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2 years ago

15

How to factor this problem?

1 answer:

ddd [48]2 years ago

6 0

Hope this can help you.

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(27 thousands 3 hundreds 5 ones) x 10

anyanavicka [17]

Answer:

Two hundred seventy three thousand and fifty .

Step-by-step explanation:

Given : (27 thousands 3 hundreds 5 ones) x 10

Solution:

27 thousands 3 hundreds 5 ones = 27,305

So, $27,305 \times 10$

$273,050$

Periods are counted from last .

The 1st period consists of ones, tens and hundred.

The 2nd period consists of thousand, 10 thousand and 100 thousands.

The 3rd period consists of million, 10 million and 100 million.

So, 273,050 is Two hundred seventy three thousand and fifty

Hence (27 thousands 3 hundreds 5 ones) x 10 is Two hundred seventy three thousand and fifty .

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

3 years ago

Y=x^2-2x-5;y=x^3-2x^2-5x-9

ehidna [41]

Are the directions asking you to solve these ?

8 0

3 years ago

Find the value of the variable.

MatroZZZ [7]

Answer:

sin 30=14/x

x=14*2

x=28

and

x=√{28²-14²}

x=14√3

7 0

3 years ago

Which of these options is not a quadratic equation in x?

il63 [147K]

Answer:

3x³ - 2x² + 1 = 0

Step-by-step explanation:

By definition, a quadratic equation cannot have an exponent higher than 2.

6 0

3 years ago

Read 2 more answers