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2 years ago

I need help please !!!!!!!!!!!!

Mathematics

1 answer:

marin [14]2 years ago

7 0

This picture shows how to think and answer.

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PLEASE HELP! EXPLAIN HOW YOU GOT THE ANSWER AND I WILL MARK BRAINLIEST!

kow [346]

3 prefer pop:4 prefer rock
3/4 prefer pop
(3/4)of 14=0.75*14=10.5 people/14people
around 11 people would prefer pop out of 14

7 0

3 years ago

What is 15-5(4x-7)=50

katovenus [111]

Simplifying

15 + -5(4x + -7) = 50

Reorder the terms:

15 + -5(-7 + 4x) = 50

15 + (-7 * -5 + 4x * -5) = 50

15 + (35 + -20x) = 50

Combine like terms: 15 + 35 = 50

50 + -20x = 50

Add '-50' to each side of the equation.

50 + -50 + -20x = 50 + -50

Combine like terms: 50 + -50 = 0

0 + -20x = 50 + -50

-20x = 50 + -50

Combine like terms: 50 + -50 = 0

-20x = 0

Solving

-20x = 0

Solving for variable 'x'.

Move all terms containing x to the left, all other terms to the right.

Divide each side by '-20'.

x = 0.0

Simplifying

x = 0.0

5 0

3 years ago

1/10 + 70/100 = help me pleas

Vikentia [17]

The answer is 0.8
I think that’s the answer

8 0

2 years ago

Read 2 more answers

The height of a tennis ball tossed into the air is modeled by h(x) = 40x â€" 16x 2, where x is elapsed time in seconds. During w

egoroff_w [7]

The time interval x >0.5 and x > 2.04 will the tennis ball be at least 15 feet above the ground.

Given that,

The height of a tennis ball tossed into the air is modeled by,

$\rm h(x) = 40x - 16x^2$

Where x is elapsed time in seconds.

We have to determine,

During what time interval will the tennis ball be at least 15 feet above the ground?

According to the question,

The height of a tennis ball tossed into the air is modeled by,

$\rm h(x) = 40x - 16x^2$

Where x is elapsed time in seconds.

Then,

When the tennis ball be at least 15 feet above the ground,

$h(x) = 15$

Substitute the value of h(x) in the equation,

$\rm h(x) = 40x - 16x^2\\\\ 15 = 40x - 16x^2\\\\ 16x^2-40x+15=0$

Factorize the equation for finding the time interval will the tennis ball be at least 15 feet above the ground is,

$\rm 16x^2-40x+15= 0\\\\x = \dfrac{-b\pm \sqrt{b^2-4ac}}{2a}\\\\x = \dfrac{-(-40)\pm \sqrt{(-40)^2-4 \times 16 \times 15}}{2 \times 16}\\\\x = \dfrac{-(-40)\pm \sqrt{1600- 960}}{32}\\\\x = \dfrac{40\pm \sqrt{640}}{32}\\\\x = \dfrac{40 + 25.29}{32} \\\\x = \dfrac{40 + 25.29}{32} \ and \ x = \dfrac{40 - 25.29}{32}\\\\x = \dfrac{65.29}{32} \ and \ x = \dfrac{14.71}{32}\\\\x = 2.04 \ and \ x = 0.45$