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Ray Of Light [21]
3 years ago
5

Dylan, Alan, Bruce, and Cecil are brothers. Dylan weighs d pounds. Alan weighs 3 pounds more than twice what Dylan weighs. Bruce

weighs 12 pounds less than three times what Dylan weighs. If Cecil weighs the same as the combined weight of Bruce and Alan, what is Cecil’s weight in terms of Dylan’s weight (d)?
Complete the steps below to solve this problem and others like it.


Part A

Alan weighs 3 pounds more than twice what Dylan weighs. Write an expression to represent Alan’s weight in terms of Dylan’s weight.
Mathematics
1 answer:
vaieri [72.5K]3 years ago
5 0

Answer:

Part A: 2D + 3 = A

Note: I have likely given you the answer to the whole problem. See below.

Step-by-step explanation:

Givens

  • Alan = A
  • Bruce = B
  • Cecil = C
  • Dylan = D

Comment

You have to pay the most attention to Alan and Bruce. Once you have equations for those two. Cecil's weight is just the sum of those two.

<em>Alan</em>

Alan = Twice*Dylan + 3 or

A = 2*D + 3

<em>Bruce</em>

Bruce = three * Dylan - 12

B = 3*D - 12

<em>Cecil</em>

Cecil = Alan + Bruce

C = A + B

Note

The last equation shows you how to combine the first two.

C = A + B

A = 2D + 3

B = 3D - 12

C = 2D + 3 + 3D - 12

C = 5D - 9

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Find the inverse please and thank you
Amanda [17]

The inverse f^{-1}(x) is such that

f\left(f^{-1}(x)\right) = x

We have

f\left(f^{-1}(x)\right) = -9\sqrt{f^{-1}(x) - 8} + 5 = x

Solve for the inverse.

-9\sqrt{f^{-1}(x)-8} + 5 = x \\\\ -9\sqrt{f^{-1}(x)-8} = x-5 \\\\ \sqrt{f^{-1}(x)-8} = -\dfrac{x-5}9 \\\\ \left(\sqrt{f^{-1}(x)-8}\right)^2 = \left(-\dfrac{x-5}9\right)^2 \\\\ f^{-1}(x) - 8 = \dfrac{(x-5)^2}{81} \\\\ \boxed{f^{-1}(x) = 8 + \dfrac{(x-5)^2}{81}}

5 0
2 years ago
The perimeter of a square mirror is 88 inches. What is the length of each side of the mirror in inches?
Anton [14]
P=4xL
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6 0
3 years ago
Please help!! What is the solution to the quadratic inequality? 6x2≥10+11x
fredd [130]

Answer:

The solution of the inequation 6\cdot x^{2} \geq 10 + 11\cdot x is \left(-\infty,-\frac{2}{3}\right]\cup\left[\frac{5}{2},+\infty\right).

Step-by-step explanation:

First of all, let simplify and factorize the resulting polynomial:

6\cdot x^{2} \geq 10 + 11\cdot x

6\cdot x^{2}-11\cdot x -10 \geq 0

6\cdot \left(x^{2}-\frac{11}{6}\cdot x -\frac{10}{6} \right)\geq 0

Roots are found by Quadratic Formula:

r_{1,2} = \frac{\left[-\left(-\frac{11}{6}\right)\pm \sqrt{\left(-\frac{11}{6} \right)^{2}-4\cdot (1)\cdot \left(-\frac{10}{6} \right)} \right]}{2\cdot (1)}

r_{1} = \frac{5}{2} and r_{2} = -\frac{2}{3}

Then, the factorized form of the inequation is:

6\cdot \left(x-\frac{5}{2}\right)\cdot \left(x+\frac{2}{3} \right)\geq 0

By Real Algebra, there are two condition that fulfill the inequation:

a) x-\frac{5}{2} \geq 0 \,\wedge\,x+\frac{2}{3}\geq 0

x \geq \frac{5}{2}\,\wedge\,x \geq-\frac{2}{3}

x \geq \frac{5}{2}

b) x-\frac{5}{2} \leq 0 \,\wedge\,x+\frac{2}{3}\leq 0

x \leq \frac{5}{2}\,\wedge\,x\leq-\frac{2}{3}

x\leq -\frac{2}{3}

The solution of the inequation 6\cdot x^{2} \geq 10 + 11\cdot x is \left(-\infty,-\frac{2}{3}\right]\cup\left[\frac{5}{2},+\infty\right).

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3 years ago
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kirill [66]

Answer:

- 1 4/5

Step-by-step explanation:

-1 1/5 + - 3/5

Since the signs are the same, we add and take the sign

1 1/5 + 3/5

1 4/5

The sign is negative

- 1 4/5

On the number line start at - 1 1/5 and go to the left 3/5( since the sign is negative)  you will end up at - 1 4/5

8 0
3 years ago
<img src="https://tex.z-dn.net/?f=x%283x%20%2B%207%29%20-%203x%20%3D%200" id="TexFormula1" title="x(3x + 7) - 3x = 0" alt="x(3x
sineoko [7]

Answer:

X=0

Step-by-step explanation:

7 0
3 years ago
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