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garik1379 [7]
2 years ago
14

A rectangle has a base of x squared y

Mathematics
1 answer:
jekas [21]2 years ago
8 0

Answer:

i don’t know

Step-by-step explanation:

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SOMEOne PLS EXPLAIN VERTICAL ASYMPTOTES TO ME IM DYING- i need to know how u get the asymptote(explanation) and answer T^TTTTTTT
Kitty [74]
A vertical asymptote is what you get when you try to divide by 0. To find where you get these, you need to look at the denominator and what values of x will make the denominator equal to 0.

In your denominator, you have (x+7)(x-5)(x-3).
What values of x makes (x+7)(x-5)(x-3)=0?
If x = -7, if x = 5, or if x = 3, then that entire expression will equal zero. (Same idea as when you solve equations by factoring.

Now the only place this can get trickier is if one of those factors — one of (x+7), (x-5), or (x-3) — also appears in the numerator. If that happens, then it’s more involved whether you have an asymptote or not. But that doesn’t happen in this example.

So the short version: Asymptotes happen when you try to divide by zero. Dividing by zero is not a good thing. So you just ask yourself, “What will make the denominator 0?”
4 0
2 years ago
What is the solution to q + (–9) = 12? (1 point)
Klio2033 [76]
To solve q+(-9)=12 we do the following.
q+(-9)=12
     +9     +9
q=21

Giving us are end answer of q=21.

Hope this helps!=)
4 0
3 years ago
What is the measure of <DCF
Shtirlitz [24]
129 degrees you just add DCE and ECF together
5 0
3 years ago
Read 2 more answers
The braking distance, in feet of a car a Travling at v miles per hour is given.
irakobra [83]

The braking distance is the distance the car travels before coming to a stop after the brakes are applied

a. The braking distances are as follows;

  • The braking distance at 25 mph, is approximately <u>63.7 ft.</u>
  • The braking distance at 55 mph,  is approximately <u>298.35 ft.</u>
  • The braking distance at 85 mph,  is approximately <u>708.92 ft.</u>

b. If the car takes 450 feet to brake, it was traveling with a speed of 98.211 ft./s

Reason:

The given function for the braking distance is D = 2.6 + v²/22

a. The braking distance if the car is going 25 mph is therefore;

25 mph = 36.66339 ft./s

D = 2.6 + \dfrac{36.66339^2}{22} = 63.7 \ ft.

At 25 mph, the braking distance is approximately <u>63.7 ft.</u>

At 55 mph, the braking distance is given as follows;

55 mph = 80.65945  ft.s

D = 2.6 + \dfrac{80.65945^2}{22} \approx 298.35 \ ft.

At 55 mph, the braking distance is approximately <u>298.35 ft.</u>

At 85 mph, the braking distance is given as follows;

85 mph = 124.6555 ft.s

D = 2.6 + \dfrac{124.6555^2}{22} \approx 708.92 \ ft.

At 85 mph, the braking distance is approximately <u>708.92 ft.</u>

b. The speed of the car when the braking distance is 450 feet is given as follows;

450 = 2.6 + \dfrac{v^2}{22}

v² = (450 - 2.6) × 22 = 9842.8

v = √(9842.2) ≈ 98.211 ft./s

The car was moving at v ≈ <u>98.211 ft./s</u>

Learn more here:

brainly.com/question/18591940

8 0
2 years ago
Rewrite the expressions in each pair so that they have the same base. c) (1 /2)^2x and (1/ 4)^x - 1
just olya [345]
I'm not completely sure but this is what I would do.
evaluate <span>(1/ 4)^x - 1 </span>as is. But change the (1 /2)^2x to (2/4)^2x. This way both fractions have the same denominator and in this sense, the same base. The 2/4 base still evaluates into 1/2 so nothing, mathematically, is being broken here.
7 0
3 years ago
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