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solmaris [256]
2 years ago
8

Fraction question...pls help out!!!

Mathematics
2 answers:
pentagon [3]2 years ago
8 0
The answer is D because, as decimals, 4 1/3 would be 4.33(a repeating decimal) and 4 3/5 would be 4.6, and 4.4 is between those two numbers. I hope this helps.
Semenov [28]2 years ago
5 0

Answer:

d. 4.4

Step-by-step explanation:

you can turn your mixed numbers into improper fractions

4 1/3= 13/3      4 3/5=23/5

from her you want common denominators, so multiply the left by 5 and the right by 3

65/15 and 69/15 is what you have now. divide 65 by 15 and your first number is 4.3, divide 69 by 15 and you get 4.6. Therefore 4.4 is the only possible answer

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F(x)=3x-1 and g(x)=x+2 find (f+g)
Doss [256]

Answer:

(f+g)(x) = 4x + 1

Step-by-step explanation:

(f+g)(x) = f(x) + g(x) = 3x -1  + x + 2

Combining the like terms:

3x + x - 1 + 2

= 4x + 1

Hence,

(f+g)(x) = 4x + 1

<em>Hope this helps and be sure to have a wonderful time ahead at Brainly! :D</em>

5 0
2 years ago
G(x) = x2 - 2<br> h(x) = 3x + 5<br> Find (g- h)(x)
marusya05 [52]

Answer:

Brainliest PLS

g=x2−2/x

h=3x+5/x          

x=5/h−3

Step-by-step explanation:

5 0
3 years ago
Prove tan(x - (π / 4)) = (sin x – cos x) / (cos x + sin x) by filling in the reasons in the table below.
Mrac [35]

No way to know what reasons you're supposed to choose from...

By definition of tangent,

\tan\left(x-\dfrac\pi4\right)=\dfrac{\sin\left(x-\frac\pi4\right)}{\cos\left(x-\frac\pi4\right)}

The angle sum identities give

\tan\left(x-\dfrac\pi4\right)=\dfrac{\sin x\cos\frac\pi4-\cos x\sin\frac\pi4}{\cos x\cos\frac\pi4+\sin x\sin\frac\pi4}

cos\dfrac\pi4=\sin\dfrac\pi4=\dfrac1{\sqrt2}, so we can cancel those terms to get

\tan\left(x-\dfrac\pi4\right)=\dfrac{\sin x-\cos x}{\sin x+\cos x}

as required.

8 0
3 years ago
Answer fast plzzzzzzzz
Paul [167]

Answer:

0.5

Step-by-step explanation:

It is less than 0.6.

8 0
3 years ago
Read 2 more answers
Daeshawn earns $45 for working 2.5 hours. At this rate, how much does she earn for working 10 hours?​
OLga [1]
I don’t know but I’m here for it
3 0
2 years ago
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