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Stells [14]
3 years ago
8

What is the least common multiple of 8, 3, and 12?

Mathematics
1 answer:
bija089 [108]3 years ago
4 0

Answer:

24

Step-by-step explanation:

cause its just math

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Let a and b be real numbers satisfying a^3 - 3ab^2 = 47 and b^3 - 3a^2 b = 52. Find a^2 + b^2.
boyakko [2]

Answer: a²+b² = -99/2

Step-by-step explanation:

Since we are given two equations, this equations will be solved simultaneously to get a² and b²

a³ - 3ab² = 47 ... 1

b³ - 3a² b = 52... 2

From 1, a(a² - 3b²) = 47...3

From 2, b(b² - 3a²) = 52... 4

Adding 3 and 4, we have;

a²+b²-3b²-3a² = 99 (note that a and b will no longer be part of the equations as they have been factored out)

a²+b²-(3b²+3a²) = 99

(a²+b²) -3(b²+a²)= 99

Taking the difference we have

- 2(a²+b²) = 99

a²+b² = -99/2

8 0
3 years ago
Jack took a test and got 13 out of 17 points. He took a make-up test and got 12 out of 15 points.
blagie [28]

Answer:

4.6

Step-by-step explanation:

The answer is 4.6%, hope it helps! sorry if wrong!

3 0
3 years ago
After arianna completed some work, she figured she still had 78 21/100 pictures to paint. if she completed another 34 23/25 pict
dezoksy [38]

Subtract the number she completed from what she had left to paint originally.

The fractions have different denominators so first step is to rewrite the fractions with a common denominator.

23/25 can be rewritten as 92/100

Now you have 78 21/100 - 34 92/100

Because 21/100 is smaller than 92/100 subtract 1 whole number from 78 and rewrite 1 as 100/100 and add that to the fraction.

78 21/100 becomes 77 121/100

Now subtract :

77 121/100 - 34 92/100

77-34 = 43

And 121/100 - 92/100 = 29/100

Combine to get 43 and 29/100

3 0
3 years ago
Which measure is greater?? One inch or one centimeter
Nimfa-mama [501]

One inch is the greater measure

8 0
3 years ago
(1 point) consider the function f(t)=⎧⎩⎨⎪⎪⎪⎪0,−5,−6,6,t<00≤t<11≤t<7t≥7;f(t)={0,t<0−5,0≤t<1−6,1≤t<76,t≥7; 1. wr
sergij07 [2.7K]
f(t)=\begin{cases}0&\text{for }t

Recall that

u(t)=\begin{cases}0&\text{for }t

Take it one piece at a time. For t\ge0, we can scale u(t) by -5:

-5u(t)=\begin{cases}0&\text{for }t

If we shift the argument by 1 and scale by -5, we have

-5u(t-1)=\begin{cases}0&\text{for }t

so if we subtract this from -5u(t), we'll end up with

-5u(t)+5u(t-1)=\begin{cases}0&\text{for }t

For the next piece, we can add another scaled and shifted step like

-6u(t-1)+6u(t-7)=\begin{cases}0&\text{for }t

so that

-5u(t)+5u(t-1)-6u(t-1)+6u(t-7)=\begin{cases}0&\text{for }t

For the last piece, we add one more term:

6u(t-7)=\begin{cases}0&\text{for }t

and so putting everything together, we get f(t):

f(t)\equiv-5u(t)+5u(t-1)-6u(t-1)+6u(t-7)+6u(t-7)
f(t)\equiv-5u(t)-u(t-1)+12u(t-7)
5 0
3 years ago
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