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Tatiana [17]
4 years ago
5

Mimi wrote the equation 5c = 3k to represent a

Mathematics
1 answer:
larisa86 [58]4 years ago
5 0

Answer:

3/5=c

Step-by-step explanation:

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You had $20 to spend on seven avocados. After buying them you had $6. How much did each avocado cost?
mestny [16]
They each cost 2 dollars

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3 years ago
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Find the volume of the pyramid
Scorpion4ik [409]
The answer is V=384.
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Which of the following situations describes a function? Remember you are looking for an independent variable and a dependent var
BlackZzzverrR [31]

The situation that is a function is: C. The weight of a package and the cost of postage.

<h3>What is a Function?</h3>

A function can be described as a relation whereby there can only be one exact possible output value (dependent variable) for every input value (independent variable) in the relation. This means, no input value has two different output values. However, two different input values can have the same output value.

In a function, the output value is dependent on the input value.

For example, the cost of postage (output) is dependent on the weight of a package (input).

Therefore, the situation that is a function is: C. The weight of a package and the cost of postage.

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brainly.com/question/15602982

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6 0
2 years ago
Simplify by combining like terms: -18+3(5x-2)+8=46
Dmitriy789 [7]

-18 +3(5x-2)+8=46

-18 + 15x-6 +8 = 46

-18 + 15x +2 =46

-16 +15x =46

15x = 62

x = 62/15 = 4.1333

5 0
4 years ago
David drops a soccer ball off a building. The building is 75 meters tall.
Nezavi [6.7K]

a) 70.1 m

The ball is moving by uniformly accelerated motion, with constant acceleration g=9.8 m/s^2 (acceleration due to gravity) towards the ground. The height of the ball at time t is given by the equation

h(t)=h_0 - \frac{1}{2}gt^2

where h_0 = 75 m is the height of the ball at time t=0. Substituting t=1 s, we can find the height of the ball 1 seconds after it has been dropped:

h(1 s)=75 m - \frac{1}{2}(9.8 m/s^2)(1 s)^2=70.1 m

b) 3.9 s

We can still use the same equation we used in the previous part of the problem:

h(t)=h_0 - \frac{1}{2}gt^2

This time, we want to find the time t at which the ball hits the ground, which means the time t at which h(t)=0. So we have

0=h_0 - \frac{1}{2}gt^2

And solving for t we find

t=\sqrt{\frac{2h_0}{g}}=\sqrt{\frac{2(75 m)}{9.8 m/s^2}}=3.9 s

8 0
4 years ago
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