Mathematics, 10.02.2020 03:36 maysen2001

5(5c-1)-1=23c+6 solve the equation

Answers: 3

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Answers

Answer from: mariasoledad1

C=4

Step-by-step explanation:

5(5c-1)-1 = 23c + 6

1. Using PEMDAS. You know to do parenthesis first which is 5(5c-1)

Using distributive property you will get 25c-1-1 = 23c + 6

2. Simply on both sides and get 25c-2=23c+6

3. Using the rule that whatever you do on one side of the equal sign, you do on the other you can subtract 23c from both sides and get 2c-2=6. Then add 2 on both sides and get 2c=8

4. Divide both sides by 2 to get rid of the 2 and you get c=4

Answer from: skoolroks

Solution for 5(5c-1)-1=23c+4 equation:

Simplifying
5(5c + -1) + -1 = 23c + 4

Reorder the terms:
5(-1 + 5c) + -1 = 23c + 4
(-1 * 5 + 5c * 5) + -1 = 23c + 4
(-5 + 25c) + -1 = 23c + 4

Reorder the terms:
-5 + -1 + 25c = 23c + 4

Combine like terms: -5 + -1 = -6
-6 + 25c = 23c + 4

Reorder the terms:
-6 + 25c = 4 + 23c

Solving
-6 + 25c = 4 + 23c

Solving for variable 'c'.

Move all terms containing c to the left, all other terms to the right.

Add '-23c' to each side of the equation.
-6 + 25c + -23c = 4 + 23c + -23c

Combine like terms: 25c + -23c = 2c
-6 + 2c = 4 + 23c + -23c

Combine like terms: 23c + -23c = 0
-6 + 2c = 4 + 0
-6 + 2c = 4

Add '6' to each side of the equation.
-6 + 6 + 2c = 4 + 6

Combine like terms: -6 + 6 = 0
0 + 2c = 4 + 6
2c = 4 + 6

Combine like terms: 4 + 6 = 10
2c = 10

Divide each side by '2'.
c = 5

Simplifying
c = 5

Answer from: Quest

the transverse axis of a hyperbola is the line that contains the two vertices and the two focuses.

in this example (hyperbola of equation x^2/2 - y^2/4 = 1), the transverse axis is the x-axis .

step-by-step explanation:

Which of the following statements best describes the transverse axis of a hyperbola?

Answer from: Quest

a. geometricb. g[h] = (3/4)·g[h-1]; g[1] = 384c. g[h] = 384·(3/4)^(h-1)d. g[12] ≈ 16.2183 gallonse. g[12] ≈ 64.8732 quarts

step-by-step explanation:

part a. the differences are decreasing, so we know the sequence is not arithmetic. the ratios are constant (=3/4), so we know the sequence is geometric.

part b. from part a, we know the common ratio is 3/4, so each term is 3/4 of the one before. the initial term is the number of gallons at h=1.

g[h] = (3/4)·g[h-1]

g[1] = 384 . . . . initial value required for a recursive definition

part c. the general formula for a geometric sequence is

a[n] = a[1]·r^(n-1)

we have a[1] = 384, r = 3/4, and the term number given by h. the formula that follows is

g[h] = 384·(3/4)^(h-1)

part d. filling in the value and doing the arithmetic, we have

g[12] = 384·(3/4)^11 ≈ 384·0.042235136 ≈ 16.2183 . . gallons

part e. each gallon is 4 quarts, so

16.2183 gal × 4 qt/gal = 64.8732 qt

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