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ololo11 [35]
3 years ago
11

How many liters of a 10% antifreeze solution and a 40% antifreeze solution must be mixed to make 15 liters of a 25% solution? Ro

und to two decimal places, as needed
Mathematics
1 answer:
butalik [34]3 years ago
8 0

Answer:

7.5 liters of a 10% antifreeze solution AND

7.5 liters of a 40% antifreeze solution

Step-by-step explanation:

According to the question,we were asked to find out how many liters of 10% anti freeze solution and 40% anti freeze each to make a 15 liters 25% mixture

Let’s call

the amount of liters needed of our 10% solution “x”. So, how

many liters do we need of the 40% solution? Well, there are 15

liters in total, x liters have been spoken for, so what remains

of our allotted 10 liters then, is 15-x.

Now let's solve for "x"

10% of x + 40% of (15 - x) = 25% of 15 liters

0.1x + 0.4(15 - x) = 0.25(15)

0.1x + 6 - 0.4x = 3.75

Collect the like terms and we have

0.3x = 2.25

x = 2.25 × 3

X = 7.5(Liters for the 10% anti freeze solution)

The 40% solution= 15 - 7.5 = 7.5 liters

Since "X" was used to fill for the unknown amount of 10% anti freeze solution,we have 7.5 liters of the 10% solution and another 7.5 liters for the 40% solution to end up with 15 liters of the desired 25% solution.

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3.5

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. <br><br><br> Solve the equation. <br><br> 2/3 x – 6 = 9
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Hey there!

Okay let's get cracking :D

The equation is....  (2x/3) - 6 = 9

2x/3 = 15
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Find the equation, (f(x) = a(x - h)2 + k), for a parabola containing point (2, -1) and having (4, -3) as a vertex. What is the s
Nataliya [291]

Answer:

f(x)=\frac{1}{2}x^2-4x+5

Step-by-step explanation:

A parabola is written in the form

f(x)=a((x-h)^2+k) (1)

where:

h is the x-coordinate of the vertex of the parabola

ak is the y-coordinate of the vertex of the parabola

a is a scale factor

For the parabola in the problem, we know that the vertex has  coordinates (4,-3), so we have:

h=4 (2)

ak=-3

From this last equation, we get that a=\frac{-3}{k} (3)

Substituting (2) and (3) into (1) we get the new expression:

f(x)=-\frac{3}{k}((x-4)^2+k) = -\frac{3}{k}(x-4)^2 -3 (4)

We also know that the parabola  contains the point (2,-1), so we can substitute

x = 2

f(x) = -1

Into eq.(4) and find the value of k:

-1=-\frac{3}{k}(2-4)^2-3\\-1=-\frac{3}{k}\cdot 4 -3\\2=-\frac{12}{k}\\k=-\frac{12}{2}=-6

So we also get:

a=-\frac{3}{k}=-\frac{3}{-6}=\frac{1}{2}

So the equation of the parabola is:

f(x)=\frac{1}{2}((x-4)^2 -6) (5)

Now we want to rewrite it in the standard form, i.e. in the form

f(x)=ax^2+bx+c

To do that, we simply rewrite (5) expliciting the various terms, we find:

f(x)=\frac{1}{2}((x^2-8x+16)-6)=\frac{1}{2}(x^2-8x+10)=\frac{1}{2}x^2-4x+5

6 0
3 years ago
Find the perimeter of the triangle.
Arturiano [62]

The base is from 2 to 7, which is 5 units long.

The height is from 1 to 8, which is 7 units tall.

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8.6


Perimeter = 5 + 7 + 8.6 = 20.6


The answer is C.

4 0
3 years ago
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