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3 years ago

Y is such that 4y _ 7<= 3y and 3y <= 5y +8. What range of values of y satisfies both inequalities

Mathematics

1 answer:

fgiga [73]3 years ago

6 0

Answer:

Step-by-step explanation:

Given the inequality expressions

4y - 7 ≤ 3y and 3y≤5y+8

For 4y - 7 ≤ 3y

Collect like terms

4y - 3y ≤ 7

y ≤ 7

For 3y≤5y+8

Collect like terms

3y - 5y ≤ 8

-2y ≤ 8

y ≥ 8/-2

y ≥ -4

Combining both solutions

-4≤y≤7

<em>Hence the range of values of y that satisfies both inequalities is -4≤y≤7</em>

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A diameter of a circle has endpoints P(-10,-2) and Q(4,6).

Lesechka [4]

Check the picture below.

$\bf ~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ P(\stackrel{x_1}{-10}~,~\stackrel{y_1}{-2})\qquad Q(\stackrel{x_2}{4}~,~\stackrel{y_2}{6}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left( \cfrac{4-10}{2}~~,~~\cfrac{6-2}{2} \right)\implies \left( \cfrac{-6}{2}~,~\cfrac{4}{2} \right)\implies \stackrel{\textit{center}}{(-3~,~2)} \\\\[-0.35em] ~\dotfill$

$\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ \stackrel{\textit{center}}{(\stackrel{x_1}{-3}~,~\stackrel{y_1}{2})}\qquad Q(\stackrel{x_2}{4}~,~\stackrel{y_2}{6})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ \stackrel{radius}{r}=\sqrt{[4-(-3)]^2+[6-2]^2}\implies r=\sqrt{(4+3)^2+(6-2)^2} \\\\\\ r=\sqrt{49+16}\implies r=\sqrt{65} \\\\[-0.35em] ~\dotfill$

$\bf \textit{equation of a circle}\\\\ (x- h)^2+(y- k)^2= r^2 \qquad center~~(\stackrel{-3}{ h},\stackrel{2}{ k})\qquad \qquad radius=\stackrel{\sqrt{65}}{ r} \\[2em] [x-(-3)]^2+[y-2]^2=(\sqrt{65})^2\implies (x+3)^2+(y-2)^2=65$

5 0

3 years ago

Pls help me. lol Im not smart

satela [25.4K]

Answer:

45/8

Step-by-step explanation:

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2 years ago

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Kevin designs a sprinkler system for his yard. One rotation of the the sprinkler waters a circle with an area of 255pie square f

pantera1 [17]

The area of a circle can be found using:
pi x r^2 (where r is the radius)

If the area of the circle is 255pi, then:
pi x r^2 = 255pi

Now we need to make r the subject:

r^2 = 255 (cancel the pi on both sides)

r = square root of 255

r = 15.96871942

Answer: 15.96871942 feet
Hope the answer helps :)

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Step-by-step explanation:

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