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lora16 [44]
3 years ago
12

Solve the expression for a=12 , b=6 and c=3

Mathematics
1 answer:
iragen [17]3 years ago
7 0

Answer:

12(6^2-6x3)= 216

3^2 = 9

Step-by-step explanation:

216/9 = 24

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I don’t understand please help?
Lerok [7]

Answer:

D

Step-by-step explanation:

Cuz sqr root of 20 = 2 sqrt 5, -2 sqrt 5

7 0
3 years ago
What is the base representation of the number 219
andrezito [222]
The base representation is 200, because it’s the base number

Hope this helps:)
4 0
3 years ago
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Solve for the base.<br> 32 is 0.50% of ?
inn [45]

Step-by-step explanation:

6400

this i your answers

plz mark me as brainlist

8 0
3 years ago
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Jeremy's father drives him to school in rush hour traffic in 20 minutes. One day there is no traffic, so his father can drive hi
nexus9112 [7]
Hello.

Let's call the initial speed 'x'. When driving at that speed, Jeremy's father takes 20 minutes to arrive to the school. When driving 18 mph faster (x + 18), he gets there in 12 minutes. Therefore:

20 - x
12 - x + 18

As they're inversely proportional:

12 - x
20 - x + 18

\frac{12}{20} = \frac{x}{x+18}
12x + 216 = 20x
216 = 8x
x = \frac{216}{8}
x = 27

That means that the initial speed is of 27mph. Since we have the time and the speed, we can find the distance:

Average Speed = Total Distance Travelled / Time Interval

20 minutes = 0.33h

27 = \frac{Distance}{0.33}
Distance = 27.0.33
Distance = 12.21 miles

Hope I helped and good luck.
3 0
3 years ago
Read 2 more answers
Can someone help me
defon
Answer for problem 46 is choice A
Answer for problem 47 is choice B
Answer for problem 48 is choice E

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Work Shown

Problem 46
Equation 1: 3x+y = 17
Equation 2: x+3y = -1
Add equation 1 to equation 2 to get 4x+4y = 16. Divide every term by 4 to get x+y = 4. Then finally multiply both sides by 3 to get 3x+3y = 12
That shows why the answer is choice A

---------------------------------
Problem 47)

If y hours pass by, then y-(2/3)y=y/3 is the time value (2/3)y hours ago

So,
Distance = rate*time
d = r*t
d = x*(y/3)
d = (xy)/3
That's why the answer is choice B

---------------------------------
Problem 48)
Let L1,L2,L3 be the three lists where
L1 = {a1,a2,a3,...,ak} there are k scores here
L2 = {a1,a2,...,a10} there are 10 scores here
L3 = {a11,a12,...,ak} the remaining k-10 scores
S(L1) = sum of the scores in list L1
M(L1) = mean of L1 = 20 = S(L1)/k
M(L2) = mean of L2 = 15 = S(L2)/10
S(L1) = 20k
S(L2) = 150
S(L1) = S(L2)+S(L3)
M(L1) = [S(L2)+S(L3)]/k
20 = [150+S(L3)]/k
20k = 150+S(L3)
S(L3) = 20k-150
M(L3) = [S(L3)]/(k-10)
M(L3) = (20k-150)/(k-10)
So that shows why the answer is choice E
3 0
4 years ago
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