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3 years ago

Find the area of the shaded region. Round to the nearest hundredth where

Mathematics

1 answer:

Svetllana [295]3 years ago

6 0

Answer:

192.42m²

Step-by-step explanation:

The larger circle: pi(10.5²)≈346.36

The smaller circle: pi(7²)≈153.94

Then subtract

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My age is the difference between twice my age in 4 years and twice my age 4 years ago how old am I?

Llana [10]

Answer: 16

Step-by-step explanation:

let a = your present age

a = 2(a+4) - 2(a-4)

a = 2a + 8 - 2a + 8

a = 2a - 2a + 8 + 8

a = 16 yrs is your age

6 0

2 years ago

5(4x - 1.5x) + 12 = 4x - 2

MAXImum [283]

Answer:

-1.647058824

Step-by-step explanation:

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7 0

2 years ago

Read 2 more answers

students surveyed boys and girls separately to determine which sport was enjoyed the most. After completing the boys survey, it

Anit [1.1K]

Im sorry i do not know the answer but i hope you have a wonderful day and never give up on things i almost gave up on a test i said i cant do it but i did i know this might sound pufft yeah sure its fake well its not

4 0

3 years ago

Suppose you are climbing a hill whose shape is given by the equation z = 1300 − 0.005x2 − 0.01y2, where x, y, and z are measured

Lisa [10]

Answer:

a) "ascend"

Step-by-step explanation:

The equation is $z = 1300-0.005x^2-0.01y^2$

This question is asking for the value of the directional derivative in the direction of -j [since south that's y negative]

So, this is the negative of partial derivative of y. Thus we have:

$z = 1300-0.005x^2-0.01y^2\\z_y=-0.02y$

Now we are at (100,120,1106), we take y value of 120 and put it in the partial:

$z_y=-0.02y\\z_y=-0.02(120)\\z_y=-2.4$

We take the negative, that is -(-2.4) = 2.4

Since this is positive, we can say the hill is sloping upward at this point, so you will start to ascend.

4 0

3 years ago

Which of the following expressions is a factor of the polynomial x 2 +3/2x-1

Elodia [21]

Answer:

$\large \boxed{\sf \ \ \ (x+2)(x-\dfrac{1}{2}) \ \ \ }$

Step-by-step explanation:

Hello,

let's solve

$x^2+\dfrac{3}{2}x-1=0\\\\ 2x^2+3x-2=0 \ \text{multiply by 2}\\\\\\$

$\Delta=b^2-4ac=9+4*2*2=9+16=25$

There are two solutions

$x_1=\dfrac{-3-\sqrt{25}}{4}\\\\x_1=\dfrac{-3-5}{4}\\\\x_1=\dfrac{-8}{4}\\\\\boxed{x_1=-2}$

And

$x_2=\dfrac{-3+\sqrt{25}}{4}\\\\x_2=\dfrac{-3+5}{4}\\\\x_2=\dfrac{2}{4}\\\\\boxed{x_2=\dfrac{1}{2}}$

So we can write

$x^2+\dfrac{3}{2}x-1\\\\=(x-x_1)(x-x_2)\\\\=(x+2)(x-\dfrac{1}{2})$

Hope this helps.

Do not hesitate if you need further explanation.

Thank you

5 0

3 years ago