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hodyreva [135]
3 years ago
14

Which of the following is the factored form of 9y2-121​

Mathematics
1 answer:
lukranit [14]3 years ago
4 0
It's the first one.

If you work backwords, I'd say it's easier. 3y swuared is 9y2, and 11 squared is 121. Then just keep the sign the same
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A rectangle has a perimeter of 36 inches and a width of 10 inches. What is the length of the rectangle?
leonid [27]

Answer:

8

Step-by-step explanation:

8 0
3 years ago
On a coordinate plane, rhombus W X Y Z is shown. Point W is at (7, 2), point X is at (5, negative 1), point Y is at (3, 2), and
Otrada [13]

Answer:

P = 4\sqrt{13}

Step-by-step explanation:

Given

W = (7, 2)

X = (5, -1)

Y = (3, 2)

Z =(5, 5)

Required

The perimeter

To do this, we first calculate the side lengths using distance formula

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

So, we have:

WX = \sqrt{(5- 7)^2 + (-1 - 2)^2

WX = \sqrt{13}

XY = \sqrt{(3-5)^2 + (2--1)^2}

XY = \sqrt{13}

YZ = \sqrt{(5-3)^2 + (5-2)^2}

YZ = \sqrt{13}

ZW = \sqrt{(7 - 5)^2 + (2 - 5)^2}

ZW = \sqrt{13}

The perimeter is:

P = WX + XY + YZ + ZW

P = \sqrt{13}+\sqrt{13}+\sqrt{13}+\sqrt{13}

P = 4\sqrt{13}

5 0
3 years ago
Read 2 more answers
Linda goes water-skiing one sunny afternoon. After skiing for 15 min, she signals to the driver of the boat to take her back to
trasher [3.6K]
60 = a * (-30)^2
a = 1/15
So y = (1/15)x^2


abc)
The derivative of this function is 2x/15. This is the slope of a tangent at that point.
If Linda lets go at some point along the parabola with coordinates (t, t^2 / 15), then she will travel along a line that was TANGENT to the parabola at that point.
Since that line has slope 2t/15, we can determine equation of line using point-slope formula:
y = m(x-x0) + y0
y = 2t/15 * (x - t) + (1/15)t^2
Plug in the x-coordinate "t" that was given for any point.


d)
We are looking for some x-coordinate "t" of a point on the parabola that holds the tangent line that passes through the dock at point (30, 30).
So, use our equation for a general tangent picked at point (t, t^2 / 15):
y = 2t/15 * (x - t) + (1/15)t^2
And plug in the condition that it must satisfy x=30, y=30.
30 = 2t/15 * (30 - t) + (1/15)t^2
t = 30 ± 2√15 = 8.79 or 51.21
The larger solution does in fact work for a tangent that passes through the dock, but it's not important for us because she would have to travel in reverse to get to the dock from that point.
So the only solution is she needs to let go x = 8.79 m east and y = 5.15 m north of the vertex.
4 0
3 years ago
Jacob leaves his summer cottage and drives home. After
krok68 [10]

Answer:

A linear relationship can be written as:

y = a*x + b

Where a is the slope and b is the y-intercept.

If this line passes through the points (a, b) and (c, d) then the slope can be written as:

a = (a - c)/(b - d)

Here y will represent the distance between Jacob and his house, and the variable x represents the time that he has ben driving.

In this case, we know that after driving for 5 hours, he is 112km from home.

Then we can write this point as (5h, 112km)

We also know that after 7 hours he is 15km from home.

Then we can write this point as (7h, 15km)

Then the slope of this function will be:

a = (15km - 112km)/(7h - 5h) = -48.5 km/h

Then the equation is:

y = -(48.5 km/h)*x + b

To find the value of b, we can replace the values of one of the points, for example in the point (7h, 15km)

This means that we need to replace x by 12h, and y by 15km, then we get:

15km = -( 48.5 km/h)*7h + b

15km + ( 48.5 km/h)*7h = b = 354.5 km

then the equation will be:

y = (-48.5 km/h)*x + 354.5 km

Now we want to answer: How long had Jacob been driving when he was 209 km from  home?

Then we need to only replace y by 209km, and solve for x:

209km = (-48.5 km/h)*x + 354.5 km

209km - 354.5 km = (-48.5 km/h)*x

-145.5km =  (-48.5 km/h)*x

-145.5km/( -48.5 km/h) = x = 3h

So he is 209km away from his home after driving for 3 hours.

6 0
3 years ago
In a for loop, is it possible to use indexing?
kenny6666 [7]

Answer:

Yes

Step-by-step explanation:

A for loop basically relies on repeating the same code for a pre-set number of times. During which you can make any code repeat inside of it, including indexing through a type of list. Many times a for-loop will use the indexes in a list to calculate the number of times that the loop has to repeat. This is usually done in order to search or apply changes in an array of other types of data structures that have countless values stored in it.

4 0
3 years ago
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