weight. How Manu calories will a jogger who weighs 151 pounds in
Alex777 [14]

This is a rate*unit problem.


So, 4 calories per pound x 151 pounds

4\times151=604

He will burn 604 calories in one hour

6 0
3 years ago
What is m∠A to the nearest degree?
Anuta_ua [19.1K]
So this is a right triangle so you can use sin, cosine, or tangent. for A, you have the sides opposite and hypotenuse, which means you use sin*
sin(A) = opposite / hypotenuse
sin(A) = 56 / 75
arcsin(56/75) = A
A= about 48.3
so to the nearest degree, your answer is 48

*soh cah toa (if you dont know what that is, i can explain it)
4 0
3 years ago
Read 2 more answers
A cable car begins its trip by moving up a hill. As it moves up, it gains elevation at a constant rate of 50 feet/minute until i
asambeis [7]

Answer:

he cable car's elevation will be 750 feet after 15 minutes

Step-by-step explanation:

Given:

elevation, e =2,000 feet

rate, I = 50 feet/minute

The equation that models the car's elevation, e, after t minutes is e = |t

The cable car's elevation will be 750 feet after how many minutes?

since, e = It

for 2,000 feet = 50 feet/minute  * t

t = 2000/50 =40minutes

using the same constant rate, I for 750ft, T =

since, e = It

for 750 feet = 50 feet/minute  * t

t = 750/50 =15minutes

3 0
3 years ago
Kyle solved 18 of 24 puzzles ina puzzle book. He says that he can use an equivalent fraction to find the percent of puzzles in t
Rashid [163]

\frac{18}{24}  =  \frac{9}{12}  =   \frac{3}{4}
The percentage is 75℅
8 0
3 years ago
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