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2 years ago

8

The radius of a circle is 8 miles. What is the area of a sector bounded by a 135° arc?

1 answer:

Jlenok [28]2 years ago

7 0

Answer:

13.5* pi sq miles

Step-by-step explanation:

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A total of 525 records of a hospital were reviewed and 175 of the patients had a heart problem, 150 a respiratory problem, 125 s

Tcecarenko [31]

Answer:

a

Step-by-step explanation:

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3 years ago

$800 deposit over 10 years with an interest rate of 2%

TiliK225 [7]

Answer:

the interest is 960.

Step-by-step explanation:

the formula: I=prt i=interest p= pricipal [the money you start with]

r= rate t= time

I=? I= 800 (0.02) (10)

P= $800 I= 160+ 800 {because $160 is added}

R= 2% --> 0.02 [as a decimal] I= $960

T= 10 years

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3 years ago

A baseball has a 48 cm diameter. What is the volume of the contents of the ball?

Andrew [12]

Answer:

201cm

Step-by-step explanation:

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2 years ago

Evaluate each power.<br><br> a. (−7)0 b. (43)0 c. 10 d. (0.5)0

Sunny_sXe [5.5K]

Answer: The answer is two

Step-by-step explanation:

3 0

1 year ago

Solve the following system of equation

krok68 [10]

Answer:

x = 6 , y = -4 , Z = -1

Step-by-step explanation:

Solve the following system:

{2 x + 3 y - Z = 1 | (equation 1)

3 x + y + 2 Z = 12 | (equation 2)

-3 + x + 2 y = -5 | (equation 3)

Express the system in standard form:

{2 x + 3 y - Z = 1 | (equation 1)

3 x + y + 2 Z = 12 | (equation 2)

x + 2 y+0 Z = -2 | (equation 3)

Swap equation 1 with equation 2:

{3 x + y + 2 Z = 12 | (equation 1)

2 x + 3 y - Z = 1 | (equation 2)

x + 2 y+0 Z = -2 | (equation 3)

Subtract 2/3 × (equation 1) from equation 2:

{3 x + y + 2 Z = 12 | (equation 1)

0 x+(7 y)/3 - (7 Z)/3 = -7 | (equation 2)

x + 2 y+0 Z = -2 | (equation 3)

Multiply equation 2 by 3/7:

{3 x + y + 2 Z = 12 | (equation 1)

0 x+y - Z = -3 | (equation 2)

x + 2 y+0 Z = -2 | (equation 3)

Subtract 1/3 × (equation 1) from equation 3:

{3 x + y + 2 Z = 12 | (equation 1)

0 x+y - Z = -3 | (equation 2)

0 x+(5 y)/3 - (2 Z)/3 = -6 | (equation 3)

Multiply equation 3 by 3:

{3 x + y + 2 Z = 12 | (equation 1)

0 x+y - Z = -3 | (equation 2)

0 x+5 y - 2 Z = -18 | (equation 3)

Swap equation 2 with equation 3:

{3 x + y + 2 Z = 12 | (equation 1)

0 x+5 y - 2 Z = -18 | (equation 2)

0 x+y - Z = -3 | (equation 3)

Subtract 1/5 × (equation 2) from equation 3:

{3 x + y + 2 Z = 12 | (equation 1)

0 x+5 y - 2 Z = -18 | (equation 2)

0 x+0 y - (3 Z)/5 = 3/5 | (equation 3)

Multiply equation 3 by 5/3:

{3 x + y + 2 Z = 12 | (equation 1)

0 x+5 y - 2 Z = -18 | (equation 2)

0 x+0 y - Z = 1 | (equation 3)

Multiply equation 3 by -1:

{3 x + y + 2 Z = 12 | (equation 1)

0 x+5 y - 2 Z = -18 | (equation 2)

0 x+0 y+Z = -1 | (equation 3)

Add 2 × (equation 3) to equation 2:

{3 x + y + 2 Z = 12 | (equation 1)

0 x+5 y+0 Z = -20 | (equation 2)

0 x+0 y+Z = -1 | (equation 3)

Divide equation 2 by 5:

{3 x + y + 2 Z = 12 | (equation 1)

0 x+y+0 Z = -4 | (equation 2)

0 x+0 y+Z = -1 | (equation 3)

Subtract equation 2 from equation 1:

{3 x + 0 y+2 Z = 16 | (equation 1)

0 x+y+0 Z = -4 | (equation 2)

0 x+0 y+Z = -1 | (equation 3)

Subtract 2 × (equation 3) from equation 1:

{3 x+0 y+0 Z = 18 | (equation 1)

0 x+y+0 Z = -4 | (equation 2)

0 x+0 y+Z = -1 | (equation 3)

Divide equation 1 by 3:

{x+0 y+0 Z = 6 | (equation 1)

0 x+y+0 Z = -4 | (equation 2)

0 x+0 y+Z = -1 | (equation 3)

Collect results:

Answer: {x = 6 , y = -4 , Z = -1

6 0

3 years ago