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Aneli [31]
3 years ago
12

(3x^3 - 7x^2 + 12) - (3x^3 + 6x^2 + 10x)

Mathematics
1 answer:
ad-work [718]3 years ago
5 0
The answer should be B. you combine the like terms. when it is a negative sign in front of the parenthesis, when removing the parenthesis, every term inside the parenthesis need to change sign, negative to positive, positive to negative.
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The length of a rectangle is 4 more than 3 times the width if the perimeter of the rectangle is 18.4 cm what is the area of the
garik1379 [7]
Let's call the width: w

the lenght is then 3w+4 ("4 more than 3 times the width")

and the parameter would be 2(w+3w+4), that is 2*(4w+4), that is 8w+8.

this is also equal to 18.4:

8w+8=18.4
8w=10.4
w=1.3

this is the width, and the lenght is:
4+3*1,3=4+3.9=7.9

and the area is their product:
1.3*7.9=10.27
6 0
3 years ago
Which of these equations does not have any solutions?
Readme [11.4K]

the answer is C........

7 0
3 years ago
You’ve observed the following returns on Crash-n-Burn Computer’s stock over the past five years: 14 percent, –9 percent, 16 perc
MatroZZZ [7]

Answer:

a. 9%

b. 0.01445

c. 12.02%

Step-by-step explanation:

n = 5 years

Return rates = 14%, –9%, 16%, 21%, and 3%

a. Arithmetic average return

A= \frac{14-9+16+21+3}{5} \\A=9.00

b. Historical variance (to 5 decimal places)

V= \frac{\sum (X_i-A)^2}{n-1} \\V= \frac{(0.14-0.09)^2+(-0.09-0.09)^2+(0.16-0.09)^2(0.21-0.09)^2(0.03-0.09)^2}{5-1}\\ V=0.01445

c. Standard deviation.

s = \sqrt{V} \\s= \sqrt{0.01445}\\s=0.1202 = 12.02\%

8 0
3 years ago
2.
mihalych1998 [28]
Answer:
False

Explanation:
No, a triangle cannot be constructed with sides of 2 in., 3 in., and 6 in.

For three line segments to be able to form any triangle you must be able to take any two sides, add their length and this sum be greater than the remaining side.
3 0
3 years ago
Read 2 more answers
Quadrilateral BCDE is inscribed in circle A. The measure of ⏜ is 73°. The measure of ⏜ is 53° and the measure of ⏜ is 102°. Dete
Anton [14]

Answer: c

Step-by-step explanation:

4 0
3 years ago
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