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2 years ago

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A mixture contains 40oz of glycol and water, and it is 10% glycol. The mixture is to be strengthened to 25% by adding glycol. Ho

w much glycol is in the original mixture?

1 answer:

Mazyrski [523]2 years ago

5 0

I found 50, making a rule of three where
40__10
x___25

x=10
40+10= 50
( I hope I'm right, but from what I understand this would be the resolution)

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Mr peterson will buy 80 feet of fencing. The pieces of fence come in whole numbers of feet. His enclosure will be a rectangle an

DIA [1.3K]

The perimeter is:
P = 2x + 2y = 80
The area is:
A = x * y
The area as a function of a variable is:
A (x) = x * (40-x)
Rewriting we have:
A (x) = 40x-x ^ 2
We derive the function:
A '(x) = 40-2x
We equal zero and clear x:
40-2x = 0
2x = 40
x = 40/2
x = 20 feet
The other dimension is:
y = 40 - x
y = 40 - 20
y = 20 feet
The area will be:
A = x * y
A = 20 * 20
A = 400 feet ^ 2
Answer:
The dimensions are:
x = 20 feet
y = 20 feet
The area is:
A = 400 feet ^ 2

5 0

3 years ago

Helppppp plsss!!!!!!!!!!!

bagirrra123 [75]

Answer:

22 inch is the diameter because d =2r

8 0

2 years ago

A radioactive substance has a continuous decay rate of 0.056 per minute. How many grams of a 120 gram sample will remain radioac

sammy [17]

Answer:

The mass of the radioactive sample after 40 minutes is 12.8 g.

Step-by-step explanation:

The mass of the sample can be found by using the exponential decay equation:

$N_{t} = N_{0}e^{-\lambda t}$

Where:

N(t): is the amount of the sample at time t =?

N₀: is the initial quantity of the sample = 120 g

t = 40 min

λ: is the decay constant = 0.056 min⁻¹

Hence, the mass of the sample after 40 min is:

$N_{t} = N_{0}e^{-\lambda t} = 120g*e^{-0.056min^{-1}*40 min} = 12.8 g$

Therefore, the mass of the radioactive sample after 40 minutes is 12.8 g.

I hope it helps you!

5 0

3 years ago

How do I know what the circle and triangle equal to?

Brrunno [24]

Answer:

circle= 5, triangle= 3

Step-by-step explanation:

Please see attached picture for full solution.

8 0

2 years ago

Find the coordinates of the image of a triangle with vertices A(0, – 3), B(3, 0), and

lora16 [44]

Answer:

A'(-3,0), B'(0,-3) and C'(4,7)

Step-by-step explanation:

We are given that the vertices of triangle are A(0,-3), B(3,0) and C(-7,4).

We have to find the coordinates of the image of triangle under a rotation of 90° clockwise about the origin.

90° clockwise about the origin

Rule: $(x,y)\rightarrow (y,-x)$

Using the rule

The coordinates of A'

$A(0,-3)\rightarrow A'(-3,0)$

The coordinates of B'

$B(3,0)\rightarrow B'(0,-3)$

The coordinates of C'

$C(-7,4)\rightarrow C'(4,7)$

Hence, the vertices of image of triangle is given by

A'(-3,0), B'(0,-3) and C'(4,7)

8 0

2 years ago