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V125BC [204]
4 years ago
15

56xw+49xk2-24yw-21yk2

Mathematics
1 answer:
nekit [7.7K]4 years ago
8 0
The answer smiplified is 56xw + 49xk2 - 24wy - 21k2y 
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SOS HELPPPP
Hunter-Best [27]

Answer:

18

Step-by-step explanation:

5 0
3 years ago
If f(x) = 3x + 10 and g(x) = 2x - 4, find (f - g)(x).
torisob [31]
<h3>Answer: Choice A)  x+14</h3>

=======================================

Work Shown:

(f-g)(x) = f(x) - g(x)

(f-g)(x) = (f(x)) - (g(x))

(f-g)(x) = (3x+10) - (2x-4)

(f-g)(x) = 3x+10 - 2x+4

(f-g)(x) = (3x-2x) + (10+4)

(f-g)(x) = x+14

4 0
4 years ago
Which of the following is equivalent to 0. 71 mL? A. 0. 00071 L B. 0. 0071 L C. 71 L D. 710 L.
hichkok12 [17]

By using relation between ml and litre we got that 0.71 ml is equal to 0.00071 L

<h3>What is relation between ml and litre?</h3>

Relation between ml and litre is that 1 litre =1000 ml

we know that litre and ml are two units of volume and we can convert data from one to another using their relation

Now here given data is in ml

and data is 0.71 ml

We can convert 0.71 ml in litre as

71ml = 0.71/1000 litre

71ml= 0.00071 litre =0.00071 L

By using relation between ml and litre we got that 0.71 ml is equal to 0.00071 L

To know more about litre visit : brainly.com/question/277659

4 0
3 years ago
Krishawn adds two rational expressions, and then simplifies the answer. His simplified answer is (x-6)-(8+2)/x-2 Which expressio
arlik [135]
The answer is C.

You're welcome ;)
5 0
3 years ago
Margee thinks she can use logs to solve 56 = x^8, since logs seem to make exponents disappear. Unfortunately, Margee is wrong. E
aniked [119]

Answer:

We use log to solve only those equations in which we have our variables in power form.

Step-by-step explanation:

Given : Margee thinks she can use logs to solve 56=x^8, since logs seem to make exponents disappear but Margee is wrong

We have to explain the difference between equations like 2=(1.04)^x  and 56=x^8

We use log to solve only those equations in which we have our variables in power form.

Out of given equation only 2=(1.04)^x has x in power form so we can apply log for solving the equation as,

2=(1.04)^x\\\\\ \text{Taking ln both sides},\\\ln (2)=\ln (1.04)^x\\\\\\\\text{Using idenity} \ln a^b=b\ln a\\\\\\ln (2)=\ln (1.04)^x\\\\\text{on solving, we get}\\\\x=17.673

While solving other equation ,

56=x^8, we can directly take  8 root both side, 56=x^8\\\\\\56^{\frac{1}{8}}=x^{(8\times{\frac{1}{8}})

Thus,  We use log to solve only those equations in which we have our variables in power form.

7 0
4 years ago
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