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Lostsunrise [7]
3 years ago
12

PLEASEEEEE HELPPP ILL GIVE 30 POINTS PLSSS HELP

Mathematics
1 answer:
beks73 [17]3 years ago
8 0
Answer 4

4multiplyed by 4 is 16
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1/2 e = 1/4 help I don't know what this means
marissa [1.9K]
E is 1/2, because 1/2 of 1/2 is 1/4
4 0
3 years ago
Please at least help me with one of them cause I have no idea
Fudgin [204]

Answer:

Solving v = \frac{4}{3}\pi r^3 for r gives us: r=\sqrt[3]{\frac{3v}{4\pi}}

Solving a_{ave} = \frac{v_2-v_1}{t_2-t_1} for v2 gives us: v_2 = (t_2-t_1)a_{ave}+v_1

Step-by-step explanation:

Solving an equation for a variable or constant means that we have to isolate the value on one side of the equation or write the whole equation in terms of that variable or constant.

Now,

Solving v = \frac{4}{3}\pi r^3 for r

v = \frac{4}{3}\pi r^3

Multiplying whole equation by 3/4

\frac{3}{4}.v = \frac{3}{4}.\frac{4}{3} \pi r^3

\frac{3}{4}v = \pi r^3

Dividing by Pi on both sides

\frac{3v}{4\pi} = \frac{\pi r^3}{\pi}\\\frac{3v}{4\pi} = r^3

Taking cube root on both sides

\sqrt[3]{r^3} = \sqrt[3]{\frac{3v}{4\pi}}  \\r = \sqrt[3]{\frac{3v}{4\pi}}

Now

Solving a_{ave} = \frac{v_2-v_1}{t_2-t_1} for v2

Multiplying both sides by (t2-t1)

(t_2-t_1)a_{ave} = \frac{v_2-v_1}{t_2-t_1}(t_2-t_1)\\(t_2-t_1)a_{ave} = v_2-v_1

Adding v1 on both sides

(t_2-t_1)a_{ave}+v_1 = v_2-v_1+v_1\\(t_2-t_1)a_{ave}+v_1 = v_2

Hence,

Solving v = \frac{4}{3}\pi r^3 for r gives us: r=\sqrt[3]{\frac{3v}{4\pi}}

Solving a_{ave} = \frac{v_2-v_1}{t_2-t_1} for v2 gives us: v_2 = (t_2-t_1)a_{ave}+v_1

6 0
3 years ago
A random number generator is used to create a list of 300 single digit numbers. Of those 300 numbers, 146 are odd and 154 are ev
mario62 [17]
0.44<<< vitamin CCC^<>^<>^<><span />
6 0
3 years ago
Read 2 more answers
Simplify -4.5+4.2+5.6-7.3
kumpel [21]

Answer:

-2

Step-by-step explanation:

Hope you do well on your assignment!

6 0
3 years ago
Read 2 more answers
I have a question about polynomial division it is in the picture
Cloud [144]

To answer this question, we have to do the long division process for polynomials. We can do the operation as follows:

To do this division process, we have:

1. Divide the first term of the dividend by the first element of the divisor. They are:

\frac{-4x^3}{4x^2}=-x

2. Now, we have to multiply this result by the divisor, and the result will change its sign since we have to subtract that result from the dividend as follows:

-x\cdot(4x^2_{}-4x-4)=-4x^3+4x^2+4x

And since we to subtract this result from the dividend, we end up with:

-(-4x^3+4x^2+4x)=4x^3-4x^2-4x

3. Then we have the following algebraic addition:

\frac{\begin{cases}-4x^3+24x^2-15x-15 \\ 4x^3-4x^2-4x\end{cases}}{20x^2-19x-15}

4. Again, we need to divide the first term of the dividend by the first term of the divisor as follows:

\frac{20x^2}{4x^2}=5

5. And we have to multiply 5 by the divisor, and the result will be subtracted from the dividend:

5\cdot(4x^2-4x-4)=20x^2-20x-20

Since we have to subtract this from the dividend, we have:

-(20x^2-20x-20)=-20x^2+20x+20

6. And we have to add this algebraically to the dividend we got in the previous step:

\frac{\begin{cases}20x^2-19x-15 \\ -20x^2+20x+20\end{cases}}{x+5}

And this is the remainder of the division, x + 5.

As we can see from the division process, we got as:

1. The quotient: -x + 5

q=-x+5

2. The remainder: x + 5.

R=x+5

Since we have that the dividend = divisor * quotient + remainder.

Therefore, the result for this division is:

-4x^3+24x^2-15x-15=(4x^2-4x-4)\cdot(-x+5)+(x+5)

4 0
1 year ago
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