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3 years ago

13

Andrea jumped 5 times every 45 minutes. At that rate, how many

1 answer:

elena55 [62]3 years ago

4 0

Answer:

6 times

Step-by-step explanation:

If Andrea jumps 5 times every 45 minutes, she will jump 1 time every 9 minutes. 54 divided by 9=6 times. Hope it helps!

You might be interested in

Find the difference quotient <img src="https://tex.z-dn.net/?f=%5Cfrac%7Bf%28x%20%2B%20h%29%20-%20f%28x%29%7D%7Bh%7D" id="TexFor

AlladinOne [14]

Answer:

$-2x - h - 3$

Step-by-step explanation:

Step 1: Define

Difference Quotient: $\frac{f(x+h)-f(x)}{h}$

f(x) = -x² - 3x + 1

f(x + h) means that x = (x + h)

f(x) is just the normal function

Step 2: Find difference quotient

<u>Substitute:</u> $\frac{[-(x+h)^2-3(x+h)+1]-(-x^2-3x+1)}{h}$
<u>Expand and Distribute:</u> $\frac{[-(x^2+2hx+h^2)-3x-3h+1]+x^2+3x-1}{h}$
<u>Distribute:</u> $\frac{-x^2-2hx-h^2-3x-3h+1+x^2+3x-1}{h}$
<u>Combine like terms:</u> $\frac{-2hx-h^2-3h}{h}$
<u>Factor out </u><em><u>h</u></em><u>:</u> $\frac{h(-2x-h-3)}{h}$
<u>Simplify:</u> $-2x - h - 3$

8 0

3 years ago

Read 2 more answers

Wha) is the image of the point (9,2) after a rotation of 90° counterclockwise about the origin?

ElenaW [278]

Given the point:

(x, y) ==> (9, 2)

Let's find the new point of the image after a rotation of 90 degrees counterclockwise about the origin.

To find the image of the point after a rotation of 90 degrees counterclockwise, apply the rules of rotation.

After a rotation of 90 degrees counterclockwise, the point (x, y) changes to (-y, x)

Thus, we have the point after the rotation:

(x, y) ==> (-y, x)

(9, 2) ==> (-2, 9)

Therefore, the image of the points after a rotation of 90 degrees counterclockwise is:

(-2, 9)

ANSWER:

(-2, 9)

5 0

1 year ago

CAN SOMEONE HELP ME WITH THIS PLEASE AND THANK YOU.

Aloiza [94]

If you convert the fractions to improper, you get 9/2 times 7/4. you can then just multiply the tops and bottoms and simplify! hope this helps

6 0

3 years ago

Read 2 more answers

Write point-slope form of the line through points (3,-5) and (-5,-5)

ratelena [41]

Y-5=-2(x-3)

The formula is y-y1=m(x-x1)
1. Find the slope
2. Replace y1 & x1 with the first point

8 0

3 years ago

Find an equation of the tangent to the curve at the given point by both eliminating the parameter and without eliminating the pa

konstantin123 [22]

Answer:

$y=2x-8$

Step-by-step explanation:

The given parametric equation is;

$x=6+ln(t),y=t^2+3$

<h3><u>BY ELIMINATING THE PARAMETER</u></h3>

To eliminate the parameter we make $t$ the subject in one equation and put it inside the other.

We make $t$ the subject in $x=6+ln(t)$ because it is easier.

$\Rightarrow x-6=ln(t)$

$\Rightarrow {e}^{x-6}=e^{ln(t)}$

$\Rightarrow {e}^{x-6}=t$

Or

$t={e}^{x-6}$

We now substitute this into $y=t^2+3$ .

This gives us;

$y=(e^{x-6})^2+3$ .

$\Rightarrow y=e^{2(x-6)}+3$ .

We have now eliminated the parameter.

The equation of the tangent at (6,4) is given by;

$y-y_1=m(x-x_1)$

where the gradient function is given by;

$\frac{dy}{dx}=2e^{2(x-6)}$

We substitute $x=6$ into the gradient function to obtain the gradient.

$\Rightarrow m=2e^{2(6-6)}$

$\Rightarrow m=2e^0$

$\Rightarrow m=2$

The equation of the tangent becomes

$y-4=2(x-6)$

We simplify to obtain

$y=2x-12+4$

$y=2x-8$

<h3><u>WITHOUT ELIMINATING THE PARAMETER</u></h3>

The given parametric equation is;

$x=6+ln(t),y=t^2+3$

For $x=6+ln(t)$

$\frac{dx}{dt}=\frac{1}{t}$

For $y=t^2+3$

$\frac{dy}{dt}=2t$

The slope is given by;

$\frac{dy}{dx}=\frac{\frac{dy}{dt} }{\frac{dx}{dt} }$

$\frac{dy}{dx}=\frac{2t }{\frac{1}{t} }$

$\frac{dy}{dx}=2t^2$

At the point, (6,4), we plug in any of the values into the parametric equation and find the corresponding value for $t$ .

Notice that

When $x=6$ , $6=6+\ln(t)$

$6-6=\ln(t)$

$0=\ln(t)$

$e^0=e^\ln(t)$

$1=t$

when $y=4$ , $4=t^2+3$

$4-3=t^2$

$1=t^2$

$t=\pm1$

But the slope is the same when we plug in any of these values for t.

$\frac{dy}{dx}=2(\pm1)^2=2$

The equation of the tangent becomes

$y-4=2(x-6)$

We simplify to obtain

$y=2x-12+4$

$y=2x-8$

7 0

3 years ago