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Levart [38]
2 years ago
7

What is the common denominator of 2/4 and 7/8

Mathematics
1 answer:
Ivenika [448]2 years ago
5 0

Answer:

8

Step-by-step explanation:

\dfrac 24 = \dfrac{ 2 \times 2}{4 \times 2} = \dfrac 4 8

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The high temperatures for 8 days are shown. 48,42,44,49,58,31,58,46. Which measures of central tendency BEST describes the data
Troyanec [42]

Answer:

47 and the median

Step-by-step explanation:

5 0
3 years ago
PLEASE HELP ME ASAP I WILL MARK YOU AS BRILLIANT<br>please help with these three
GalinKa [24]

Answer:

26:  x=34

28: x=17

30: m=30

Step-by-step explanation:

<u>26:</u> 2x+22=90

subtract 22

2x=68

divide by 2

x=34

<u>28:</u> 18+3x+21=90

add 18 and 21

3x+39=90

subtract 39

3x=51

divide by 3

x=17

<u>30:</u> 43+87+m+20=180

add 43 and 87 and 20

m+150=180

subtract 150

m=30

6 0
3 years ago
Triangle JKL is dilated by a scale factor of 5 to form triangle J'K'L'. Side K'L'
nikdorinn [45]
13, I believe? Sorry if you get it wrong
8 0
2 years ago
Put 3/8, 2/5, and 0.38 from least to greatest
Vesnalui [34]
3/8=0.375
2/5=0.4
0.38=0.38

0.375<0.38<0.4

therefore: 3/8,0.38,2/5

*Remember that you can convert fractions to decimals by dividing the numerator by the denominator! This makes it much easier to compare :)
5 0
2 years ago
Read 2 more answers
If f(1) = 0, what are all the roots of the function f(x)=x^3+3x^2-x-3? Use the Remainder Theorem.
Sophie [7]
There's no if about it, 

f(x)=x^3+3x^2-x-3&#10;

has a zero f(1)=0 so x-1 is a factor.   That's the special case of the Remainder Theorem; since f(1)=0 we'll get a remainder of zero when we divide f(x) by x-1.

At this point we can just divide or we can try more little numbers in the function.  It doesn't take too long to discover f(-1)=0 too, so  x+1 is a factor too by the remainder theorem.  I can find the third zero as well; but let's say that's out of range for most folks.

So far we have 

x^3+3x^2-x-3 = (x-1)(x+1)(x-r)

where r is the zero we haven't guessed yet.  Again we could divide f(x) by (x-1)(x+1)=x^2-1 but just looking at the constant term we must have

-3 = -1 (1)(-r) = r

so

x^3+3x^2-x-3 = (x-1)(x+1)(x+3)

We check f(-3)=(-3)^3+3(-3)^2 -(-3)-3 = 0 \quad\checkmark

We usually talk about the zeros of a function and the roots of an equation; here we have a function f(x) whose zeros are

x=1, x=-1, x=-3

8 0
2 years ago
Read 2 more answers
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