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ANEK [815]
3 years ago
15

Is this correct? H^3•h^4•k=h^12•k

Mathematics
2 answers:
andriy [413]3 years ago
7 0
No it is incorrect try rearranging things
jenyasd209 [6]3 years ago
5 0

Answer:

No

Step-by-step explanation:

We can use the rule [ x^y * x^z = x^y+z ]

h^3 * h^4

h^3+4

h^7

Best of Luck!

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Help pleaseeeeee thanks so much
aleksklad [387]

Answer:

The answer is n = <u>31</u>

Step-by-step explanation:

Equation:   <u>2n + 13</u> = 75

2n = 75 - 13

2n = 62

n = 62/2

n = 31

check:

2*31 + 13 = 75

62 + 13 = 75

4 0
3 years ago
Read 2 more answers
Justin is measuring his yard for a fence. He needs to figure out how much wood it will take. What unit of measure does Justin ne
sineoko [7]

Answer:

i think it is miles yay

6 0
2 years ago
What is the length of leg s of the triangle below?
Alchen [17]

The question is not well formatted. A ell formatted version of the question type is attached and solved below :

Answer:

10

Step-by-step explanation:

Using trigonometry :

Taking the angle 45°

From trigonometry :

Sin θ = opposite / hypotenus

Opposite = s ; hypotenus = 10√2

Sin 45 = s / 10√2

Recall :

Sin 45 = 1/√2

We have ;

1/√2 = s / 10√2

s = 10√2 * 1/√2

s = 10√2 / √2

s = 10

6 0
2 years ago
I absolutely suck in math so please just help me
agasfer [191]

All you have to do is substitute the values of x.

h(0) = |-3(0) + 5|

h(0) = |5|

h(0) = 5

The answer for h(0) is 5.

h(5) = |-3(5) +5|

h(5) = |-15 + 5|

h(5) = |-10|

h(5) = 10

The answer to h(5) is 10.

I hope this helps!

5 0
3 years ago
ASAP!! Three consecutive numbers have a sum of 156. What are the 3 numbers?
ryzh [129]

For this problem, all you need to do is find the three #'s that add up to 156.

So, lets look at the answers and add them up.

A. 50, 52, 54

50 + 52 + 54 = 156

B. 51,52,53

51 + 52 + 53 = 156

C. 49,50,51

49 + 50 + 51 = 150

D. 49,51,53

49 + 51 + 53 = 153

We get the answers (50,52,54) and (51,52,53)

Now, consecutive numbers are numbers that in order, like 1,2,3.

Therefore, the answer is (51,52,53)

3 0
3 years ago
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