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3 years ago

13

Five times a number ,subtracted from. Four is triple the number find the number

1 answer:

svlad2 [7]3 years ago

4 0

That is the answer I got Hope I Helped

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Represent root 3 on the number line

Vedmedyk [2.9K]

Answer:

first represent root 2

6 0

3 years ago

For each graph, write the equation of the given translation

jasenka [17]

Answer:

Step-by-step explanation:

if you modify the parent function to go <u>up and down</u>

for example parent function is y=|x| to go <u>down</u> 3 you do y=|x|-3

if you modify the parent function to go <u>left and right</u>

for example parent function is y=|x| to go <u>right </u>3 you do y=|x-3|

a.goes down 3, and 2 to the right

y=|x-2|-3

b. 3 down and 1 to the left

y=(x+1)²-3

c. 5 up and 2 to the right

d. 1 up and 5 to the right

6 0

3 years ago

What is the distance between -36 and -4 on a number line?

Amanda [17]

-4 - (-36) = -4 + 36 = 32 units

4 0

3 years ago

Read 2 more answers

Please solve i will give brainiest 100 point question ****** do the whole page please need to pass or i will fail its my final t

dmitriy555 [2]

Answer:

1. Find the difference between the areas.

<u>Area of the small rectangle</u>: $(x+2)(x+7)=x^2+7x+2x+14=x^2+9x+14$

<u>Area of the big rectangle</u>: $(x+9)(x+11)=x^2+11x+9x+99=x^2+20x+99$

The difference is: $11x+85$

$( x^2+20x+99)- (x^2+9x+14)=x^2+20x+99-x^2-9x-14=11x+85$

2.

You can solve this question just by looking at the graph.

a) The height is 4 meters.

$f(d)=h=-2d^2+7d+4$

To find the height of the bleachers, we should consider the moment before the shoot, when the distance is equal to 0.

$f(0)=h=-2(0)^2+7(0)+4$

$h=4$

The height is 4 meters.

b) 9 meters.

For $d=1$

$f(1)=h=-2(1)^2+7(1)+4$

$f(1)=h=-2+7+4$

$h=9$

b) The ball travels 4 meters.

But to calculate it, it is when $h=0$

$0=-2d^2+7d+4$

Using the quadratic formula:

$d=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$d=\frac{-7 \pm \sqrt{7^2-4\left(-2\right)4}}{2\left(-2\right)}$

$d=\frac{-7\pm\sqrt{81}}{-4}$

$d=\frac{-7\pm9}{-4}$

It will give us to solutions, once it is a quadratic equation, but we are talking about a positive distance.

$d=-\frac{1}{2} \text{ or }d=4$

3.

In this question, we have to find the area of the cylinder and the sphere.

From the information given, we have

a = 5mm and d = 5mm, therefore the radius is 2.5 mm.

The volume of a cylinder:

$V=\pi r^2h$

$V=\pi (2.5)^2 \cdot 5$

$V=31.25 \pi$

$V_{c} \approx 98.17 \text{ m}^3$

The volume of the sphere:

$V=\frac{4}{3} \pi r^2$

$V_{s} \approx 65.4 \text{ m}^3$

The volume of the capsule is approximately $163.57 \text{ m}^3$

3 0

4 years ago

-1/6 + 2/3 (9 - 3/4) - 1/2<br><br> I’m confused if someone can help

emmainna [20.7K]

Answer:

4.83333333333

Step-by-step explanation:

I'm sorry if that is wrong but I believe it is right

7 0

3 years ago