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nlexa [21]
3 years ago
5

The table shows a proportional relationship between x

Mathematics
1 answer:
lakkis [162]3 years ago
7 0
What’s the question it’s not showing up for me or you just didn’t type it
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Can someone please help first answer gets brainlist
OLga [1]

Answer:

3/5

Step-by-step explanation:

0.6 is actually 6/10 as a fraction. You divide bot the 6 and the 10 by 2 to get 3/5

3 0
3 years ago
Can someone please explain how to simplify expressions that have negative or positive exponents
solniwko [45]

Step-by-step explanation:

I guess you just slove the exponent can I see the actual question

8 0
3 years ago
b. Sixty-five pounds of candy was divided into four different boxes. The second box contained twice the amount of the first box.
Ainat [17]

First box = 14 pounds

Second box = 2x = 28 pounds

Third Box = x+2= 16 pounds

Fourth box = x/2 = 7 pounds

<u>Step-by-step explanation:</u>

Here we have , Sixty-five pounds of candy was divided into four different boxes. The second box contained twice the amount of the first box. The third box contained two more pounds than the first box. The last box contained one-fourth the amount in the second box. We need to find How much candy was in each box. Let's find out:

We have a total of 65 pounds of candy ! Let in first box we have x pounds so , second box contained twice the amount of the first box i.e.

⇒ 2x

The third box contained two more pounds than the first box i.e.

⇒ x+2

The last box contained one-fourth the amount in the second box i.e.

⇒ (\frac{1}{4})2x = \frac{x}{4}

Therefore , Sum of pounds of candy are :

⇒ \frac{x}{2} +x+2+2x+x=65

⇒ \frac{x}{2} +4x=63

⇒ \frac{9x}{2}=63

⇒ x=63(\frac{2}{9} )

⇒ x=14

Therefore , Candy in each box is :

First box = 14 pounds

Second box = 2x = 28 pounds

Third Box = x+2= 16 pounds

Fourth box = x/2 = 7 pounds

8 0
4 years ago
What is 302.135 rounded to the nearest tenth
xz_007 [3.2K]
Well, it would be 302.1 as three is lowere than five, and when its lower you just leave it. if its five or higher than you round up. hope that helps! :)
4 0
3 years ago
Read 2 more answers
Find (f • g) when f(x) = x^2 + 5x + 6 and g(x) = 1/x+3
Nikitich [7]

Answer:

option D

Step-by-step explanation:

f(x) = x^2 + 5x + 6

g(x)= \frac{1}{x+3}

(fog)(x) = f(g(x))

Plug in g(x) in f(x)

We plug in 1/x+3 in the place of x  in f(x)

f(g(x))= f(\frac{1}{x+3})= (\frac{1}{x+3})^2 + 5(\frac{1}{x+3}) + 6

To simplify it we take LCD

LCD is (x+3)(x+3)

\frac{1}{(x+3)(x+3)}+5\frac{1*(x+3)}{(x+3)(x+3)}+\frac{6(x+3)(x+3)}{(x+3)(x+3)}

\frac{1}{x^2+6x+9}+\frac{(5x+15)}{x^2+6x+9}+\frac{6x^2+36x+54}{x^2+6x+9}

All the denominators are same so we combine the numerators

\frac{1+5x+15+6x^2+36x+54}{x^2+6x+9}

\frac{6x^2+41x+70}{x^2+6x+9}

Option D is correct

8 0
3 years ago
Read 2 more answers
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