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babymother [125]
3 years ago
10

Find the area of a circle with diameter 14 in.

Mathematics
2 answers:
VashaNatasha [74]3 years ago
7 0

Answer: 153.86

Step-by-step explanation:

Area of circle = r^2*3.14

r=7

7^2*3.14= 153.86

trasher [3.6K]3 years ago
7 0

Answer:

153.86 squared inches

Step-by-step explanation:

Area of a circle = pi r^2

Given pi = 3.14

Diameter = 14inches

r = radius

r = diameter/2

= 14/2

= 7 inches

Therefore area = 3.14 x (7)^2

= 3.14 x 7 x 7

3.14 x 49

153.86 squared inches

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The formula for glue says add 45 mL of Hardner to each container of resin how much Hardner should be added to 18 containers of r
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Answer:

810 mL of hardener

Step-by-step explanation:

You need 45 mL of hardener for each container and you have 18 containers so:

45 x 18 = 810 mL

7 0
3 years ago
If NTM is congruent to OUM, SP is congruent to SQ, S is the center of the circle, and OM = 18, find NT. Round the answer to the
Aleks [24]
OM=18, so OQ=QM=18/2=9.
 Given QU=8
from figure OQU is a right angled triangle , so OU^2=OQ^2 + QU^2
OU^2 = 9*9 + 8*8 = 81+72=153;
OU=sqrt(153) = 12.37 =13(approx);
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4 0
3 years ago
I really need help! This is last minute and its 1 am. I'm tired so I'm going to leave it to you guys to solve my problems and pr
Illusion [34]

Answer:

<h2><u>question 1</u></h2>

n^{2} - 20n -96 = 0

use product and sum method

product = -96

sum = -20

numbers needed = ( -24 , 4)

n - 24 = 0

n + 4 = 0

hence <u>n = 24 and n = -4 </u>

<u></u>

<h2><u>Question 2 </u></h2>

<u />x^{2} + 12 x = 48<u />

in the form ax^{2}  +bx +c = 0

= x^{2} +12x - 48

make use of the formula :

\frac{-b+-\sqrt{b^{2} -4ac} }{2a}

replace values to make 2 equations :

1.\frac{-12+\sqrt{12^{2} -4*1*-48} }{2*1} = 3.17

2.\frac{-12-\sqrt{12^{2} -4*1*-48} }{2*1} = -15.2

hence <u>x = 3.17 and x = -15.2</u>

<u />

<h2><u>Question 3 </u></h2>

<u />x^{2} -14x+40=0<u />

use product and sum method

product = 40

sum = -14

numbers needed = (-10 , -4)

x - 10 = 0

x - 4 = 0

hence<u> x = 10 and x = 4</u>

<u />

<h2><u>Question 4 </u></h2>

<u />5b^{2} -20b-18 = 7<u />

in the form ax^{2}  +bx +c = 0

this becomes 5b^{2} -20b-18-7

= 5b^{2} -20b-25

can simplify by 5

= b^{2} -4b-5 =0\\

use product and sum method

product = -5

sum = -4

numbers needed (-5 , 1)

b-5 = 0

b + 1 = 0

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5 0
3 years ago
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Answer:

The attachment shows ΔBAC ~ ΔBDA

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You want segment AB to be part of two similar, but not congruent, triangles. One way to do that is to make AB the hypotenuse of one triangle and the leg of another.

It is convenient to construct these triangles using point M as the arbitrary midpoint of the hypotenuse of the larger triangle. (We don't know the coordinates of M—we just know it is on the perpendicular bisector of AB.) BC is a diameter of circle M, and AD is the altitude of ΔABC.

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The value of a car is $15,000 and depreciates at a rate of 8% per year. What is the
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Sorry for tje late answer but it is 150 * 0.08^x

4 0
2 years ago
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