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mr Goodwill [35]

3 years ago

11

This is the 6th time i've asked this question. i'm so desperate, please help me!

1 answer:

OverLord2011 [107]3 years ago

4 0

Answer:

Look at explanation.

Step-by-step explanation:

For the first four, take the first term and add the common difference to it to get the second term. Then add the common difference to the second term to get the third term, and so on until the 5th term. If the common difference is negative, you subtract it instead of adding it.

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– 2y – 10 = x<br> - 10x - y = -14

fredd [130]

Answer:

hey there guys ^^

Step-by-step explanation:

5 0

3 years ago

PLEASE ANSWER ASAP!!!!!!! WILL GIVE BRAINLIEST!!!!!!!!

Misha Larkins [42]

Answer:

Based on the data, this is an example of causation.

Step-by-step explanation:

i got it right on the test .

6 0

2 years ago

Read 2 more answers

a person bought 8 items at a market and received $8 in change. if each item was priced at $4, how much did the person give the v

mariarad [96]

If a person bought 8 items and each item was priced at $4, that means the total amount spent was $32.

$(8)(4)=32$

The person received 8 dollars in change. This means that the person payed 8 dollars more than what they spent. Well, we just found out how much they spent, so we can find out how much money the person gave the vendor.

Let p = amount of money paid by the person.

$32+8=p$

$p=40$

The person gave the vendor $40.

3 0

2 years ago

The students lockers at a college are to be numbered consecutively form 1 to 800 using plastic digits. Each digit cost 10c. What

KIM [24]

Answer would be eight because you divide eight hundred by ten cents to get eight.

6 0

3 years ago

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

3 years ago