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lora16 [44]
3 years ago
12

I WILL GIVE BRAINLEST!!!!!!

Mathematics
1 answer:
IgorLugansk [536]3 years ago
4 0
Can Smart people Answer this plz.??????? I WILL GIVE BRAINLEST!!!!!!


Frank earned x dollars the first week of his new job. He earned 5% more the
second week than the first week. Which expression represents the total amount of
money Frank earned for both weeks?
A. 1.50x
B. 1.05x
C. 2.00x
D. 2.05x


PLSSS SHOW YOUR WORK PLSSSS
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What is the surface area of the equilateral triangular prism? A) 51 cm2 B) 56 cm2 C) 62 cm2 D) 67 cm2
ELEN [110]
You have to give some info about the prism
3 0
4 years ago
Read 2 more answers
Which of the fractions is in lowest terms and is equivalent to -1.25
hjlf

Answer:

  The lowest terms equivalent to - 1.25  =  - \frac{5}{4}

Step-by-step explanation:

- 1.25 = - 1\frac{25}{100}

         = - ( 1 × 100 +25) ÷ 100

         =  -( \frac{125}{100} ) × (\frac{5}{5}

        =  - \frac{25}{20}

        =  - (\frac{25}{20}) × (\frac{5}{5}

        =  - \frac{5}{4}

7 0
3 years ago
What is the greatest common factor "GCF" of 24x^2 y^3 z^4 + 18x^6 z - 36x^3 y^2
Leokris [45]

Answer:

6x^2

Step-by-step explanation:

4 0
3 years ago
Consider the given function and the given interval. f\(x\) = (x - 7)**2 text(, ) [5 text(, ) 11] (a) Find the average value fave
Jet001 [13]

a. f has an average value on [5, 11] of

f_{\rm ave}=\displaystyle\frac1{11-5}\int_5^{11}(x-7)^2\,\mathrm dx=\frac{(x-7)^3}{18}\bigg|_5^{11}=\frac{4^3-(-2)^3}{18}=4

b. The mean value theorem guarantees the existence of c\in(5,11) such that f(c)=f_{\rm ave}. This happens for

(c-7)^2=4\implies c-7=\pm2\implies c=9\text{ or }c=5

8 0
3 years ago
Does anyone know how to find the side length of this square.if so then, i will mark brainlist.
kvv77 [185]
Hello!

To find the side length you use the equation

a =  \sqrt{2}  \frac{d}{2}

a is side length
d is diagonal

Put in the number we know

a =  \sqrt{2} *  \frac{27}{2}

Divide

a =  \sqrt{2} * 13.5

Multiply

a = 19.091

The answer is 19.09

Hope this helps!
8 0
4 years ago
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