All categories

3 years ago

The radius of a circle is 5 m. Find its area to the nearest tenth.

Mathematics

1 answer:

OLga [1]3 years ago

3 0

The area would be 78.54

Step-by-step explanation:

Here's why:

By using the area of a circle formula, you should plug in the radius and get the following of:

π5^2 = 78.539...

If you round to the nearest tenth you would get the answer of 78.54.

You might be interested in

find the component equation of the plane which is normal to the vector -2i+5j+k and which contains the point (-10;7;5).

maksim [4K]

Given:

A plane is normal to the vector = -2i+5j+k

It contains the point (-10,7,5).

To find:

The component equation of the plane.

Solution:

The equation of plane is

$a(x-x_0)+b(y-y_0)+c(z-z_0)=0$

Where, $(x_0,y_0,z_0)$ is the point on the plane and $\left< a,b,c\right>$ is normal vector.

Normal vector is -2i+5j+k and plane passes through (-10,7,5). So, the equation of the plane is

$-2(x-(-10))+5(y-7)+1(z-5)=0$

$-2(x+10)+5y-35+z-5=0$

$-2x-20+5y-35+z-5=0$

$-2x+5y+z-60=0$

$-2x+5y+z=60$

Therefore, the equation of the plane is $-2x+5y+z=60$ .

4 0

3 years ago

Solve this equation 3x - 5 = -26

Dmitry_Shevchenko [17]

3x - 5 = -26

step 1:

subtract +5 on each side because you have to do the inverse operation

step 2:

3x= -21

step 3:

now divide 3 on each side

the answer is -7

Hope this helped!

TheOneAndOnlyLara~

5 0

2 years ago

Read 2 more answers

Select the correct answer from the drop-down menu.

ELEN [110]

Missing term = –2xy

Solution:

Let us first find the quotient of $-8x^2y^3 \div xy$ .

$-8x^2y^3 \div xy=\frac{-8x^2y^3 }{xy}$

$=\frac{-8\times x\times x\times y\times y\times y}{xy}$

Taking common term xy outside in the numerator.

$=\frac{xy(-8\times x\times y\times y)}{xy}$

Both xy in the numerator and denominator are cancelled.

$=-8xy^2$

Thus, the quotient of $-8x^2y^3 \div xy$ is $-8xy^2$ .

Given the quotient of $-8x^2y^3 \div xy$ is same as the product of 4xy and ____.

$-8xy^2=4xy$ × missing term

Divide both sides by 4xy, we get

⇒ missing term = $\frac{-8xy^2}{4xy}$

Cancel the common terms in both numerator and denominator.

⇒ missing term = –2xy

Hence the missing term of the product is –2xy.

4 0

3 years ago

Below is the data of ages at the local amusement park on a certain day.

nignag [31]

Answer:

Step-by-step explanation:

To find the intervals you will need to find the lowest and highest numbers, in this case, it would be 1 and 35. A general rule would be 5-7 intervals, I will use 5.

Here are the intervals with the number of people that were in each:

1-7 (9)

8-14 (14)

15-21 (8)

22-28 (3)

29-35 (5)

14+9+8+3+5=39

6 0

4 years ago

Solve the following C^2-14=5c

kipiarov [429]

Answer:

c = - 2, c = 7

Step-by-step explanation:

Given

c² - 14 = 5c ( subtract 5c from both sides )

c² - 5c - 14 = 0 ← in standard form

To factorise the quadratic

Consider the factors of the constant term (- 14) which sum to give the coefficient of the c- term (- 5)

The factors are - 7 and + 2, since

- 7 × 2 = - 14 and - 7 + 2 = - 5, hence

(c - 7)(c + 2) = 0

Equate each factor to zero and solve for c

c - 7 = 0 ⇒ c = 7

c + 2 = 0 ⇒ c = - 2

8 0

3 years ago