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garik1379 [7]
3 years ago
9

The total snowfall in Burke, Virginia, this January was 7 1/4 in. In February, 20% more snow fell than in January.

Mathematics
2 answers:
Elis [28]3 years ago
7 0
I think the answer is 7 2/9
sergiy2304 [10]3 years ago
3 0
The awnser is 7 2/9 i agree with jesuschild777
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Is this the correct answer?
alexgriva [62]

Answer:

Step-by-step explanation:

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3 years ago
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Let t = 5.<br><br> What is the value of 5t – 5?
yKpoI14uk [10]
If t=5, you just need to substitute 5 for t in the equation and solve, so

5(5) - 5
25 - 5 = 20

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3 years ago
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If the area of the region bounded by the curve y^2 =4ax and the line x= 4a is 256/3 Sq units, using integration find the value o
almond37 [142]

If the area of the region bounded by the curve y^2 =4ax  and the line x= 4a  is \frac{256}{3} Sq units, then the value of  a will be 2 .

<h3>What is area of the region bounded by the curve ?</h3>

An area bounded by two curves is the area under the smaller curve subtracted from the area under the larger curve. This will get you the difference, or the area between the two curves.

Area bounded by the curve  =\int\limits^a_b {x} \, dx

We have,

y^2 =4ax  

⇒ y=\sqrt{4ax}

x= 4a,

Area of the region  =\frac{256}{3}  Sq units

Now comparing both given equation to get the intersection between points;

y^2=16a^2

y=4a

So,

Area bounded by the curve  =   \[ \int_{0}^{4a} y \,dx \] ​

 \frac{256}{3} =\[  \int_{0}^{4a} \sqrt{4ax}  \,dx \]

\frac{256}{3}=   \[\sqrt{4a}  \int_{0}^{4a} \sqrt{x}  \,dx \]

 \frac{256}{3}=   \[2\sqrt{a}  \int_{0}^{4a} x^{\frac{1}{2} }   \,dx \]                                            

\frac{256}{3}= 2\sqrt{a}  \left[\begin{array}{ccc}\frac{(x)^{\frac{1}{2}+1 } }{\frac{1}{2}+1 }\end{array}\right] _{0}^{4a}

\frac{256}{3}= 2\sqrt{a}  \left[\begin{array}{ccc}\frac{(x)^{\frac{3}{2} } }{\frac{3}{2} }\end{array}\right] _{0}^{4a}

\frac{256}{3}= 2\sqrt{a} *\frac{2}{3}  \left[\begin{array}{ccc}(x)^{\frac{3}{2}\end{array}\right] _{0}^{4a}

On applying the limits we get;

\frac{256}{3}= \frac{4}{3} \sqrt{a}   \left[\begin{array}{ccc}(4a)^{\frac{3}{2}  \end{array}\right]

\frac{256}{3}= \frac{4}{3} \sqrt{a} *\sqrt{(4a)^{3} }

\frac{256}{3}= \frac{4}{3} \sqrt{a} *  8 *a^{2}   \sqrt{a}

\frac{256}{3}= \frac{4}{3} *  8 *a^{3}

⇒ a^{3} =8

a=2

Hence, we can say that if the area of the region bounded by the curve y^2 =4ax  and the line x= 4a  is \frac{256}{3} Sq units, then the value of  a will be 2 .

To know more about Area bounded by the curve click here

brainly.com/question/13252576

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8 0
2 years ago
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A spinner has 14 equal sectors with different letters for each sector, as shown below:
LuckyWell [14K]
The correct answer would be 1/14
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3 years ago
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1 7/8 divided by 2 2/5
Elan Coil [88]

Answer:1 23/62

Step-by-step explanation:

17

8

22

5

=1

23

62

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