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3 years ago

Solve for x. + 6 = 10 4 1 16 64

Mathematics

2 answers:

RUDIKE [14]3 years ago

5 0

ANSWER: x = 16

SOLUTION:

1) Multiply both sides of the equation by 4:

x/4 + 6 = 10 ------> x + 24 =40

2) Move the constant (+ 24) to the right-hand side and change its sign:

x = 40 - 24

3) Subtract the numbers:

x + 16 (simplified)

Alecsey [184]3 years ago

4 0

Answer: First, turn the mixed fraction 16 1/4 into a improper fraction. It would become 65/4

x+65/4=64

Next, I would multiply the entire equation by 4 to get rid of the fraction

It would become 4x+65=256

Now solve for X

4x=191

x=191/4

x=47.75

Step-by-step explanation:

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If you place a 33-foot ladder against the top of a building and the bottom of the ladder is 26 feet from the bottom of the build

Firdavs [7]

Answer:

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2 years ago

A sequence of 7 numbers starts with 8. The second number in the sequence is unknown and may be positive or negative. The third t

scoray [572]

Answer:

Explanation:

$a_1=8,a_7=0$

Since the third term is the sum of the two previous terms

$a_3=a_2+a_1$

$a_7=a_6+a_5$

$a_7=a_5+a_4+a_4+a_3$

Continuing in like manner

$a_7=5a_3+3a_2$

Since $a_3=a_2+a_1$

$a_7=5(a_2+a_1+3a_2$

$a_7=5a_2+5a_1+3a_2$

$a_7=8a_2+5a_1$

Recall: $a_1=8,a_7=0$

$0=8a_2+5*8$

$8a_2=-40$

$a_2=-5$

The sequence is therefore:

8,-5,3,-2,1,-1,0

The sum of the seven numbers is 4.

8 0

3 years ago

What is the difference? StartFraction 2 x 5 Over x squared minus 3 x EndFraction minus StartFraction 3 x 5 Over x cubed minus 9

Fiesta28 [93]

The expression which is equivalent to the difference of the given algebraic fraction is,

$\dfr\dfrac{(x+5)(x+2)}{x(x-3)(x+3)}$

<h3>What is the difference? </h3>

Difference is the mathematical operation in which the one number is subtract from another number.

To find the difference of two fraction numbers, we find the least common factor of denominator and write the all numbers as one fraction.

The given expression in the problem is,

$\dfrac{2x+5}{x^2-3x}-\dfrac{3x+5}{x^3-9x}-\dfrac{x+1}{x^2-9}$

Factorize the denominator of each term as,

$\dfrac{2x+5}{x(x-3)}-\dfrac{3x+5}{x(x-3)(x+3)}-\dfrac{x+1}{(x-3)(x+3)}$

Here, the least common factor of the denominators is x(x-3)(x+3), therefore, the expression can be written as,

$\dfrac{(x+3)(2x+5)-(3x+5)-(x)(x+1)}{x(x-3)(x+3)}\\\dfr\dfrac{2x^2+5x+6x+15-3x-5-x^2-x}{x(x-3)(x+3)}\\\dfr\dfrac{x^2+7x+10}{x(x-3)(x+3)}\\\dfr\dfrac{(x+5)(x+2)}{x(x-3)(x+3)}$