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spayn [35]
3 years ago
10

How do I simplify this?

Mathematics
1 answer:
chubhunter [2.5K]3 years ago
6 0
You can multiply each top and bottom by sqrt 15
then it will be 
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An airplane is flying at a speed of 500 mi/h at an altitude of one mile. The plane passes directly above a radar station at time
d1i1m1o1n [39]

Answer:

a)s=\sqrt{1+d^2}

b)d(t)=500t

c)s(t) =\sqrt{1+250000t^2}

Step-by-step explanation:

d = Horizontal distance

s = the distance between the plane and the radar station

The horizontal distance (d), the one mile altitude, and s form a right triangle.

So, use Pythagoras theorem

Hypotenuse^2=Perpendicular^2+Base^2

s^2=1^2+d^2

a) s=\sqrt{1+d^2}

(b) Express d as a function of the time t (in hours) that the plane has flown.

distance = speed \times time

d(t)=500t

(c) Use composition to express s as a function of t.

s(t) =\sqrt{1+d^2}

using b

s(t) =\sqrt{1+(500t)^2}\\s(t) =\sqrt{1+250000t^2}

6 0
3 years ago
Hi do any of you know how to do this?
Oksi-84 [34.3K]

Answer: yes, it's a quadratic equation, the formula is x = −b ± √(b2 − 4ac) 2a. a=8m2 b=-2m and c =0. This only works if the other side is 0. The answer would be m=1!!!!! can i plz get a brainliest!

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
Donut $2.50
anastassius [24]
Bro don’t click the link ☝️
4 0
3 years ago
What is the answer to this question?<br> Y=x^2-1 with the input as 5
mixas84 [53]

We can see that, when the input is 5, the output is 24.

<h3>How to evaluate an equation?</h3>

Here we have the equation:

y = x^2 - 1

Where y is the output and x is the input.

We want to evaluate it with the input as 5, so we need to replace the variable x by the number 5, we will get:

y = 5^2 - 1 = 24

Then we can see that, when the input is 5, the output is 24.

If you want to learn more about evaluating functions:

brainly.com/question/1719822

#SPJ1

4 0
2 years ago
A bag of chocolates is labeled to contain 0.384 pounds of chocolates. The actual weight of the chocolates is 0.3798 pounds. Are
Marta_Voda [28]

Based on the information given the chocolates stated on the label is 0.0042-ounce lighter.

Weight:

Using this formula

Chocolate weight=Current weight-Actual weight

Where:

Current weight=0.384

Actual weight=0.3798

Let plug in the formula

Chocolate weight=0.384-0.3798

Chocolate weight=0.0042 ounce

In conclusion, the chocolates stated on the label are 0.0042-ounce lighter.

6 0
2 years ago
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