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earnstyle [38]
3 years ago
8

If

Mathematics
1 answer:
klasskru [66]3 years ago
8 0

Answer:

f:h = 1 : 10

Step-by-step explanation:

\frac{f}{g}=\frac{1}{4}\\\\Cross multiply,\\\\4f = g\\\\g = 4f   ------------(I)\\\\\frac{g}{h}=\frac{2}{5}\\\\\frac{4f}{h}=\frac{2}{5}       from(I)\\\\\frac{f}{h}=\frac{2}{5*4}\\\\\frac{f}{h}=\frac{1}{5*2}\\\\\frac{f}{h}=\frac{1}{10}

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Word Questions - Fractions and Percents<br><br> HELP PLEASE
svlad2 [7]

Answer:

4a. 15 wear braces

b. 20%

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b. 35%

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b. $45  - $90 / 2 = $45

Step-by-step explanation:

pls mark brainliest! <3

7 0
3 years ago
For the graph of the function h(t) below determine h(153)
34kurt

Answer:

1

Step-by-step explanation:

As can be seen, for h(t) where t is a multiple of 6, we get -6. Since 150 is a multiple of 6, 153 is 3 more and since the graph visually repeats every 6 units, we can see that h(t) when t is a (multiple of 6) + 3 = 1

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3 0
3 years ago
How do you know how many places are in fifty million without writing the number in standard form
Sunny_sXe [5.5K]
The millions place is befor the second comma form left to right and fifty means tens place so skip to the left to times
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7 0
4 years ago
Identify each function that has a remainder of -3 when divided x+6
Sergeu [11.5K]

Answer:

D

Step-by-step explanation:

According to remainder theorem, you can know the remainder of these polynomials if you plug in x = -6 into them.

<em>So we will plug in -6 into x of all the polynomials ( A through D) and see which one equals -3.</em>

<em />

<em>For A:</em>

x^5 + 2x^2 - 30x + 30\\=(-6)^5 + 2(-6)^2 - 30(-6) + 30\\=-7494

For B:

x^4 + 4x^3 - 21x^2 - 53x + 12\\=(-6)^4 + 4(-6)^3 - 21(-6)^2 - 53(-6) + 12\\=6

For C:

x^3 - 10x^2 - 7\\=(-6)^3 - 10(-6)^2 - 7\\=-583

For D:

x^4 + 6x^3 - 10x - 63\\=(-6)^4 + 6(-6)^3 - 10(-6) - 63\\=-3

The only function that has a remainder of -3 when divided by x + 6 is the fourth one, answer choice D.

5 0
3 years ago
Help and you shall receive 20 points
alukav5142 [94]

Answer:

its D   7 2/5

Step-by-step explanation:

6 0
3 years ago
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