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Margaret [11]
3 years ago
10

Which phrase describes the variable expression y-5

Mathematics
1 answer:
ruslelena [56]3 years ago
7 0
To translate y - 5 into words:

a number y is decreased by 5

i hope this helps!!
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Which equation demonstrates the additive identity property?
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first you have to add

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What equivalent of 4:25
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Answer: 16 percent

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Which of the following is a proportional relationship?
Solnce55 [7]

The two equations represent the proportional relationship.

y=3x and y=12x are proportional relation ship equations

proportion equations can be defined as

If we change x the y will change in the same proportion.

<h3>What is the proportional relationship?</h3>

Proportional relationships are relationships between two variables where their ratios are equivalent.

Another way to think about them is that, in a proportional relationship, one variable is always a constant value time the other.

That constant is known as the constant of proportionality.

proportional relationship equation contain (0,0) points

If we put x=0

This  will give us,y=0

If we put x=0, in y=12x

It will give y=0

put if we put x=0 in

y=3x it will give us y=0

hence these two equations represent the proportional relationship.

To learn more about the equation visit:

brainly.com/question/2972832

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6 0
2 years ago
Please help 6th grade math i will mark brainliest to 1 answer select all that are equivalent to 15%
Triss [41]

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7 0
3 years ago
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The height h (in feet) of an object dropped from a ledge after x seconds can be modeled by h(x)=−16x2+36 . The object is dropped
kakasveta [241]

Check the picture below.

\bf ~~~~~~\textit{initial velocity in feet} \\\\ h(t) = -16t^2+v_ot+h_o \quad \begin{cases} v_o=\textit{initial velocity}&\\ \qquad \textit{of the object}\\ h_o=\textit{initial height}&\\ \qquad \textit{of the object}\\ h=\textit{object's height}&\\ \qquad \textit{at "t" seconds} \end{cases}

so the object hits the ground when h(x) = 0, hmmm how long did it take to hit the ground the first time anyway?

\bf h(x)=-16x^2+36\implies \stackrel{h(x)}{0}=-16x^2+36\implies 16x^2=36 \\\\\\ x^2=\cfrac{36}{16}\implies x^2 = \cfrac{9}{4}\implies x=\sqrt{\cfrac{9}{4}}\implies x=\cfrac{\sqrt{9}}{\sqrt{4}}\implies x = \cfrac{3}{2}~~\textit{seconds}

now, we know the 2nd time around it hit the ground, h(x) = 0, but it took less time, it took 0.5 or 1/2 second less, well, the first time it took 3/2, if we subtract 1/2 from it, we get 3/2 - 1/2  = 2/2 = 1, so it took only 1 second this time then, meaning x = 1.

\bf ~~~~~~\textit{initial velocity in feet} \\\\ h(x) = -16x^2+v_ox+h_o \quad \begin{cases} v_o=\textit{initial velocity}&0\\ \qquad \textit{of the object}\\ h_o=\textit{initial height}&\\ \qquad \textit{of the object}\\ h=\textit{object's height}&0\\ \qquad \textit{at "t" seconds}\\ x=\textit{seconds}&1 \end{cases} \\\\\\ 0=-16(1)^2+0x+h_o\implies 0=-16+h_o\implies 16=h_o \\\\[-0.35em] ~\dotfill\\\\ ~\hfill h(x) = -16x^2+16~\hfill

quick info:

in case you're wondering what's that pesky -16x² doing there, is gravity's pull in ft/s².

4 0
3 years ago
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