All categories

3 years ago

Plz help im almost done just a few more hw questions I'm stuck on

Mathematics

1 answer:

sesenic [268]3 years ago

8 0

The third one should be -2(2x-6)=12-4x
Last one should be 2(2x-6)+4x=8x-12

You might be interested in

1) Solve quadratic equations by factoring. 2x^2 + 5x - 12 = 0

Hoochie [10]

(2x-3)(x+4) —> x= 3/2 x= -4

3 0

3 years ago

Read 2 more answers

ANSWER ASAP THANKS ...............

mylen [45]

Answer:

250

Step-by-step explanation:

5*5=25

25*10=250

+$++$$+$

6 0

3 years ago

Evaluate the expressions 12(8-5)

saveliy_v [14]

Answer:

Step-by-step explanation:

First you would solve what's in the parenthesis because the Order of Pemdas.

Second you would multiply the answer of what's in the parenthesis which is 3 to 12.

Third you would get the answer which is 36!

8 0

4 years ago

Read 2 more answers

Hi boys can you help me pls!!!!!!!

goblinko [34]

Answer: OPTION A.

Step-by-step explanation:

Given the following function:

$h(x)=-\frac{1}{4}x^2+\frac{1}{2}x+\frac{1}{2}$

You know that it represents the the height of the ball (in meters) when it is a distance "x" meters away from Rowan.

Since it is a Quadratic function its graph is parabola.

So, the maximum point of the graph modeling the height of the ball is the Vertex of the parabola.

You can find the x-coordinate of the Vertex with this formula:

$x=\frac{-b}{2a}$

In this case:

$a=-\frac{1}{4}\\\\b=\frac{1}{2}$

Then, substituting values, you get:

$x=\frac{-\frac{1}{2}}{(2)(-\frac{1}{4}))}\\\\x=1$

Finally, substitute the value of "x" into the function in order to get the y-coordinate of the Vertex:

$h(1)=y=-\frac{1}{4}(1)^2+\frac{1}{2}(1)+\frac{1}{2}\\\\y=0.75$

Therefore, you can conclude that:

The maximum height of the ball is 0.75 of a meter, which occurs when it is approximately 1 meter away from Rowan.

7 0

3 years ago

Suppose that the functions r and s are defined for all real numbers x as follows.

mixas84 [53]

ANSWER

Given

$r(x) = x - 1$

and

$s(x) = 3 {x}^{2}$
ANSWER TO QUESTION 1

$(r + s)(x) = r(x) + s(x)$

$(r + s)(x) = x - 1 + 3 {x}^{2}$

$(r + s)(x) = 3 {x}^{2} + x - 1$

ANSWER TO QUESTION 2

$(r \times s)(x) = r(x) \times s(x)$

$(r \times s)(x) = (x - 1) \times 3 {x}^{2}$

$(r \times s)(x) = 3 {x}^{3} - 3 {x}^{2}$

ANSWER TO QUESTION 3

$( r - s)(x) = r(x) - s(x)$

$( r - s)(x) = x - 1 - 3 {x}^{2}$

$( r - s)( - 3) = (- 3) - 1 - 3 ({ - 3}^{2} )$

$( r - s)( - 3) = - 4 - 3 \times 9$

$( r - s)( - 3) = - 4 - 27 = - 31$

4 0

3 years ago