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valentina_108 [34]
3 years ago
7

H is equal to?????????????????? ​

Mathematics
2 answers:
Doss [256]3 years ago
5 0
It would be h=-Tcubes over 9+3r. Please give me a like
Naya [18.7K]3 years ago
5 0

Answer:

H=3r-T³/9

Step-by-step explanation:

8^⅓=2

Thus we have

T=6/2×(r-H/3)^⅓=3(r-H/3)^⅓

T³=27(r-H/3)=27r-9H

9H=27r-T³

H=3r-T³/9

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HELP MEEEEEEEEEE WILL MARK BRAINLESTI HAVE OVERDUE ASSIGNMENTS
ki77a [65]

Answer:

17

Step-by-step explanation:

2x - 7 = 27

2x = 34

x = 17

8 0
4 years ago
Which table represents a linear function?
Viktor [21]

Answer:

table C represents a linear function.

8 0
3 years ago
4> Solve by using Laplace transform: y'+5y'+4y=0; y(0)=3 y'(o)=o
harina [27]

Answer:

y=3e^{-4t}

Step-by-step explanation:

y''+5y'+4y=0

Applying the Laplace transform:

\mathcal{L}[y'']+5\mathcal{L}[y']+4\mathcal{L}[y']=0

With the formulas:

\mathcal{L}[y'']=s^2\mathcal{L}[y]-y(0)s-y'(0)

\mathcal{L}[y']=s\mathcal{L}[y]-y(0)

\mathcal{L}[x]=L

s^2L-3s+5sL-3+4L=0

Solving for L

L(s^2+5s+4)=3s+3

L=\frac{3s+3}{s^2+5s+4}

L=\frac{3(s+1)}{(s+1)(s+4)}

L=\frac3{s+4}

Apply the inverse Laplace transform with this formula:

\mathcal{L}^{-1}[\frac1{s-a}]=e^{at}

y=3\mathcal{L}^{-1}[\frac1{s+4}]=3e^{-4t}

7 0
3 years ago
118 {121÷(11×11)-(-4)-(3-7)}​
Rama09 [41]

<u>Q</u><u>uest</u><u>ion</u><u>:</u>

To Simplify:

118 {121÷(11×11)-(-4)-(3-7)}

<u>Solu</u><u>tion</u>:

↠118 {121÷121+4-(-4)}

↠118 {1+4+4}

↠118 {5+4}

↠118{9}

↠118×9

↠1062

▬▬▬▬▬▬▬▬▬▬▬▬

The upper Solution is done by applying BODMAS

<u>Abou</u><u>t</u><u> </u><u>BODMAS</u><u>:</u>

B→ Bracket

O→ Of

D→ Division

M→ Multiplication

A→ Addition

S→ Subtraction

5 0
3 years ago
Read 2 more answers
PLEASE I NEED HELP IF IT'S WRONG I WILL REPORT
Alinara [238K]

Answer:

what do you need help wth ?

Step-by-step explanation:

3 0
3 years ago
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