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PIT_PIT [208]
3 years ago
5

Concentrated juice boxes 8 boxes = 16 1/4 Keith would like to bring enough concentrated juice in order to have 214 cups availabl

e per day. How much juice does he need and is 8 boxes Please answer soon I really need this
Mathematics
1 answer:
wolverine [178]3 years ago
4 0

Answer:

15 3/4

Yes

Step-by-step explanation:

Given that :

8 boxes = 16 1/4 cups

Required cup of concentrated juice per day = 2 1/4 cups

Total cups that will be required for the entire 7 days :

Number of cups per day * number of days

2 1/4 * 7

= 15 3/4 cups

Since 8 boxes = 16 1/4 ; then 8 boxes will be enough

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Find angle <QPS in the diagram​
marin [14]

Answer:

40°

Step-by-step explanation:

Because triangle QSR is isosceles ∠SQR=∠SRQ=35°. The sum of the angles in a triangle is 180°, so ∠QSR=180°-35°-35°=110°. The measure of a straight line is 180°, so ∠PSQ=180°-110°=70°. Because triangle PSQ is also isosceles ∠PSQ=∠PQS=70°. Then, ∠QPS=180°-70°-70°=40°.

4 0
2 years ago
parallel lines r and s are cut by two transversals, parallel lines t and u. Which angles are corresponding angles with angle 8?
Anuta_ua [19.1K]

Option A is correct answer

4 0
3 years ago
Write the equation of the line with a slope of 3/4 and a y'-intercept of -7 in Slope-Intercept and Poin Slope Form.
Shtirlitz [24]
Slope intercept form: y = 3/4 x - 7
Point slope form: y + 7 = 3/4 (x+0)
4 0
2 years ago
Petes boat can travel 48 miles upstream in 4 hours. The return trip takes 3 hours. Find the speed of the boat in still water and
tankabanditka [31]
So... hmm bear in mind, when the boat goes upstream, it goes against the stream, so, if the boat has speed rate of say "b", and the stream has a rate of "r", then the speed going up is b - r, the boat's rate minus the streams, because the stream is subtracting speed as it goes up

going downstream is a bit different, the stream speed is "added" to boat's
so the boat is really going faster, is going b + r

notice, the distance is the same, upstream as well as downstream
thus   \bf \begin{cases}
b=\textit{rate of the boat}\\
r=\textit{rate of the river}
\end{cases}\qquad thus
\\\\\\

\begin{array}{lccclll}
&distance&rate&time(hrs)\\
&----&----&----\\
upstream&48&b-r&4\\
downstream&48&b+4&3
\end{array}
\\\\\\

\begin{cases}
48=(b-r)(4)\to 48=4b-4r\\\\
\frac{48-4b}{-4}=r\\
--------------\\
48=(b+r)(3)\\
-----------------------------\\\\
thus\\\\
48=\left[ b+\left(\boxed{\frac{48-4b}{-4}}\right) \right] (3)
\end{cases}

solve for "r", to see what the stream's rate is

what about the boat's? well, just plug the value for "r" on either equation and solve for "b"
5 0
3 years ago
Use f(x)=1/2x and f^-1(x)=2x to solve the problems f^1(-2), f(-4), f(f^-1(-2)?
neonofarm [45]
F^1(-2) = 2(-2) = -4
f(-4) = 1/2 * -4 =  -2

f(f^-1(x)) = x    so answer is  -2
8 0
3 years ago
Read 2 more answers
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