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Rzqust [24]
3 years ago
15

What is the nth term for 1 7 15 25 37

Mathematics
1 answer:
Tju [1.3M]3 years ago
3 0
51 hoped this helped since it goes plus two after adding 6
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On a test of 76 ​items, Pedro got 93​% correct.​ (There was partial credit on some​ items.) How many items did he get​ correct?
Firlakuza [10]
You would use the “is/of” method.

93 / 100 = x / 76

Cross multiply and divide:

7068 / 100 = 70.68

This means Pedro got around 70 to 71 questions correct and 5 or 6 questions incorrect.
6 0
3 years ago
The triangle T has vertices at (-2, 1), (2, 1) and (0,-1). (It might be an idea to
Firdavs [7]

Rewrite the boundary lines <em>y</em> = -1 - <em>x</em> and <em>y</em> = <em>x</em> - 1 as functions of <em>y </em>:

<em>y</em> = -1 - <em>x</em>  ==>  <em>x</em> = -1 - <em>y</em>

<em>y</em> = <em>x</em> - 1  ==>  <em>x</em> = 1 + <em>y</em>

So if we let <em>x</em> range between these two lines, we need to let <em>y</em> vary between the point where these lines intersect, and the line <em>y</em> = 1.

This means the area is given by the integral,

\displaystyle\iint_T\mathrm dA=\int_{-1}^1\int_{-1-y}^{1+y}\mathrm dx\,\mathrm dy

The integral with respect to <em>x</em> is trivial:

\displaystyle\int_{-1}^1\int_{-1-y}^{1+y}\mathrm dx\,\mathrm dy=\int_{-1}^1x\bigg|_{-1-y}^{1+y}\,\mathrm dy=\int_{-1}^1(1+y)-(-1-y)\,\mathrm dy=2\int_{-1}^1(1+y)\,\mathrm dy

For the remaining integral, integrate term-by-term to get

\displaystyle2\int_{-1}^1(1+y)\,\mathrm dy=2\left(y+\frac{y^2}2\right)\bigg|_{-1}^1=2\left(1+\frac12\right)-2\left(-1+\frac12\right)=\boxed{4}

Alternatively, the triangle can be said to have a base of length 4 (the distance from (-2, 1) to (2, 1)) and a height of length 2 (the distance from the line <em>y</em> = 1 and (0, -1)), so its area is 1/2*4*2 = 4.

6 0
3 years ago
P(x)=x(2x+5)(x+1)What are the zeros of this polynomial
erastova [34]

Answer:

(0,-5/2,-1)

Step-by-step explanation:

x(2x+5)(×+1)=0

x=0 2x+5=0 x+1=0

2x=-5 x=-1

x=-5/2

5 0
2 years ago
Please help!! I’ll mark brainliest!
elena-14-01-66 [18.8K]

Answer:

Just from looking at the picture, I'm assuming that we're trying to find the terms that are like to the terms in the blue section.

1. 8c since both are in terms of c

2. 7t since both are in terms of t

3. 8r since both are in terms of r

4. -4n since both are in terms of n

5. xy since both are in terms of xy

6. 16 since both don't have any variables

7. 4sp since both are in terms of sp

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
Must do this to get the other question answer
weeeeeb [17]
So what exactly is the question

3 0
3 years ago
Read 2 more answers
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