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Bumek [7]
4 years ago
13

I need the inverse operation for questions 1,3,7,9

Mathematics
1 answer:
Veseljchak [2.6K]4 years ago
7 0

3. g(x) = (x-3)^2 -5

Replace g(x) with y

y = (x-3)^2 -5

Switch the x and the Y:

x = (y-3)^2 -5

Solve for y:

Add 5 to both sides:

x + 5 = (y-3)^2

Take the square root of both sides:

+ /- SQRT(x+5) = y -3

Add 3 to both sides:

y = +/- SQRT(x +5) +3

The solution is both the positive and negative:

SQRT(x +5) +3 and -SQRT(x+5) +3

7. Following the same steps as 3

The answer is SQRT(x-2) -6 and -SQRT(X-2) -6

9. Following thw same steps as 3.

The answer is SQRT(2(x+4)) +2 and - SQRT(2(x+4))

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2. The volume of a cylinder is 628 cm. The height is 2 em. What is the diameter?
miv72 [106K]
The volume is simply just 628 / 2 then you divide that by two, easy sorry if it’s wrong
7 0
3 years ago
1. Consider an athlete running a 40-m dash. The position of the athlete is given by , where d is the position in meters and t is
sasho [114]

There is some information missing in the question, since we need to know what the position function is. The whole problem should look like this:

Consider an athlete running a 40-m dash. The position of the athlete is given by d(t)=\frac{t^{2}}{6}+4t where d is the position in meters and t is the time elapsed, measured in seconds.

Compute the average velocity of the runner over the intervals:

(a) [1.95, 2.05]

(b) [1.995, 2.005]

(c) [1.9995, 2.0005]

(d) [2, 2.00001]

Answer

(a) 6.00041667m/s

(b) 6.00000417 m/s

(c) 6.00000004 m/s

(d) 6.00001 m/s

The instantaneous velocity of the athlete at t=2s is 6m/s

Step by step Explanation:

In order to find the average velocity on the given intervals, we will need to use the averate velocity formula:

V_{average}=\frac{d(t_{2})-d(t_{1})}{t_{2}-t_{1}}

so let's take the first interval:

(a) [1.95, 2.05]

V_{average}=\frac{d(2.05)-d(1.95)}{2.05-1.95}

we get that:

d(1.95)=\frac{(1.95)^{3}}{6}+4(1.95)=9.0358125

d(2.05)=\frac{(2.05)^{3}}{6}+4(2.05)=9.635854167

so:

V_{average}=\frac{9.6358854167-9.0358125}{2.05-1.95}=6.00041667m/s

(b) [1.995, 2.005]

V_{average}=\frac{d(2.005)-d(1.995)}{2.005-1.995}

we get that:

d(1.995)=\frac{(1.995)^{3}}{6}+4(1.995)=9.30335831

d(2.005)=\frac{(2.005)^{3}}{6}+4(2.005)=9.363335835

so:

V_{average}=\frac{9.363335835-9.30335831}{2.005-1.995}=6.00000417m/s

(c) [1.9995, 2.0005]

V_{average}=\frac{d(2.0005)-d(1.9995)}{2.0005-1.9995}

we get that:

d(1.9995)=\frac{(1.9995)^{3}}{6}+4(1.9995)=9.33033358

d(2.0005)=\frac{(2.0005)^{3}}{6}+4(2.0005)=9.33633358

so:

V_{average}=\frac{9.33633358-9.33033358}{2.0005-1.9995}=6.00000004m/s

(d) [2, 2.00001]

V_{average}=\frac{d(2.00001)-d(2)}{2.00001-2}

we get that:

d(2)=\frac{(2)^{3}}{6}+4(2)=9.33333333

d(2.00001)=\frac{(2.00001)^{3}}{6}+4(2.00001)=9.33339333

so:

V_{average}=\frac{9.33339333-9.33333333}{2.00001-2}=6.00001m/s

Since the closer the interval is to 2 the more it approaches to 6m/s, then the instantaneous velocity of the athlete at t=2s is 6m/s

8 0
3 years ago
The length of a rectangle is 4 units more than its width. The area of the rectangle is 25 more than 4 times the width.
never [62]

Answer:

B or D ( answer = 5 => Two options is one! One of the options should be 5)

Note: The width of a rectangle will never be negative.

6 0
4 years ago
Read 2 more answers
I need help with this
Vladimir [108]

Answer:

{4,5} value makes the inequality

Step-by-step explanation:

4+1>4

5+1>4

4 0
3 years ago
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Someone please help me!!
densk [106]
So for a you need to add up all the numbers and for b you need to multiply all the numbers

Can you help me I posted a new question that is worth 50 points
7 0
4 years ago
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