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2 years ago

15

What is 38475 x 304944 ? My teacher won't let me use a calculator!

1 answer:

bearhunter [10]2 years ago

7 0

Answer:

1,059,680,400

Step-by-step explanation:

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Help me i actually need this asap...Only right answer ..No spammer.No wrong answer.

Leona [35]

$\sqrt[3]{8} \: - \sqrt{3}$

7 0

2 years ago

Read 2 more answers

PLEASE HELP WITH THIS I HAVE NO IDEA HOW TO SOLVE!!!!

VARVARA [1.3K]

Write the vertex form of the equation and find the necessary coefficient to make it work.
.. y = a*(x +3)^2 -2
.. = ax^2 +6ax +9a -2

You require the y-intercept to be 7. So, for x=0, you have
.. 9a -2 = 7
.. 9a = 9
.. a = 1

The equation you seek is
.. y = x^2 +6x +7

4 0

2 years ago

2/5 x 1 2/3<br> i know its easy

worty [1.4K]

Answer:

2/3

Step-by-step explanation:

Convert any mixed numbers to fractions.

Then your initial equation becomes:

25×53

Applying the fractions formula for multiplication,

=2×55×3

=10/15

Simplifying 10/15, the answer is

=2/3

7 0

2 years ago

The polar curve $r = 1 + \cos \theta$ is rotated once around the point with polar coordinates $(2,0).$ What is the area of the r

mash [69]

Answer:

$Area = -2.3147$

Step-by-step explanation:

Given

$r = 1 + \cos \theta$

Required

Determine the area with coordinates $(2,0)$

The area is represented as:

$Area = \frac{1}{2}\int\limits^b_a {r^2} \, d\theta$

Where

$r = 1 + \cos \theta$

and

$(a,b) = (2,0)$

Substitute values for r, a and b in

$Area = \frac{1}{2}\int\limits^b_a {r^2} \, d\theta$

$Area = \frac{1}{2}\int\limits^0_2 {(1 + cos\theta)^2} \, d\theta$

Expand

$Area = \frac{1}{2}\int\limits^0_2 {(1 + cos\theta)(1 + cos\theta)} \, d\theta$

$Area = \frac{1}{2}\int\limits^0_2 {(1 + 2cos\theta+cos^2\theta} )\, d\theta$

By integratin the above, we get:

$Area = \frac{1}{2}*\frac{(cos(\theta) + 4)sin(\theta) + 3\theta}{2}[0,2]$

$Area = \frac{(cos(\theta) + 4)sin(\theta) + 3\theta}{4}[0,2]$

Substitute 0 and 2 for $\theta$ one after the other

$Area = \frac{(cos(0) + 4)sin(0) + 3*0}{4} - \frac{(cos(2) + 4)sin(2) + 3*2}{4}$

$Area = \frac{(cos(0) + 4)sin(0)}{4} - \frac{(cos(2) + 4)sin(2) + 6}{4}$

$Area = \frac{(1 + 4)*0}{4} - \frac{(cos(2) + 4)sin(2) + 6}{4}$

$Area = - \frac{(cos(2) + 4)sin(2) + 6}{4}$

$Area = \frac{-sin(2)(cos(2) + 4) - 6}{4}$

Get sin(2) and cos(2) in radians

$Area = \frac{-0.9093 * (-0.4161 + 4) - 6}{4}$

$Area = \frac{-9.2588}{4}$

$Area = -2.3147$

3 0

2 years ago

Please help! This is overdue and I don’t have time!!!☹️

Alecsey [184]

the answer will be around 53.398

8 0

3 years ago