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jonny [76]
3 years ago
15

What is the product of the polynomials below?

Mathematics
1 answer:
Allisa [31]3 years ago
8 0
<span>(7x2 + 5x + 7)(4x - 6)
= 28x^3 - 42x^2 + 20x - 30x + 28x  - 42
= 28x^3 - 22x^2 - 2x - 42

It's  D</span>
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Arc is 41 and angle is 40. I have no other given information but I have to find x
tiny-mole [99]

First of all I want to point out you drew the diagram a little wrong. The Arc is 41 doesn't mean its 41 degrees it means it has length 41 so remove the degrees symbol.

Now for the answer the other arc have to have angle 40 too because vertical angles. And because the radius is the same, both of the length has formula 40/360*pi*2*radius which is 41 in this case. So x has to be 41 also :) Done!

7 0
3 years ago
What is the answer to 140=m+25
DaniilM [7]
You Add 140 + 25 First
7 0
3 years ago
Read 2 more answers
Please help I'm stuck on this questions thanks
polet [3.4K]
So its asking for basically the percentage of the first number out of the second. 

3. 25/50 = 50%
4. 125/75 = 167%
5. 32/28 = 114%
6. 7/10 = 70%


Hope this helped!  :)
6 0
3 years ago
If SV⊥RT, m∠RSU = (17x – 3)°, and m∠UST = (6x – 1)°, find TSV
goblinko [34]

Applying the linear pair theorem, the measure of angle TSV in the image given is: 86°.

<h3>How to Apply the Linear Pair Theorem?</h3>

Given the following angles in the image above:

Measure angle RSU = (17x - 3)°,

Measure angle UST = (6x – 1)°

To find the measure of angle TSV, we need to find the value of x in the given expressions as shown below:

m∠RSU + m∠UST = 180 degrees (linear pair]

Substitute the values

17x - 3 + 6x - 1 = 180

Solve for x

23x - 4 = 180

23x = 180 + 4

23x = 184

x = 8

m∠TSV = 180 - 2(m∠UST) [Linear Pair Theorem]

m∠TSV = 180 - 2(6x - 1)

Plug in the value of x

m∠TSV = 180 - 2(6(8) - 1)

m∠TSV = 86°

Therefore, applying the linear pair theorem, the measure of angle TSV in the image given is: 86°.

Learn more about the linear pair theorem on:

brainly.com/question/5598970

#SPJ1

6 0
1 year ago
CALCULUS: For an object whose velocity in ft/sec is given by v(t) = sin(t), what is its distance, in feet, travelled on the inte
rodikova [14]

The linked answer is wrong because that integral gives you the net displacement of the object, not the total distance.

To get the distance, you have to integrate the speed (as opposed to velocity), which involves integrating the absolute value of the velocity function.

\mathrm{distance} = \displaystyle\int_1^5 |\sin(t)| \,\mathrm dt

By definition of absolute value,

|\sin(t)|=\begin{cases}\sin(t)&\text{for }\sin(t)\ge0\\-\sin(t)&\text{for }\sin(t)

Over this particular integration interval,

• sin(<em>t</em> ) ≥ 0 for 1 ≤ <em>t</em> < <em>π</em>, and

• sin(<em>t</em> ) < 0 for <em>π</em> < <em>t</em> ≤ 5

so you end up splitting the integral at <em>t</em> = <em>π</em> as

\mathrm{distance} = \displaystyle\int_1^\pi \sin(t)\,\mathrm dt + \int_\pi^5 (-\sin(t))\,\mathrm dt

Now compute the distance:

\mathrm{distance} = -\cos(t)\bigg|_1^\pi + \cos(t)\bigg|_\pi^5

\mathrm{distance} = -(\cos(\pi) - \cos(1)) + (\cos(5) - \cos(\pi))

\mathrm{distance} = -2\cos(\pi) + \cos(1) + \cos(5) \approx 2.82

making B the correct answer.

7 0
3 years ago
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