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3 years ago

Four more than twice a number is twenty.

Mathematics

1 answer:

wariber [46]3 years ago

4 0

Answer:

Step-by-step explanation:

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Hello, can I please get help in the image below? Thank you so much!

vodomira [7]

C.
x + 4 9 + 2 8 = 1 8 0

8 0

2 years ago

Y=6x-1, y=3x-4 the solution is_____

TEA [102]

Answer:

y= -7 and x= -1

Step-by-step explanation:

To do this, line up the variables so it looks like this...

y-6x=-1

y-3x=-4

The reason why -6x and -3y changes to negative is because that it moved to the other side of the equation.

distribute negative to the bottom equation because the two equations have the same number variable which is y...

y-6x=-1

-y+3x=4

Y cancels out and it will be left with -3x = 3.

X = -1.... plug that in to any of the two equation and you get y = -7.

Hope I helped and have a nice day!!

5 0

2 years ago

Let f(x) = x2 – 4, x < 0. What is f-1?

Galina-37 [17]

Answer:

-3

Step-by-step explanation:

f(x) = x2 – 4

replace x with -1

f(-1) = (-1)^2 – 4

= -3

5 0

1 year ago

Mrs. Jones purchased 6.2 pounds of ground beef. She cooked 0.5 of the ground beef. How many pounds of ground beef did she cook?

Anvisha [2.4K]

Answer:

She cooked 3.1 pounds

Step-by-step explanation:

6.2 x 0.5 = 3.1

5 0

2 years ago

Determine what must be multiplied so that the expression below will have a common denominator. Use the backslash / to type in a

Mashutka [201]

Answer:

Multiply the first fraction by $\frac{x+4}{x+4}$

Multiply the second fraction by $\frac{x-4}{x-4}$

Step-by-step explanation:

Given

$\frac{4}{x^2 - 16} + \frac{3}{x^2 + 8x + 16}$

Required

Make the denominator equal

$\frac{4}{x^2 - 16} + \frac{3}{x^2 + 8x + 16}$

Factorize the denominator

$\frac{4}{x^2 - 4^2} + \frac{3}{x^2 + 4x + 4x+ 16}$

$\frac{4}{(x - 4)(x + 4)} + \frac{3}{(x+ 4)(x + 4)}$

Multiply the first fraction by $\frac{x+4}{x+4}$

Multiply the second fraction by $\frac{x-4}{x-4}$

$\frac{x+4}{x+4} * \frac{4}{(x - 4)(x + 4)} + \frac{3}{(x+ 4)(x + 4)} * \frac{x-4}{x-4}$

$\frac{4(x + 4)}{(x - 4)(x + 4)(x + 4)} + \frac{3(x - 4)}{(x+ 4)(x + 4)(x - 4))}$

$\frac{4(x + 4)}{(x^2 - 16)(x + 4)} + \frac{3(x - 4)}{(x^2 - 16)(x + 4))}$

<em>The expression now have the same denominator</em>

7 0

3 years ago