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sergey [27]
3 years ago
5

Factor the polynomial: a^2 − ab −8a+8b

Mathematics
1 answer:
Blizzard [7]3 years ago
8 0
Ab9 is your answer my friend
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The height of a pyramid-shaped sculpture with a square base is one half the length of each side. The volume of the sculpture is
Nimfa-mama [501]

The volume of a pyramid is V = 1/3 x base length^2 x height

Let X = base length.

We are told the height is 1/2 the length, so height would be 1/2x

Now we have:

972 = 1/3 * x^2 * 1/2x

Simplifying we get:

972 = 1/6x^3

Multiply both sides by 6:

5832 = x^3

Take the cube root of both sides:

x = 18

The length of the base is 18 inches.

The height is 1/2 that, so the height would be 18/2 = 9 inches.

4 0
3 years ago
Y''+y'+y=0, y(0)=1, y'(0)=0
mars1129 [50]

Answer:

y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+\frac{1}{\sqrt{3}}\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )

Step-by-step explanation:

A second order linear , homogeneous ordinary differential equation has form ay''+by'+cy=0.

Given: y''+y'+y=0

Let y=e^{rt} be it's solution.

We get,

\left ( r^2+r+1 \right )e^{rt}=0

Since e^{rt}\neq 0, r^2+r+1=0

{ we know that for equation ax^2+bx+c=0, roots are of form x=\frac{-b\pm \sqrt{b^2-4ac}}{2a} }

We get,

y=\frac{-1\pm \sqrt{1^2-4}}{2}=\frac{-1\pm \sqrt{3}i}{2}

For two complex roots r_1=\alpha +i\beta \,,\,r_2=\alpha -i\beta, the general solution is of form y=e^{\alpha t}\left ( c_1\cos \beta t+c_2\sin \beta t \right )

i.e y=e^{\frac{-t}{2}}\left ( c_1\cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )

Applying conditions y(0)=1 on e^{\frac{-t}{2}}\left ( c_1\cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right ), c_1=1

So, equation becomes y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )

On differentiating with respect to t, we get

y'=\frac{-1}{2}e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )+e^{\frac{-t}{2}}\left ( \frac{-\sqrt{3}}{2} \sin \left ( \frac{\sqrt{3}t}{2} \right )+c_2\frac{\sqrt{3}}{2}\cos\left ( \frac{\sqrt{3}t}{2} \right )\right )

Applying condition: y'(0)=0, we get 0=\frac{-1}{2}+\frac{\sqrt{3}}{2}c_2\Rightarrow c_2=\frac{1}{\sqrt{3}}

Therefore,

y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+\frac{1}{\sqrt{3}}\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )

3 0
3 years ago
What is the slope of the line that passes through the point 3,5 and -2,2
Nostrana [21]

Answer:

3/5

Step-by-step explanation:

Slope=

\frac{y2 - y1}{x2 - x1}

slope =

\frac{2 - 5}{ - 2 - 3}

=

\frac{3}{5}

8 0
3 years ago
39. Write a ratio and a percent for the shaded area.
Xelga [282]

Hey there! :)

Answer:

First choice. 36/100 or 36%.

Step-by-step explanation:

Total # of squares:

10 × 10 = 100 squares.

# of squares shaded:

6 × 6 = 36 squares.

shaded/total to find the percent shaded:

Fraction: 36/100

Percentage: 36/100 × 100 = 36%.

5 0
3 years ago
Read 2 more answers
The table shows the batting records for members of a softball team. Use the table to find each experimental probability.
Lerok [7]

Answer:

A. 20/31

B. 5/12

Step-by-step explanation:

A. P(Juanita hits) = 20/31

B. P(Su doesn't hit) = (24-14)/24 = 10/24 = 5/12

7 0
3 years ago
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