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dangina [55]
3 years ago
14

1. Given (x) =1/2(x + 4)(x – 11), use your graphing calculator to graph this function and

Mathematics
1 answer:
marin [14]3 years ago
6 0
Here are my answers!
the only one i’m not completely sure of is #4 but i’m assuming it’s where the function is above and below the x-axis

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Nicole buys envelopes for her home office that come in boxes of 125 envelopes. If she buys 18 boxes, how many envelopes will she
yawa3891 [41]

Answer:

➩ 2,250 envelopes.

Step-by-step explanation:

Given that, Nicole buys 125 envelopes per box for her home office.

Therefore, She will have 18*125 envelopes in all

So, 18*125 is 2,250 envelopes

6 0
3 years ago
How is the volume of the prism calculated
lukranit [14]

V=B·H

V is volume

B is Base

H is height

3 0
3 years ago
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Karina needs a total of $45 to buy her mother a birthday present. She has saved 20% of the amount so far. How much has she saved
CaHeK987 [17]

Answer:

$9

Step-by-step explanation:

20% of 45 is 9.

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3 years ago
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What is the value of |4 X -3 + 9| plzzz hurry
elixir [45]

Answer:

|4X + 6|

Step-by-step explanation:

8 0
3 years ago
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Tan^2 A/1+cot^2 A + cot^2 A/1+tan^2 A=sec^2 A cosec^2 A-3
omeli [17]
\frac{tan^2x}{1+cot^2x}+\frac{cot^2x}{1+tan^2x}=sec^2x\ cosec^2x-3\\\\L=\frac{tan^2x(1+tan^2x)+cot^2x(1+cot^2x)}{(1+cot^2x)(1+tan^2x)}=\frac{tan^2x+tan^4x+cot^2x+cot^4x}{1+tan^2x+cot^2x+tanxcotx}\\\\=\frac{tan^2x+cot^2x+tan^4x+cot^4x}{1+tan^2x+cot^2x+1}=\frac{tan^2x+2+cot^2x+tan^4x-2+cot^4x}{tan^2x+cot^2x+2}

=\frac{(tanx+cotx)^2+(tan^2x-cot^2x)^2}{(tanx+cotx)^2}=\frac{(tanx+cotx)^2}{(tanx+cotx)^2}+\frac{(tan^2x-cot^2x)^2}{(tanx+cotx)^2}\\\\=1+\frac{(tanx-cotx)^2(tanx+cotx)^2}{(tanx+cotx)^2}=1+(tanx-cotx)^2\\\\=1+tan^2x-2tanx\ cotx+cot^2x=tan^2x+cot^2x+1-2\\\\=\left(\frac{sinx}{cosx}\right)^2+\left(\frac{cosx}{sinx}\right)^2-1=\frac{sin^2x}{cos^2x}+\frac{cos^2x}{sin^2x}-1=\frac{sin^4x+cos^4x}{sin^2x\ cos^2x}-1

=\frac{(sin^2x)^2+2sin^2x\ cos^2x+(cos^2x)^2-2sin^2x\ cos^2x}{sin^2x\ cos^2x}-1\\\\=\frac{(sin^2x+cos^2x)^2-2sin^2x\ cos^2x}{sin^2x\ cos^2x}-1=\frac{1^2-2sin^2x\ cos^2x}{sin^2x\ cos^2x}-1\\\\=\frac{1}{sin^2x\ cos^2x}-\frac{2sin^2x\ cos^2x}{sin^2x\ cos^2x}-1=\frac{1}{sin^2x}\cdot\frac{1}{cos^2x}-2-1\\\\=cosec^2x\cdot sec^2x-3=sec^2x\ cosec^2x-3=R
3 0
4 years ago
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