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3 years ago

3m^3 3 + 4f + 4 + 2f + 2f^2 2 + 1

Mathematics

1 answer:

damaskus [11]3 years ago

6 0

Answer:

33m^33+2f^22+6f+5

Step-by-step explanation:

this is the answer for if you are supposed to be simplifying :)

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Any one have ideas of how to do this... haha I am so lost

Anastasy [175]

How the heck r we supposed to solve that if it says “evaluate 5!”

4 0

3 years ago

Given that <br><img src="https://tex.z-dn.net/?f=x%20%3D%205y%20-%20z" id="TexFormula1" title="x = 5y - z" alt="x = 5y - z" alig

yuradex [85]

x = 5y - z

You can think that x and z are numbers, and we have one variable z.

x -5y = 5y - 5y - z

x - 5y = - z

(x - 5y)*(-1) = - z *(-1)

z = - x + 5y

7 0

3 years ago

The altitude of a triangle is increasing at a rate of 1.5 1.5 centimeters/minute while the area of the triangle is increasing at

seraphim [82]

Answer:

The base reduces at 3.75cm/min

Step-by-step explanation:

Given

Let

$h \to altitude$

$b \to base$

$A \to Area$

So:

$\frac{dh}{dt} = 1.5cm/min$

$\frac{dA}{dt} = 1.5^2cm/min$

The area of a triangle is:

$A = \frac{1}{2}bh$

Calculate b when $A =88cm^2; h =8cm$

$A = \frac{1}{2}bh$

$88=\frac{1}{2} * b * 8$

$88 =b * 4$

Solve for b

$b = 88/4$

$b = 22$

We have:

$A = \frac{1}{2}bh$

Differentiate with respect to time

$\frac{dA}{dt} =\frac{1}{2}(h\frac{db}{dt} + b\frac{dh}{dt})$

Substitute the following values in the above equation

$\frac{dh}{dt} = 1.5cm/min$ $\frac{dA}{dt} = 1.5^2cm/min$ $b = 22$ $h = 8$

$1.5 = \frac{1}{2}(8 * \frac{db}{dt} + 22 * 1.5)$

Multiply both sides by 2

$3 = 8 * \frac{db}{dt} + 22 * 1.5$

$3 = 8 * \frac{db}{dt} + 33$

Collect like terms

$8 * \frac{db}{dt} = 3 -33$

$8 * \frac{db}{dt} = -30$

Divide both sides by 8

$\frac{db}{dt} = -\frac{30}{8}$

$\frac{db}{dt} = -3.75$

4 0

3 years ago

STAN walk 10 km to the grocery store. He stops then walks 10 km east back to his house. What distance does he cover? what was hi

Bogdan [553]

STAN walk 10 km to the grocery store.

He stops then walks 10 km east back to his house.

Let us draw this situation to better understand the problem.

As you can see, Stan first walked to the grocery store and covered a distance of 10 km then he came back to the house from the grocery store and covered a distance of 10 km.

So the total distance is 10 + 10 = 20 km

On the other hand, notice that Stan is exactly at the point where he first started that is his house, therefore, the displacement is 0 in this case.

Recall that the displacement is a vector quantity so the direction of travel also matters.

From house to grocery store = 10 km

From the grocery store to house = -10 km

Overall displacement = 10 km - 10 km = 0 km

Therefore,

Distance = 20 km

Displacement = 0 km

3 0

1 year ago

A fraternity charged $2.00 admission for dudes and $1.00 for ladies for their finals week bash. The fraternity made $100.00 and

SVETLANKA909090 [29]

Given : 2D + 1L = 100,
And D + L = 70

This is a simultaneous equation can be solved as follows

2D + L = 100
- D - L = -70
————————————-
D - 0 = 30

i.e D (number of dudes) = 30

Therefore, since we already know the fraternity sold 70 tickets in total,

70 -30 = 40 (the number of ladies attended)

3 0

3 years ago

3m^3 ​3 ​​ + 4f + 4 + 2f + 2f^2 ​2 ​​ + 1

3m^3 3 + 4f + 4 + 2f + 2f^2 2 + 1