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Andrei [34K]
3 years ago
14

If the product of two consecutive integers is 156 what type of equation do you create to solve this problem

Mathematics
2 answers:
anastassius [24]3 years ago
8 0
N + (n+1) = 156
2n + 1 = 156
2n = 156 - 1 = 155
2n = 155
n = 155/2
n = 77 r1
answer is
n = 77
n + 1 = 78
zhenek [66]3 years ago
6 0
I'm not sure, but I knew that 12×12=144 so I guessed that it was 12×13 and checked what that was and it was 156, so I guess and check probably
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The sum of three numbers is 99. The first number is 9 more than the third. The second number is 4 times the third. What are the
klemol [59]
First number=x+9
second number=4x
third number=x
4x+x+x+9=99
6x=90
x=15
 first number=24
second number=60
third number=15


6 0
3 years ago
The Miller family visited Mama's Kitchen and ordered 4 hamburgers and 3 medium fries and paid $17.35. James ordered a medium dri
Airida [17]
Know: 17.35 is equivalent to 4 burgers 3 med fries
6.09 Is equivalent to 1 burger med fries
9.34 8s equivalent to 2 burgers med fries and med drink
So what you want to do is take 6.09 and subtract it from 9.34 which is 3.25. Now 3.25 is your burger price.
What you want to do now is take your 3.25 and multiply it by 4 which would be 13. 17.35 subtract 13 would be 4.
4 divided by 3 is 1.33 which is now the price of a med fries.
now for the drink. 3.25 plus 1.33 is 4.58.
6.09 subtract 4.58 is 1.51
Now you have 3.25 as your burger
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6 0
3 years ago
How to do distributed property on 25+55<br> How to do distributive property on 25+55
lisabon 2012 [21]

Answer:

80

Step-by-step explanation:

25+55=80

5 0
3 years ago
How do you expand 3(m - 7 + 4m + 2) using distributive property?
zheka24 [161]

Answer:

3m - 21 + 12m +6

Step-by-step explanation:

Multiply everything inside the brackets by 3!

3m - 21 + 12m +6

8 0
3 years ago
If we approximate the function y=sin(x) with a0+a1 x a2 x^2 +a3 x^3, what is a0,a2,a2,a3?
Rashid [163]

The coefficients a_0,a_1,a_2,a_3 could be chosen to be the coefficients in the Maclaurin series of \sin(x).

We have

y = \sin(x) \approx a_0 + a_1 x + a_2 x^2 + a_3 x^3 \\\\ \implies y(0) = 0 = a_0

y' = \cos(x) \approx a_1 + 2a_2 x + 3a_3 x^2 \\\\ \implies y'(0) = 1 = a_1

y'' = -\sin(x) \approx 2a_2 + 6a_3 x \\\\ \implies y''(0) = 0 = 2a_2

y''' = -\cos(x) \approx 6a_3 \\\\ \implies y'''(0) = -1 = 6a_3

It follows that a_0=0, a_1=1, a_2=0, and a_3 = -\frac16.

7 0
2 years ago
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