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3 years ago

12

Find the simplified quotient. y^2+y/y^2-2y/y^2-4y+4

1 answer:

BabaBlast [244]3 years ago

7 0

Answer: y2-1/y-4y+4

if this helped please mark brainliest

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The mean lifespan of a sylvania light bulb is 1600 hours with a standard deviation of 300 hours, an iqr of 450 hours, and a rang

IgorLugansk [536]

Given that t<span>he mean lifespan of a sylvania light bulb is 1600 hours with a standard deviation of 300 hours, an iqr of 450 hours, and a range of 690 hours. the lifespan distribution is relatively symmetric.

Since, the distribution is relatively symmetric, the measure of spread that is the best measurement is the standard deviation.</span>

6 0

2 years ago

Triangle has side length 7, 10 and 12 is it a right triangle explain your reasoning

Vikentia [17]

Answer: No, because 7^2 + 10^2 is > 12^2

Step-by-step explanation: 7^2 +10^2= 49 + 100= 149 149 $\neq$ 144

5 0

2 years ago

Find the distance between (-2, 8) and (11, 2). Round to the nearest hundredth.

MakcuM [25]

Answer:

14.32

Step-by-step explanation:

We can find the distance using the distance formula:

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

We then substitute (-2,8) as $(x_1,y_1)$ and (11,2) as $(x_2,y_2)$ .

$d=\sqrt{(11--2)^2+(2-8)^2} \\d=\sqrt{(13)^2+(-6)^2} \\d=\sqrt{169+36}\\d=\sqrt{205}=14.32$

6 0

2 years ago

Please Help! 39 and 40 please!!! :D

Pavel [41]

Answer:

Look below.

Step-by-step explanation:

39.

10+ = 149 times x

3 - 9 = 143 times x

40.

2(w times 1.5 + w)

6 0

3 years ago

An airplane climbs 100 feet during the first second after takeoff. In each succeeding second it climbs 100 feet more than it cli

tigry1 [53]

Answer:

15 seconds.

Step-by-step explanation:

∵ The distance covered by plane in first second = 100 ft,

Also, in each succeeding second it climbs 100 feet more than it climbed during the previous second,

So, distance covered in second second = 200,

In third second = 300,

In fourth second = 400,

............, so on....

Thus, the total distance covered by plane in n seconds = 100 + 200 + 300 +400......... upto n seconds

$=\frac{n}{2}(2\times 100 + (n-1)100)$ ( Sum of AP )

$=\frac{n}{2}(200+100n-100)$

$=\frac{n}{2}(100+100n)$

$=50n+50n^2$

Suppose the distance covered in n seconds is 12,000 feet,

$\implies 50n+50n^2=12000$

$n+n^2=240$

$n^2+n-240=0$

$n^2+16n-15n-240=0$

$n(n+16)-15(n+16)=0$

$(n-15)(n+16)=0$

$\implies n=15\text{ or }n=-16$

∵ n can not be negative,

Hence, after 15 seconds the plane will reach an altitude of 12,000 feet above its takeoff height.

6 0

2 years ago

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