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malfutka [58]
3 years ago
5

E=VIT how would I isolate the “V” variable

Mathematics
1 answer:
Zigmanuir [339]3 years ago
6 0

Answer:

V =\frac{E}{IT}

Step-by-step explanation:

You can isolate the "V" variable by dividing by IT on both sides:

\frac{E}{IT} =\frac{VIT}{IT}

On the left, the IT from the top and bottom cancel, leaving you with just V:

V =\frac{E}{IT}

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Rewrite the following expression.
faust18 [17]

We have been given the expression

x^{\frac{9}{7}}

We have the exponent rule

x^{\frac{a}{b}}= x^{a^{(\frac{1}{b})}}

Using this rule, we have

x^{\frac{9}{7}}= x^{9^{(\frac{1}{7})}}

Now, using the fact that x^{\frac{1}{n}}=\sqrt[n]{x}, we get

x^{\frac{9}{7}}= \sqrt[7]{x^9}\\
\\
x^{\frac{9}{7}}=\sqrt[7]{x^7\times x^2}\\
\\
x^{\frac{9}{7}}=x\sqrt[7]{x^2}

D is the correct option.

6 0
3 years ago
Read 2 more answers
4x^2 y+8xy'+y=x, y(1)= 9, y'(1)=25
jarptica [38.1K]

Answer with explanation:

\rightarrow 4x^2y+8x y'+y=x\\\\\rightarrow 8xy'+y(1+4x^2)=x\\\\\rightarrow y'+y\times\frac{1+4x^2}{8x}=\frac{1}{8}

--------------------------------------------------------Dividing both sides by 8 x

This Integration is of the form ⇒y'+p y=q,which is Linear differential equation.

Integrating Factor

 =e^{\int \frac{1+4x^2}{8x} dx}\\\\e^{\log x^{\frac{1}{8}+\frac{x^2}{2}}\\\\=x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}

Multiplying both sides by Integrating Factor  

x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}\times [y'+y\times\frac{1+4x^2}{8x}]=\frac{1}{8}\times x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}\\\\ \text{Integrating both sides}\\\\y\times x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}=\frac{1}{8}\int {x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}} \, dx \\\\8y\times x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}=\int {x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}} \, dx\\\\8y\times x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}=-[x^{\frac{9}{8}}]\times\frac{ \Gamma(0.5625, -x^2)}{(-x^2)^{\frac{9}{16}}}\\\\8y\times x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}=(-1)^{\frac{-1}{8}}[ \Gamma(0.5625, -x^2)]+C-----(1)

When , x=1, gives , y=9.

Evaluate the value of C and substitute in the equation 1.

6 0
3 years ago
In rectangle PQRS, SQ = 78, PS = 30, and mPRS = 23°. Find SR.
kondor19780726 [428]
\cos (23)=\frac{sr}{78}\rightarrow78\cdot\cos (23)=SR=67.20

7 0
1 year ago
Please answer this, 25 points and will give brainliest!!!
Gekata [30.6K]

Answer:

b

Step-by-step explanation:

The median of a triangle is a line that from the vertex touches the middle lets eliminate options

a=ab starts at a vertex but doensn't touch a the middle of a line X

b=cd starts at a vertex and touches the middle of a line Ye

C= ce stars at a vertex but doesn't end at the middle of a line X

d=mb starts in the middle of the triangle and and ends a vertex X

<u>SO THE CORRECT ANSWER IS B</u>

<em>-hope this helps :)</em>

8 0
3 years ago
X=6. y=5. <br><br><br><br> 12×+3y <br>please help me​
IgorLugansk [536]

Answer:

Your answer: 87

Step-by-step explanation:

Since you are given what both variables are in the equation, this problem requires you to plug it in.

12(6)+3(5)= ?

You can either do this in a calculator, or do this one step at a time.

12(6)=72

3(5)=15

72+15=87

7 0
3 years ago
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