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1 year ago

7

Gary's salary varies directly as the number of days he works. If his salary for five days is $52.00, how much would it be for 12

days? *

1 answer:

Viktor [21]1 year ago

8 0

Answer:

124.8

Step-by-step explanation:

52 divided by 5 = 10.4

then, multiply by 12 = 124.8

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billy is creating a circular garden divided into 8 equal sections. The diameter of the garden is 12 feet. what is the area, in s

xxTIMURxx [149]

Area=πr^2

find the area then divide by 8

we know that diameter=2radius or diameter/2=radius so
12=diameter
12/2=radius=6
subsitute
area=π(6)^2
area=36π
divide 36π by 8
36π/8=18π/4=9π/2π=4.5π
area of one section is 4.5π square feet or if we aprox pito 3.14159 then we get
4.5(3.14159)=area=14.1372 square feet or 14.14 square feet

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2 years ago

Wynn needs to find the center of the data set shown on the dot plot. The dot plot has dots. The dot plot has dots to the left of

Arisa [49]

Answer:

The dot plot has 22

dots.

The dot plot has 11

dots to the left of the

center and 11

dots to the right of the center.

The center of the data set is

between 2 and 3

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Step-by-step explanation:

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2 years ago

The relationship between money earned and hours worked is linear. Joe computes the slope between (4, 30) and (12, 90), then comp

____ [38]

Finding the slope of both coordinates, you'll get 15/2. The slopes are the same

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2 years ago

Read 2 more answers

2 x 3 x n x n=54 <br><br> What is the value of n?

kozerog [31]

2*3*n*n=54

6*n^2=54

n^2=9

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4 0

2 years ago

Read 2 more answers

In △ABC, m∠A=15 °, a=10 , and b=11 . Find c to the nearest tenth.

pshichka [43]

Answer:

The answer is:

$\bold{c\approx 20.2\ units}$

Step-by-step explanation:

Given:

In △ABC:

m∠A=15°

a=10 and

b=11

To find:

c = ?

Solution:

We can use cosine rule here to find the value of third side c.

Formula for cosine rule:

$cos A = \dfrac{b^{2}+c^{2}-a^{2}}{2bc}$

Where

a is the side opposite to $\angle A$

b is the side opposite to $\angle B$

c is the side opposite to $\angle C$

Putting all the values.

$cos 15^\circ = \dfrac{11^{2}+c^{2}-10^{2}}{2\times 11 \times c}\\\Rightarrow 0.96 = \dfrac{121+c^{2}-100}{22c}\\\Rightarrow 0.96 \times 22c= 121+c^{2}-100\\\Rightarrow 21.25 c= 21+c^{2}\\\Rightarrow c^{2}-21.25c+21=0\\\\\text{solving the quadratic equation:}\\\\c = \dfrac{21.25+\sqrt{21.25^2-4 \times 1 \times 21}}{2}\\c = \dfrac{21.25+\sqrt{367.56}}{2}\\c = \dfrac{21.25+19.17}{2}\\c \approx 20.2\ units$

The answer is:

$\bold{c\approx 20.2\ units}$

4 0

2 years ago