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2 years ago

Rewrite the following equation in exponential form.

Mathematics

1 answer:

ira [324]2 years ago

5 0

Answer:

5² = 25

Step-by-step explanation:

log5 25 = 2

log(b) (x) = (y)

make b^(y) = (x)

5² = 25

I hope this helps!

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Which table represents a linear function?

s2008m [1.1K]

Step-by-step explanation:

the second table

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7 0

3 years ago

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What is the reciprocal for<br> 3 wholes and 3/7

maxonik [38]

Answer:

to my knowledge i believe it is 24/3

Step-by-step explanation:

3 0

2 years ago

Julio fills some bins with comic books. There are 8 comic books in each bin.

Nutka1998 [239]

The answer would be 6 bins.

Explanation:

Julio has a total of 48 comic books and he is able to fit 8 in one bin.
Therefore 48 divided by 8 is 6. Also 8 x 6 = 48

5 0

2 years ago

Please help! I need this paper done before my class tomorrow! Thank you!

BabaBlast [244]

Answer:

x=7 and m<LMN = 120

Step-by-step explanation:

if MO bisects LMN then 13x - 31 must be equal to x + 53

13x - x = 53 + 31

12x = 84

x = 7

and

13x - 31 + x + 53 = m<LMN

14x + 22 = m<LMN

since x is 7

14×7 + 22 = 120

3 0

3 years ago

Read 2 more answers

Let S be the solid beneath z = 12xy^2 and above z = 0, over the rectangle [0, 1] × [0, 1]. Find the value of m > 1 so that th

jonny [76]

Answer:

The answer is $\sqrt{\frac{6}{5}}$

Step-by-step explanation:

To calculate the volumen of the solid we solve the next double integral:

$\int\limits^1_0\int\limits^1_0 {12xy^{2} } \, dxdy$

Solving:

$\int\limits^1_0 {12x} \, dx \int\limits^1_0 {y^{2} } \, dy$

$[6x^{2} ]{{1} \atop {0}} \right. * [\frac{y^{3}}{3}]{{1} \atop {0}} \right.$

Replacing the limits:

$6*\frac{1}{3} =2$

The plane y=mx divides this volume in two equal parts. So volume of one part is 1.

Since m > 1, hence mx ≤ y ≤ 1, 0 ≤ x ≤ $\frac{1}{m}$

Solving the double integral with these new limits we have:

$\int\limits^\frac{1}{m} _0\int\limits^{1}_{mx} {12xy^{2} } \, dxdy$

This part is a little bit tricky so let's solve the integral first for dy:

$\int\limits^\frac{1}{m}_0 [{12x \frac{y^{3}}{3}}]{{1} \atop {mx}} \right.\, dx =\int\limits^\frac{1}{m}_0 [{4x y^{3 }]{{1} \atop {mx}} \right.\, dx$

Replacing the limits:

$\int\limits^\frac{1}{m}_0 {4x(1-(mx)^{3} )\, dx =\int\limits^\frac{1}{m}_0 {4x-4x(m^{3} x^{3} )\, dx =\int\limits^\frac{1}{m}_0 ({4x-4m^{3} x^{4}) \, dx$

Solving now for dx:

$[{\frac{4x^{2}}{2} -\frac{4m^{3} x^{5}}{5} ]{{\frac{1}{m} } \atop {0}} \right. = [{2x^{2} -\frac{4m^{3} x^{5}}{5} ]{{\frac{1}{m} } \atop {0}} \right.$

Replacing the limits:

$\frac{2}{m^{2} }-\frac{4m^{3}\frac{1}{m^{5}}}{5} =\frac{2}{m^{2} }-\frac{4\frac{1}{m^{2}}}{5} \\ \frac{2}{m^{2} }-\frac{4}{5m^{2} }=\frac{10m^{2}-4m^{2} }{5m^{4}} \\ \frac{6m^{2} }{5m^{4}} =\frac{6}{5m^{2}}$

As I mentioned before, this volume is equal to 1, hence:

$\frac{6}{5m^{2}}=1\\m^{2} =\frac{6}{5} \\m=\sqrt{\frac{6}{5} }$

3 0

3 years ago