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andrey2020 [161]
4 years ago
15

Please answer... I Got Pointsss

Mathematics
1 answer:
DaniilM [7]4 years ago
7 0

Answer:

its is 50 i am sure of it

Step-by-step explanation:

because 50 is the middle and it is directly on the middle also i have the same school program and i pased that test

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What are the values of a and b?
ivann1987 [24]

Answer:

<h2>a = 400/21, b = 580/21</h2>

Step-by-step explanation:

Small triangles are similar. Therefore, the sides are proportional

\dfrac{a}{20}=\dfrac{20}{21}               <em>cross multiply</em>

21a=(20)(20)       <em>divide both sides by 21</em>

a=\dfrac{400}{21}

For b, we can use the Pythagorean theorem:

b^2=\left(\dfrac{400}{21}\right)^2+20^2\\\\b^2=\dfrac{160000}{441}+400\\\\b^2=\dfrac{160000}{441}+\dfrac{176400}{441}\\\\b^2=\dfrac{336400}{441}\to b=\sqrt{\dfrac{336400}{441}}\\\\b=\dfrac{\sqrt{336400}}{\sqrt{441}}\\\\b=\dfrac{580}{21}

3 0
3 years ago
What is the average of 63, 84, and 102?
Ann [662]
The answer to your question is 83
5 0
3 years ago
FIND THE RADIUS OF A CIRCLE WHO'S DIAMETER IS40 FEET
iVinArrow [24]

Answer:

40 feet

radius is half of the diameter, so the radius is half of 40

radius = diameter/2

radius = 40/2

radius=20 feet

8 0
3 years ago
Read 2 more answers
How many arrangements of the seven letters in the word SYSTEMS have the E occurring somewhere before the M? How many arrangement
Snezhnost [94]

Answer: 720ways, 24ways

Step-by-step explanation:

Given the seven letter words "SYSTEMS", if E is always occurring before M it means E and M will always be together therefore they letter 'EM' will be taken as an entity to five us 6letters i.e SYST(EM)S.

This can then be arranged in 6!ways

6! = 6×5×4×3×2×1 = 720ways

Similarly, if the E somewhere before the M and the three Ss grouped consecutively, this means E and M must always be together as well as the Ss to give (SSS)YT(EM).

This means that the letters in the bracket can be taken as an entity to give a total of 4 entities. This can them be arranged in 4! ways.

4! = 4×3×2×1

4! = 24ways

5 0
3 years ago
Conrad has a collection of three types of coins: nickels,dimes,and quarters.There are five more nickels than quarters but four t
velikii [3]
N=number of nickles
d=number of dimes
q=number of quarters
use cents for everything

total value=585
5n+10d+25q=585
divide everybody by 5 to simplify
n+2d+5q=117

we have

nicles are 5 more than quarters
n=5+q

4 times as many dimes as quarters
d=4 times q
d=4q

sub those in
5+q+2(4q)+5q=117
5+6q+8q=117
5+14q=117
minus 5
14q=112
q=8

sub back
n=5+q
n=5+8
n=13

d=4q
d=4(8)
d=32


13 nickles
32 dimes
8 quarters
divide both sides by 22

8 0
3 years ago
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