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Lapatulllka [165]
3 years ago
10

40 as a product of its prime factor can be written like this 40=2x2x2x5 write 60 as a product of its prime factor write in order

smallest to larges
Mathematics
1 answer:
olga_2 [115]3 years ago
7 0

Answer:

2^2x3x5

Step-by-step explanation:

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Identify the domain of an image that has been rotated 270 degrees counter-clockwise
tensa zangetsu [6.8K]

Answer:

A, (4,-3), (1, -1), (7,-6)

Step-by-step explanation:

f(x, y) = (y, -x) there is a 90° clockwise or 270° counterclockwise rotation

8 0
3 years ago
Will Mark Brainliest If Correct Answers! Need Some Help Please
djverab [1.8K]

Answer:

Step-by-step explanation:

a) 4x¹⁰ = 2² * x¹⁰

   64x² = 2⁶ *x²

GCF = 2² * x² = 4x²

b) 4x¹⁰ - 64x² = 4x²*(x⁸ - 16)

                  =4x² *[ (x⁴)² - 4²]

                  = 4x² * (x⁴ + 4) (x⁴ - 4)

                  = 4x² * (x⁴ + 4)* [(x²)²- 2²]

                  =4x² * (x⁴ + 4) * (x² + 2)(x² - 2)

                 

7 0
3 years ago
Twice the sum of a number and 5 is equal to three times the difference of the number and 7. Find the number.
Alexandra [31]

Step-by-step explanation:

Let the required number be x.

According to the given information:

2(x + 5) = 3(x - 7) \\  \\  \therefore \: 2x + 10 = 3x - 21 \\  \\   \therefore \: 10 + 21 = 3x -2x \\  \\ \therefore \: 31 = x \\  \\  \huge \red { \boxed{\therefore \: x = 31}}

Hence the required number is 31.

5 0
4 years ago
If a city has 159 murders a year what dors that averahe in a week
aliya0001 [1]
There are 52 weeks in one year. If we divide 159 by 52 we get 3.05. Therefore, there would be an average of 3-4 murders per week.
6 0
3 years ago
How would you use the Fundamental Theorem of Calculus to determine the value(s) of b if the area under the graph g(x)=4x between
-BARSIC- [3]
The Fundamental Theorem of Calculus regarding geometry states that 
\int\limits^b_a g{(x)} \, dx = F(b)-F(a)

Where F is the indefinite integral of g(x)

The first step is to integrate g(x)
\int\ {4x} \, dx = \frac{4x^{1+1} }{1+1} = \frac{4x^{2} }{2} =2 x^{2}

Then substitute the value of b and a=1 into 2x^{2}

[2 (b)^{2}]-[2 (1)^{2}] = 240
2b^{2} -2=240
2b^{2}=240+2
2b^{2}=242
b^{2}= \frac{242}{2}
b^{2}=121
b=11

Hence the limit of the area under g(x) is between a=1 and b=11

6 0
3 years ago
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