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Zigmanuir [339]
3 years ago
12

A chemical compound requires 8 ounces of chemical A and 12 ounces of chemical B. A mixture contains 24 ounces of chemical A and

30 ounces of chemical B. How can you fix the mixture to make the chemical compound?
Mathematics
1 answer:
bixtya [17]3 years ago
8 0
Add 6oz of chemical B and remove 1/3 of the total mixture.
I might always be wrong.
Explanation:
Since 8x3=24 and since 12x3=36 you need to add 6oz of chemical B to have 24oz of chemical A and 36oz of chemical B.
Take 1/3 of the total mixture to have 8oz of chemical A and 12oz of chemical B.
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516.66 <span>km
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3 years ago
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Answer:

40 + 3x + 4 + 3x + 4 = 180

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Step-by-step explanation:

1. since the triangle is isosceles, angles A and C will have the same measures, so to solve for x, you can add another 3x + 4 or you can multiply A by 2.

2. add like terms

3. subtract

4. divide to get x

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4 0
3 years ago
What is the domain and range of <br> f(x)=2x^2+6x+2
lbvjy [14]

As is the case for any polynomial, the domain of this one is (-infinity,  +infinity).

To find the range, we need to determine the minimum value that f(x) can have.  The coefficients here are a=2, b=6 and c = 2,

The x-coordinate of the vertex is  x = -b/(2a), which here is x = -6/4 = -3/2.

Evaluate the function at x = 3/2 to find the y-coordinate of the vertex, which is also the smallest value the function can take on.  That happens to be y = -5/2, so the range is [-5/2, infinity).

3 0
3 years ago
Andres boards a Ferris wheel at the 3-o'clock position and rides the Ferris wheel for multiple revolutions. The Ferris wheel rot
Gelneren [198K]

Answer:

a) r(t) = 0.783t radians

b) h(t) = 30cos(0.783*t) feet

Step-by-step explanation:

The situation is depicted in the picture attached

<h3>(see picture) </h3>

Since the angular speed is constant, to find an expression for the angle r(t) in radians we just cross-multiply using the fact that 1 min = 60 seconds

4.7 radians __________ 60 seconds

r(t) radians ____________  t seconds

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h(t) = 30cos(0.783*t) feet

7 0
3 years ago
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