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deff fn [24]
3 years ago
5

720 pt/h=______ qt/min

Mathematics
1 answer:
Gennadij [26K]3 years ago
7 0
1 pt = 0.5 qts.....so 720 pts = (720 * 0.5) = 360 qts
1 hr = 60 minutes

360/60 = 6

720 pt/hr = 6 qts/min
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What is the value of x3 × y1 when x = 5 and y = 10​
katovenus [111]

Answer:

150

Step-by-step explanation:

To solve the problem, put 5 instead of x, so 5*3

and put 10 instead of y, so 10*1

when combined,

(5*3) * (10*1)

= 15 * 10

= 150

Hope this helps!!

Let me know if I'm wrong...

8 0
3 years ago
Omar has 3 1/14 cups of dough to make dumplings.If he uses 3/16 cup of dough for each dumpling,how many whole dumplings can Omar
Bumek [7]

Answer:

Omar can make 16 dumplings with some dough to spare

Step-by-step explanation:

you should convert each value to have the same denominator to make the equation easier

3 and 1/14 cups is the same as 43/14 cups of dough

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43/14 divided by 3/16 = 43/14 * 16/3

=688/42 = 16.3809...

so Omar can make 16 dumplings with some dough to spare

6 0
4 years ago
runs 5 laps in 30 minutes, denise runs 4 laps in 20 minutes, paola runs 6 laps in 40 minutes. if they run at a constant rate, wh
asambeis [7]
Denise is the fastest runner. 40 divided by 6 is 6.666 repeating 30 divided by 5 is 6 and 20 divided by 4 is 5 therefore denise is fastest
3 0
3 years ago
Jaime has 21 red balloons and 28 blue balloons. He wants to make balloon bouquets. Each bouquet must have the same number of bal
likoan [24]

Answer:

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5 0
3 years ago
Is square root 1 minus sine squared theta = cos Θ true? If so, in which quadrants does angle Θ terminate?
Cloud [144]

Answer:

True; quadrants I & IV

Step-by-step explanation:

We know the relation between sine and cosine function which is given by

\sin^2 \theta +\cos^2 \theta = 1

Let us solve this equation for cosine function.

\cos^2 \theta = 1-\sin^2 \theta

Take square root both sides. When ever we take square root we need to write the solution in plus minus form

\sqrt{\cos^2 \theta}=\pm\sqrt{1-\sin^2 \theta}

\cos \theta=\pm\sqrt{1-\sin^2 \theta}

\cos \theta=-\sqrt{1-\sin^2 \theta}, \sqrt{1-\sin^2 \theta}

If Θ is in quadrants I and IV then the value will be positive and if Θ is in II and III quadrant then the value is negative.

Hence, if Θ is in quadrants I & IV, then we have

\cos \theta=\sqrt{1-\sin^2 \theta}

Thus, the correct option is: True; quadrants I & IV


6 0
4 years ago
Read 2 more answers
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