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Dovator [93]
3 years ago
12

Can sumone help me #11

Mathematics
1 answer:
Maru [420]3 years ago
5 0

Answer:

25$ per month

Step-by-step explanation:

100$ was saved every 4 months so you divide that by four. You will get 25 which is your rate. (25$ per month)

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What is the sum or product of 93+(68+7)
ser-zykov [4K]
93+(68+7)

first do what's in the parentheses
(68+7)
68+7= 75

new problem: 93+75
93+75= 168

Answer is 168
6 0
3 years ago
Read 2 more answers
The area of a rectangle is 82.84 square meters. If the length is 7.6 meters what is the width?
alexandr1967 [171]
The width is 10.9. When you need to find the length or width when you have the area you divide the area by given number you have so for this problem you divide 82.84 by 7.6 to get 10.9 meters.
4 0
3 years ago
Find the range of {(3.2, –3), (7.6, 5.9), (1.4, –3), (–9.1, 8.3)}.
MrRa [10]

Answer:

{-3, 5.9, -3, 8.3}

Step-by-step explanation:

The range is always the second number of all pairs

8 0
3 years ago
3. Jenny puts two wooden blocks together. One block has a length of 5 in., a width of 4 in., and a
Mazyrski [523]

Answer:

140 inches cubed

Step-by-step explanation:

Use the formula V = length * width * height to find the volumes of each block, then add them together to get the combined volume. So, (5 * 4 * 5) + (4 * 5 * 2) = 140 inches cubed, which is the combined volume.

Hopefully this helps- let me know if you have any questions!

3 0
2 years ago
Quadratic equations and complex numbers PLEASE HELPPPP ASAP
kondor19780726 [428]

we are given

27x^3-1=0

we can also write it as

(3x)^3-(1)^3=0

now, we can use factor formula

a^3-b^3=(a-b)(a^2+ab+b^2)

we can use above formula

we get  

(3x)^3-(1)^3=(3x-1)((3x)^2+3x\times 1+(1)^2)

now, we can simplify it

27x^3-1=(3x-1)(9x^2+3x+1)

now, we can set it to 0

27x^3-1=(3x-1)(9x^2+3x+1)=0

and then we can solve for x

(3x-1)(9x^2+3x+1)=0

3x-1=0

x=\frac{1}{3}

9x^2+3x+1=0

now, we can use quadratic formula

x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}

now, we can plug values

and we get

x=\frac{-3\pm \sqrt{3^2-4\cdot \:9\cdot \:1}}{2\cdot \:9}

x=-\frac{1}{6}+i\frac{\sqrt{3}}{6},\:x=-\frac{1}{6}-i\frac{\sqrt{3}}{6}

So, we will get solution as

x=\frac{1}{3},\:x=-\frac{1}{6}+i\frac{\sqrt{3}}{6},\:x=-\frac{1}{6}-i\frac{\sqrt{3}}{6}...............Answer

5 0
3 years ago
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