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frez [133]
2 years ago
11

4h−13=−9h What is the value of h?

Mathematics
2 answers:
Lostsunrise [7]2 years ago
7 0

Solution:

<u>Subtract 4h both sides and solve for h.</u>

  • 4h − 13 = −9h
  • => -13 = -9h - 4h
  • => -13 = -13h
  • => 13 = 13h
  • => h = 1

The value of h is 1.

asambeis [7]2 years ago
5 0

Answer:

h=1

Step-by-step explanation:

4h=9h+13 ->changes sides so sign changes

4h+9h=13 ->normal adition

13h=13 ->divide

h=1

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The volume of gas in a container at a constant temperature varies inversely as the pressure. If the volume is 32 cubic centimete
aleksklad [387]

Step-by-step explanation:

Given that,

The volume of gas in a container at a constant temperature varies inversely as the pressure.

V_1=32\ cm^3\\\\P_1=8\ \text{pounds}\\\\V_2=60\ cm^3

We need to find pressure P_2.

According to question,

\dfrac{V_1}{V_2}=\dfrac{P_2}{P_1}

Putting all the values

P_2=\dfrac{P_1V_1}{V_2}\\\\P_2=\dfrac{8\times 32}{60}\\\\P_2=4.26\ \text{pounds}  

So, the pressure is 4.26 pounds.

7 0
3 years ago
H(a) = a^2 + 3a; Find h(-10)
Colt1911 [192]
H(-10)=70
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6 0
3 years ago
Write 375 as the product of prime factors .<br> Give your answer in index form
gtnhenbr [62]

Answer:

3•5•5•5

Step-by-step explanation:

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8 0
3 years ago
If m&lt;1=6x+50 and m&lt;2=4x+40 find x
Pepsi [2]
\text{Adjacent angles on a straight line sum up to 180} \textdegree

\text{Therefore }\measuredangle 1 + \measuredangle 2 = 180 \textdegree

\text {Since } \measuredangle 1 = 6x+50 \text { and } \measuredangle 2 = x+40

(6x+50)  + (4x+40) = 180

6x+50 + 4x+40 = 180

10x+90 = 180

10x = 180 - 90

10x = 90

x = 9

Answer: x = 9
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3 years ago
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She needs another quarter cup of strawberries
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