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mojhsa [17]
3 years ago
15

Solve for r. 7(r + 1) = 20.3

Mathematics
2 answers:
docker41 [41]3 years ago
4 0

Answer:

r = 1.9

Step-by-step explanation:

7r + 7 = 20.3

<u>     - 7      - 7  </u>

7r = 13.3

divide by 7

r = 1.9

Archy [21]3 years ago
4 0

Answer:

r = 1.9

Step-by-step explanation:

7(r + 1) = 20.3

<u>Multiply the parenthesis by 7:</u>

7r + 7 = 20.3

<u>Subtract 7 from each side:</u>

7r = 13.3

<u>Divide 13.3 by 7 to find r:</u>

r = 1.9

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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

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What is the slope of the line that passes through the points El 1, 4) and F(2, 6)?
Alex787 [66]

Answer:

2/3

Step-by-step explanation:

(6 -4) / (2 - (-1)) =(2) / (2+1) = 2/3

3 0
3 years ago
Please help!!! Suppose we have two functions...
denpristay [2]

Answer:

4

Step-by-step explanation:

To find f(-1)g(-1) which is f(-1)*g(-1), first, we must find what are values of f(-1) and g(-1) first which can be found by substituting x = -1 in both functions.

\displaystyle{f(-1)=(-1)^2+1}\\\\\displaystyle{f(-1)=1+1 = 2}

Now we know that f(-1) is 2, next, we find g(-1).

\displaystyle{g(-1) = 3(-1)+5}\\\\\displaystyle{g(-1)=-3+5 = 2}

Now we know that g(-1) is 2 too. Since the question wants to find f(-1)g(-1), therefore 2*2 = 4.

Hence, the answer is 4.

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What is the mean of this set {6,7,9,9,9}​
Naya [18.7K]

Answer:

8

Step-by-step explanation:

To find the mean, or average of a data set, add all the values together, then divide by the number of values.

The data set is: 6,7,9,9,9

1. Add all the values

Add 6,7,9,9 and 9

6+7+9+9+9

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2. Divide by the number of values

Count how many numbers are in the data set. In this case, there are 5 numbers.

Divide 40 by 5.

40/5

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How do I find the horizontal asymptote of h(x)=x+6/x^2-64 ?
SashulF [63]

I assume the equation described is:

( x + 6 ) / ( x^2 - 64 )

You can compare the degree of the numerator and denominator in a function that takes the form of this type of rational equation.

Here are the three rules

#1 (Correct Answer): When the degree of the numerator is smaller then the denominator the horizontal asymptote is y = 0

#2 If the degree of the numerator and denominator is the same, then you take the leading coefficient of the numerator (n) and denominator (d) to create the answer y = n / d in this equations case it would be 1 / 1 since variables technically have an invisible 1 in front of them since anything multiplied by 1 is its self, 1x = x

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Again the final answer is that the horizontal asymptote is y = 0

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3 years ago
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