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2 years ago

What is 3+9= free brainliest

Mathematics

2 answers:

kvv77 [185]2 years ago

7 0

Answer:

Step-by-step explanation:

dimulka [17.4K]2 years ago

5 0

Answer:

Step-by-step explanation:

I counted with my fingers

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What is the length of the line TR

8090 [49]

Answer:

Step-by-step explanation:

If the other triangle is part of the problem the triangle increased by 2.5

2.5 * 6 is 15

so 15 is the length of TR

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3(4*5)=(3*4)5 what is thsi

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This is the associative property. Basically if you move the parenthesis either to left or right you’ll still get the right answer. So 4 times 5 equals 20 and 20 times 3 equals 60. 3 x 4 equals 12 and 12 x 5 equals 60

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The family size bottle of sunscreen holds 12 fluid ounces of sunscreen. The regular bottle holds 75% less. How many fewer fluid

vladimir1956 [14]

9 fluid ounces
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Citrus2011 [14]

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2 years ago

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

2 years ago