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juin [17]
4 years ago
8

Have no clue what to do

Mathematics
1 answer:
WITCHER [35]4 years ago
3 0
It says that the value of a box above two boxes is found by adding the expressions in the two boxes.  I filled in the boxes.  The top box is 13x + 30, or option B.

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Use the trapezoid rule to estimate the area under the graph of y = x2 + 2 for values of x from 0 to 3.
dimulka [17.4K]
So basically you form little trapaziods to fin =d the area. It's been a long time since I took calculus so Imma have to improvise

Area or trapaziod = area of rectangle + area of triangle so you get
2*1+(3-2)*1/2+3*1+(6-3)*1/2+6*1+(11-6)*1/2=15.5
8 0
3 years ago
Last season a Major League Baseball player got 160 hits in 200 times at bat. If the player expects to bat 400 times in the entir
zimovet [89]

Answer:

He's expected to hit 320 times.

Step-by-step explanation:

To solve this problem we first need to compute the ratio at which he hit the ball in the last season, to do that we need to divide the number of hits by the number of tries. We have:

ratio = number of hits/ number of tries

ratio = 160/200 = 0.8

Since he's expected to bat 400 times if he maintains the same ratio in this season he'll hit:

number of hits = (number of tries)*ratio

number of hits = 400*0.8 = 320

He's expected to hit 320 times.

8 0
3 years ago
13. The rate of ascent for the Discovery space shuttle is 114km in 8.5 minutes
kari74 [83]

Answer: 13.4km/min

Step-by-step explanation:

Speed in km/minutes will be calculatedby rate divided by time= 114/8.5

=13.4km/min

8 0
3 years ago
Read 2 more answers
Calculate the sum and enter it below 78 + (-89)
algol [13]

78 + (-89)


78 - 89


Answer: -11


3 0
3 years ago
Read 2 more answers
Can you prove it??<br>it's hard,I tried but couldn't solve this.
BARSIC [14]

I'll abbreviate s=\sin\theta and c=\cos\theta, so the identity to prove is

\dfrac{s+c+1}{s+c-1}-\dfrac{1+s-c}{1-s+c}=2\left(1+\dfrac1s\right)

On the left side, we can simplify a bit:

\dfrac{s+c+1}{s+c-1}=\dfrac{s+c-1+2}{s+c-1}=1+\dfrac2{s+c-1}

\dfrac{1+s-c}{1-s+c}=-\dfrac{-2+1-s+c}{1-s+c}=-1+\dfrac2{1-s+c}

Then

\dfrac{s+c+1}{s+c-1}-\dfrac{1+s-c}{1-s+c}=2\left(1+\dfrac1{s+c-1}-\dfrac1{1-s+c}\right)

So the establish the original equality, we need to show that

\dfrac1{s+c-1}-\dfrac1{1-s+c}=\dfrac1s

Combine the fractions:

\dfrac{(1-s+c)-(s+c-1)}{(s+c-1)(1-s+c)}=\dfrac{2-2s}{c^2-s^2+2s-1}

We can rewrite the denominator as

c^2-s^2+2s-1=c^2+s^2-2s^2+2s-1

then using the fact that c^2+s^2=\cos^2\theta+\sin^2\theta=1, we get

1-2s^2+2s-1=2s-2s^2

so that we have

\dfrac1{s+c-1}-\dfrac1{1-s+c}=\dfrac{2-2s}{2s-2s^2}=\dfrac1s

as desired.

6 0
4 years ago
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