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trapecia [35]
4 years ago
5

The Sparkling Pool Company supplies the graph below to each of its new clients. The graph shows the relationship between the amo

unt of water pumped into a pool and the time that has elapsed.
Which of the following statements best describe the rate at which the water is pumped into a pool.
A)The water is pumped at a rate of approximately 15 gallons per hour.
B)The water is pumped at a rate of approximately 500 gallons per hour
C)The water is pumped at a rate of approximately 15 gallons per minute.
D)The water is pumped at a rate of approximately 909 gallons per minute.

Mathematics
1 answer:
Alexandra [31]4 years ago
5 0
C)The water is pumped at a rate of approximately 15 gallons per minute.
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Hank bought 5 CDs for $64.75. If each CD cost the same price, how much did each cost?
dem82 [27]
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3 years ago
-u≥4−u≥4 true. Then write an equalivalent inequality, in terms of uu. (Numbers written in order from least to greatest going acr
ziro4ka [17]

Answer:

(a)\ u \le -4

(b)\ u =\{-\infty,.....,-6,-5,-4\}

Step-by-step explanation:

Given

-u \ge 4

Solving (a): An equivalent inequality

We have:

-u \ge 4

Multiply both sides by -1 (this changes the inequality)

-u*-1 \le 4 * -1

u \le -4

Solving (b): Values of u from least to greatest

u \le -4 implies that u ends at -4, starting from negative infinity

So, the list is:

u =\{-\infty,.....,-6,-5,-4\}

8 0
3 years ago
Two ropes, AD and BD, are tied to a peg on the ground at point D. The other ends of the ropes are tied to points A and B on a fl
Lisa [10]

Answer: 5(√3-1) unit.

Step-by-step explanation:

Since, By the below diagram,

For triangle BDC,

We can write,

tan 30^{\circ}=\frac{BC}{DC}

⇒ \frac{1}{\sqrt{3}}=\frac{BC}{5\sqrt{3}}}

⇒  \sqrt{3}=\frac{5\sqrt{3}}{BC}

⇒ BC=\frac{5\sqrt{3} }{\sqrt{3} }

⇒ BC=5\text{ unit}

Now, In triangle ADC,

tan45^{\circ}=\frac{AC}{DC}

⇒ 1=\frac{AB+BC}{DC}

⇒ 1=\frac{DC}{AB+BC}

⇒ 1=\frac{5\sqrt{3}}{AB+5}

⇒ AB+5=5\sqrt{3}

⇒ AB=5\sqrt{3}-5=5(\sqrt{3}-1)\text{ unit}

8 0
3 years ago
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Find the area of a regular hexagon with a 48-inch perimeter.
kiruha [24]
A= \cfrac{3 \sqrt{3} }{2}*a^2 \ \ \ \ [a=sidelength]  \\   \\ a=48/6=8 \ in \\ \\ \\ A= \cfrac{3 \sqrt{3} }{2}*(8)^2=A= \cfrac{3 \sqrt{3} }{2}*64=3 \sqrt{3}*32= 96 \sqrt{3} \ \ in^2
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