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Grace [21]
3 years ago
15

On Tuesday, Gloria spent 25 minutes running. She rode her scooter for 20 minutes and then read a book for 10 minutes. Gloria the

n fixed herself a snack. She started fixing herself a snack at 3:30 p.m. What time did Gloria start her run?
Mathematics
2 answers:
pogonyaev3 years ago
8 0
It would be 2:35 p.m
lisov135 [29]3 years ago
4 0
Gloria started her run at 2:35 p.m.

hope this helped

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Jada and Diego are practicing the piano for an upcoming rehearsal. The number of minutes Jada practiced is listed below:
katen-ka-za [31]

Answer:

20

Step-by-step explanation:

First, rewrite the numbers in numerical order:

8, 10, 10, 15, 15, 20, 20, 20, 20, 25, 25, 25, 35

Then find the middle number: (note there's 6 numbers on each side of <u>20</u>)

<em>8 10 10 15 15 20 </em><u>20</u> <em>20 20 25 25 25 35</em>

Hope this helps!

7 0
3 years ago
John bought a crate of floor tiles $100. The crate had 5 boxes of floor tiles. Each box contained 40 floor tiles. What is the co
mash [69]
$0.50.
There are a total of 200 tiles (5 x 40)
since John paid $100 for the 200 tiles,
We have to do $100 divided by 200.
Which is .5
So $.50
4 0
2 years ago
Read 2 more answers
Ugh I need to know number ten right now!
erastovalidia [21]
Uh, rude. The answer to this is that Rico should've moved the decimal to the right twice, not the left.
3 0
3 years ago
Read 2 more answers
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
A line has a slope of Negative three-fourths and passes through the point (–5, 4). Which is the equation of the line?
balandron [24]

Answer:

y=-3/4x

Step-by-step explanation:

the form is "y=mx(+/-b). m is slope. b is y-intercept.

8 0
3 years ago
Read 2 more answers
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