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Shalnov [3]
3 years ago
12

?? I'm stuck and I have no idea where to go from here

Mathematics
1 answer:
Alenkasestr [34]3 years ago
4 0
\cfrac{4(3(-2)-6)}{1+(-2)}= \cfrac{4(-6-6)}{1-2}= \cfrac{4(-12)}{-1}= \cfrac{-48}{-1}  =48
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How to solve -3 1/2 + 5 4/5?
monitta
Hi. This is a bit lengthy so I am attaching an image that shows it worked out step by step. I hope it helps.

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Diana

6 0
3 years ago
Use the net to find the lateral area of the prism.<br> 15 m<br> 3m 6m 3m 6m
azamat

Answer:

12cm

Step-by-step explanation

for each said biutrh

3 0
3 years ago
Write an equation of the line that passes through the points (0,4)(0,4) and (2,10)(2,10). $y=
lutik1710 [3]
I am going to guess you want to find a line which passes through the points (0,4) and (2,10)...

y₂-y₁ / x₂-x₁
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3 0
3 years ago
Dos motos, A y B, toman la salida en una carrera de 90 km. La moto A realiza el recorrido con una velocidad media inferior en 20
aleksandr82 [10.1K]

Answer:

motorcycle A: 180 km/h

motorcycle B: 200 km/h

Step-by-step explanation:

To solve this question we need to write a system of equations for A and B, using the equation:

distance = speed * time

The speed of A is 20 less than the speed of B, so:

speedA = speedB - 20

And the time A traveled is 3 minutes (0.05 hours) more than B's time, so:

timeA = timeB + 0.05

Then, using the distance equation, we have that:

distanceB = speedB * timeB

90 = speedB * timeB

distanceA = speedA * timeA

90 = (speedB - 20) * (timeB + 0.05)

90 = speedB * timeB + 0.05 * speedB - 20*timeB - 1

90 = 90 + 0.05 * speedB - 20*timeB - 1

20*timeB = 0.05 * speedB - 1

timeB = (speedB - 20)/400

using this timeB in the distance equation, we have:

90 = speedB * (speedB - 20)/400

90 * 400 = speedB^2 - 20*speedB

speedB^2 - 20*speedB - 36000 = 0

Solving this quadratic equation, we have speedB = 200 km/h

And the speedA is:

speedA = speedB - 20 = 200 - 20 = 180 km/h

6 0
2 years ago
What is the slope of the line????
Lisa [10]
Y= -4/5x -1.5

I believe this right!
5 0
2 years ago
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