All categories

3 years ago

A point is chosen at random in the circle.

Mathematics

1 answer:

HACTEHA [7]3 years ago

3 0

Answer:

78.5%

Step-by-step explanation:

the percent of time the point will land in the square can be determined by dividing the area of the square by the area of the circle

A(circle) = 3.14(1)²

A(circle) = 3.14

A(square) = side squared

if radius of square is 1 inch then each side is 2 inches

A(square) = 2²

A(square) = 4 inches

3.14/4 = 78.5%

You might be interested in

Question is in picture

stiv31 [10]

Answer:

27.5

Step-by-step explanation:

pythagorean theorem. a^2+b^2=c^2.

so 12^2+b^2=30^2

meaning 144+b^2=900

900-144=756

square root 756

b=27.5

5 0

3 years ago

Elaina just began working as an occupational therapist assistant for $21.00 per hour. For each paycheck she receives, 12% is ded

lora16 [44]

Answer:

Elaina just began working as an occupational therapist assistant for $21.00 per hour. For each paycheck she receives, 12% is deducted from the gross pay for Federal Withholding Tax. In addition, 1.45% is deducted for Medicare and 6.2% for Social Security. Below is Elaina’s paycheck with missing information.

Elaina’s gross pay for a two week period is $1680.00, what is her net pay?

Step-by-step explanation:

3 0

3 years ago

A pizza shop sells pizzas that are 10 inches (in diameter) or larger. A 10-inch cheese pizza cost $8. Each additional inch cost

Katyanochek1 [597]

A pizza shop sells pizzas that are 10 inches orr larger. A 10-inch cheese pizza costs $8 and each additional costs $1.50 and each additional topping costs $0.75.

The equation that represents the cost of a pizza is:

P = $10 + $1.50a + $0.75b

Where 'a' represents the number additional inches and 'b' represents the number of additional toppings.

4 0

3 years ago

in exercises 15-20 find the vector component of u along a and the vecomponent of u orthogonal to a u=(2,1,1,2) a=(4,-4,2,-2)

Nimfa-mama [501]

Answer:

The component of $\vec u$ orthogonal to $\vec a$ is $\vec u_{\perp\,\vec a} = \left(\frac{9}{5},\frac{6}{5},\frac{9}{10},\frac{21}{10}\right)$ .

Step-by-step explanation:

Let $\vec u$ and $\vec a$ , from Linear Algebra we get that component of $\vec u$ parallel to $\vec a$ by using this formula:

$\vec u_{\parallel \,\vec a} = \frac{\vec u \bullet\vec a}{\|\vec a\|^{2}} \cdot \vec a$ (Eq. 1)

Where $\|\vec a\|$ is the norm of $\vec a$ , which is equal to $\|\vec a\| = \sqrt{\vec a\bullet \vec a}$ . (Eq. 2)

If we know that $\vec u =(2,1,1,2)$ and $\vec a=(4,-4,2,-2)$ , then we get that vector component of $\vec u$ parallel to $\vec a$ is:

$\vec u_{\parallel\,\vec a} = \left[\frac{(2)\cdot (4)+(1)\cdot (-4)+(1)\cdot (2)+(2)\cdot (-2)}{4^{2}+(-4)^{2}+2^{2}+(-2)^{2}} \right]\cdot (4,-4,2,-2)$

$\vec u_{\parallel\,\vec a} =\frac{1}{20}\cdot (4,-4,2,-2)$

$\vec u_{\parallel\,\vec a} =\left(\frac{1}{5},-\frac{1}{5},\frac{1}{10},-\frac{1}{10} \right)$

Lastly, we find the vector component of $\vec u$ orthogonal to $\vec a$ by applying this vector sum identity:

$\vec u_{\perp\,\vec a} = \vec u - \vec u_{\parallel\,\vec a}$ (Eq. 3)

If we get that $\vec u =(2,1,1,2)$ and $\vec u_{\parallel\,\vec a} =\left(\frac{1}{5},-\frac{1}{5},\frac{1}{10},-\frac{1}{10} \right)$ , the vector component of $\vec u$ is:

$\vec u_{\perp\,\vec a} = (2,1,1,2)-\left(\frac{1}{5},-\frac{1}{5},\frac{1}{10},-\frac{1}{10} \right)$

$\vec u_{\perp\,\vec a} = \left(\frac{9}{5},\frac{6}{5},\frac{9}{10},\frac{21}{10}\right)$

The component of $\vec u$ orthogonal to $\vec a$ is $\vec u_{\perp\,\vec a} = \left(\frac{9}{5},\frac{6}{5},\frac{9}{10},\frac{21}{10}\right)$ .

4 0

3 years ago

Help please thank you giving 15 pts

slega [8]

Answer:

x=22

Step-by-step explanation:

2x-40+8x=180

10x=220

x=22

8 0

3 years ago

Read 2 more answers