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2 years ago

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Question 8 A water tank initially contained 51 liters of water. It is then drained at a constant rate of 2.5 liters per minute.

How many minutes have passed, m, when the tank contains 42 liters? Round to the nearest tenth.

1 answer:

Annette [7]2 years ago

6 0

Do 51 and 2.5 then when you get that add 42 to your answer

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HELP HELP HELP HELP help help

SVETLANKA909090 [29]

Answer:

r = 6

Step-by-step explanation:

To get the value of r we will use the pythagoras theorem a shown;

PR² = QR² + PQ²

(4+r)² = 8² + r²

Expand

16+8r+r² = 64 + r²

16 + 8r = 64

8r = 64 - 16

8r = 48

r = 48/8

r = 6

Hence the value of r is 6

7 0

3 years ago

A parabola can be drawn given a focus of

Volgvan

Answer:

$y=-\frac{1}{4}(x-2)^{2}+9$

Step-by-step explanation:

Any point on a given parabola is equidistant from focus and directrix.

Given:

Focus of the parabola is at $(2,8)$ .

Directrix of the parabola is $y=10$ .

Let $(x,y)$ be any point on the parabola. Then, from the definition of a parabola,

Distance of $(x,y)$ from focus = Distance of $(x,y)$ from directrix.

Therefore,

$\sqrt{(x-2)^{2}+(y-8)^{2}}=|y-10|$

Squaring both sides, we get

$(x-2)^{2}+(y-8)^{2}=(y-10)^{2}\\(x-2)^{2}=(y-10)^{2}-(y-8)^{2}\\(x-2)^{2}=(y-10+y-8)(y-10-(y-8))...............[\because a^{2}-b^{2}=(a+b)(a-b)]\\(x-2)^{2}=(2y-18)(y-10-y+8)\\(x-2)^{2}=2(y-9)(-2)\\(x-2)^{2}=-4(y-9)\\y-9=-\frac{1}{4}(x-2)^{2}\\y=-\frac{1}{4}(x-2)^{2}+9$

Hence, the equation of the parabola is $y=-\frac{1}{4}(x-2)^{2}+9$ .

4 0

3 years ago

Eli packed 1,920 peanuts into 8 sacks. He packed an equal number of peanuts in each sack. How many peanuts did he pack into each

svlad2 [7]

D because one thousand nine hundred twenty divided by 8 equals two hundred and forty

8 0

3 years ago

Read 2 more answers

Find the limit, if it exists, or type dne if it does not exist.

Phantasy [73]

$\displaystyle\lim_{(x,y)\to(0,0)}\frac{\left(x+23y)^2}{x^2+529y^2}$

Suppose we choose a path along the $x$ -axis, so that $y=0$ :

$\displaystyle\lim_{x\to0}\frac{x^2}{x^2}=\lim_{x\to0}1=1$

On the other hand, let's consider an arbitrary line through the origin, $y=kx$ :

$\displaystyle\lim_{x\to0}\frac{(x+23kx)^2}{x^2+529(kx)^2}=\lim_{x\to0}\frac{(23k+1)^2x^2}{(529k^2+1)x^2}=\lim_{x\to0}\frac{(23k+1)^2}{529k^2+1}=\dfrac{(23k+1)^2}{529k^2+1}$

The value of the limit then depends on $k$ , which means the limit is not the same across all possible paths toward the origin, and so the limit does not exist.

8 0

3 years ago

Write an inequality for the following statement.<br> c is less than or equal to 5

tresset_1 [31]

Answer:

c ≤5

Step-by-step explanation:

less than or equal is ≤

c ≤5

3 0

3 years ago

Read 2 more answers