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2 years ago

I need the answer !!!

Mathematics

1 answer:

bonufazy [111]2 years ago

4 0

I'm guessing you are looking for the answer on the first question, because I cannot read the rest of the second one lol.
So start with the volume of cone formula: V=3.14(5)^2 (6/3)
157.08
157
H is the answer

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Solve simultaneously <br>2 log y=log 2 + log x <br>2 to power y = 4 to power x<br>

skad [1K]

Answer:

Step-by-step explanation:

hello : here is an solution

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3 years ago

What is the slope of the line? y + 5=2(x+1)

mel-nik [20]

Answer:

Step-by-step explanation:

So we have the equation:

$y+5=2(x+1)$

This is in the format point-slope form, where:

$y-y_1=m(x-x_1)$

Here, m is the slope.

In our original equation, 2 replaces m.

Therefore, our slope is 2.

5 0

3 years ago

Read 2 more answers

6 times the difference of a number and 4 equals 7

Nonamiya [84]

6( x - 4) = 7
6 x - 24 = 7
6 x = 7 + 24
6 x = 31
x = 31/6

6 ( 31/6 - 4) = 7
186/6 - 24 = 7
186/6 = 144/6 = 7

42/6 = 7
7 = 7

4 0

3 years ago

What is the decibel level of the noise from a fire alarm with intensity 10^-1 watts per square inch

prisoha [69]

Answer:

110 db

Step-by-step explanation:

6 0

2 years ago

Find the partial fraction decomposition of the rational expression with prime quadratic factors in the denominator

SpyIntel [72]

$\dfrac{5x^4-7x^3-12x^2+6x+21}{(x-3)(x^2-2)^2}=\dfrac{a_1}{x-3}+\dfrac{a_2x+a_3}{x^2-2}+\dfrac{a_4x+a_5}{(x^2-2)^2}$
$\implies 5x^4-7x^3-12x^2+6x+21=a_1(x^2-2)^2+(a_2x+a_3)(x-3)(x^2-2)+(a_4x+a_5)(x-3)$

When $x=3$ , you're left with

$147=49a_1\implies a_1=\dfrac{147}{49}=3$

When $x=\sqrt2$ or $x=-\sqrt2$ , you're left with

$\begin{cases}17-8\sqrt2=(\sqrt2a_4+a_5)(\sqrt2-3)&\text{for }x=\sqrt2\\17+8\sqrt2=(-\sqrt2a_4+a_5)(-\sqrt2-3)\end{cases}\implies\begin{cases}-5+\sqrt2=\sqrt2a_4+a_5\\-5-\sqrt2=-\sqrt2a_4+a_5\end{cases}$

Adding the two equations together gives $-10=2a_5$ , or $a_5=-5$ . Subtracting them gives $2\sqrt2=2\sqrt2a_4$ , $a_4=1$ .

Now, you have

$5x^4-7x^3-12x^2+6x+21=3(x^2-2)^2+(a_2x+a_3)(x-3)(x^2-2)+(x-5)(x-3)$
$5x^4-7x^3-12x^2+6x+21=3x^4-11x^2-8x+27+(a_2x+a_3)(x-3)(x^2-2)$
$2x^4-7x^3-x^2+14x-6=(a_2x+a_3)(x-3)(x^2-2)$

By just examining the leading and lagging (first and last) terms that would be obtained by expanding the right side, and matching these with the terms on the left side, you would see that $a_2x^4=2x^4$ and $a_3(-3)(-2)=6a_3=-6$ . These alone tell you that you must have $a_2=2$ and $a_3=-1$ .

So the partial fraction decomposition is

$\dfrac3{x-3}+\dfrac{2x-1}{x^2-2}+\dfrac{x-5}{(x^2-2)^2}$

7 0

3 years ago