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2 years ago

9

which equation represents a proportional relationship that has a constant of proportionality equal to ?

1 answer:

Ede4ka [16]2 years ago

6 0

I don’t think is wrong I don’t know what I

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A bakery charges two different prices for desserts. The table shows the number of cupcakes and the number of cookies sold in the

andreev551 [17]

Answer:

hope this helps

Step-by-step explanation:do not need one really

8 0

3 years ago

Stacie had a meal that cost $36.00. If she tipped the waited 18% of the total bill for the service, How much was the tip?

Dovator [93]

18% = 0.18

Tip = $36.00 x 0.18 = $6.48

Total bill = $36.00 + $6.48 = $42.48

Answer:

Total bill = $42.48

Tip = $6.48

6 0

3 years ago

If I can walk 9673.6 meters in 5 hours. How far can I walk in 70 minutes. ( I need the work)

den301095 [7]

First, convert 5 hours into minutes:

5 hours 60 min

------------ * --------------- = 300 min

1 1 hr

Next, find the unit rate in m/min:

9673.6 m

--------------- = 32.24 m/min

300 min

Now find the distance you can walk in 70 minutes at this rate:

32.24 m

------------- * 70 min = 2247 meters (answer)

1 min

You could walk 2247 meters (to the nearest meter) in 70 minutes.

6 0

3 years ago

PLEASE HELP I WILL MARK BRAINLIEST

Lelu [443]

Answer:

$27 \: {km}^{2}$

Step-by-step explanation:

Given is the shape of a trapezoid.

Therefore,

Area of the trapezoid

$= \frac{1}{2} (2 + 7) \times 6 \\ \\ = 9 \times 3 \\ \\ = 27 \: {km}^{2}$

5 0

3 years ago

The hypotenuse of a right triangle has endpoints A(4, 1) and B(–1, –2). On a coordinate plane, line A B has points (4, 1) and (n

GarryVolchara [31]

Answer:

$(-1,1),(4,-2)$

Step-by-step explanation:

Given: The hypotenuse of a right triangle has endpoints A(4, 1) and B(–1, –2).

To find: coordinates of vertex of the right angle

Solution:

Let C be point $(x,y)$

Distance between points $(x_1,y_1),(x_2,y_2)$ is given by $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

$AC=\sqrt{(x-4)^2+(y-1)^2}\\BC=\sqrt{(x+1)^2+(y+2)^2}\\AB=\sqrt{(4+1)^2+(1+2)^2}=\sqrt{25+9}=\sqrt{34}$

ΔABC is a right angled triangle, suing Pythagoras theorem (square of hypotenuse is equal to sum of squares of base and perpendicular)

$34=\left [ (x-4)^2+(y-1)^2 \right ]+\left [ (x+1)^2+(y+2)^2 \right ]$

Put $(x,y)=(-1,1)$

$34=\left [ (-1-4)^2+(1-1)^2 \right ]+\left [ (-1+1)^2+(1+2)^2 \right ]\\34=25+9\\34=34$

which is true. So, $(-1,1)$ can be a vertex

Put $(x,y)=(4,-2)$

$34=\left [ (4-4)^2+(-2-1)^2 \right ]+\left [ (4+1)^2+(-2+2)^2 \right ]\\34=9+25\\34=34$

which is true. So, $(4,-2)$ can be a vertex

Put $(x,y)=(1,1)$

$34=\left [ (1-4)^2+(1-1)^2 \right ]+\left [ (1+1)^2+(1+2)^2 \right ]\\34=9+4+9\\34=22$

which is not true. So, $(1,1)$ cannot be a vertex

Put $(x,y)=(2,-2)$

$34=\left [ (2-4)^2+(-2-1)^2 \right ]+\left [ (2+1)^2+(-2+2)^2 \right ]\\34=4+9+9\\34=22$

which is not true. So, $(2,-2)$ cannot be a vertex

Put $(x,y)=(4,-1)$

$34=\left [ (4-4)^2+(-1-1)^2 \right ]+\left [ (4+1)^2+(-1+2)^2 \right ]\\34=4+25+1\\34=30$

which is not true. So, $(4,-1)$ cannot be a vertex

Put $(x,y)=(-1,4)$

$34=\left [ (-1-4)^2+(4-1)^2 \right ]+\left [ (-1+1)^2+(4+2)^2 \right ]\\34=25+9+36\\34=70$

which is not true. So, $(-1,4)$ cannot be a vertex

So, possible points for the vertex are $(-1,1),(4,-2)$

7 0

3 years ago

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