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3 years ago

The sum of the squares of two consecutive integers is 145. Find the two integers.

Mathematics

1 answer:

worty [1.4K]3 years ago

6 0

Answer: The two integers are 8 and 9.

Step-by-step explanation: Let n represent the number.

n^2 + (n + 1)^2 = 145

By foiling, this can be rearranged to:

n^2 + n^2 + 2n + 1 = 145

Subtract 145 from both sides:

2n^2 + 2n - 144 = 0

Divide both sides by 2:

n^2 + n - 72 = 0

Factor the polynomial:

(n + 9)(n - 8) = 0

Solve for n:

n = -9, n = 8

Since they need to be consecutive, make the 9 positive. It will be positive either way because it is being squared.

Now, n = 9 and n = 8.

These are your two integers. Hope this helps!

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CAN SOMEONE PLEASE SHOW WORK FOR THESE PROBLEMS WILL GIVE BAINLIEST!!!!!!!!!EMERGENCY

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Answer:

Part 1) $x\geq10$

Part 2) $m\leq -9$

Part 3) $p\geq 5$

Part 4) $x$

Part 5) $b$

Part 6) $n$

Part 7) $n$

Part 8) $r\leq 4$

Part 9) $x\geq 7$

Part 10) $p\leq 0$

Part 11) $x$

Part 12) $a > 24$

Step-by-step explanation:

Part 1) $2x+4\geq24$

Subtract 4 both sides

$2x\geq24-4$

$2x\geq20$

Divide by 2 both sides

$x\geq10$

the solution is the interval ------> [10,∞)

The solution is the shaded area to the right of the solid line at number 10 (closed circle).

see the attached figure

Part 2) $\frac{m}{3}-3\leq -6$

Adds 3 both sides

$\frac{m}{3}\leq -6+3$

$\frac{m}{3}\leq -3$

Multiply by 3 both sides

$m\leq -9$

the solution is the interval ------> (-∞,-9]

The solution is the shaded area to the left of the solid line at number -9 (closed circle).

see the attached figure

Part 3) $-3(p+1)\leq -18$

applying the distributive property left side

$-3p-3\leq -18$

adds 3 both sides

$-3p\leq -18+3$

$-3p\leq -15$

Multiply by -1 both sides

$3p\geq 15$

Divide by 3 both sides

$p\geq 5$

the solution is the interval ------> [5,∞)

The solution is the shaded area to the right of the solid line at number 5 (closed circle).

see the attached figure

Part 4) $-4(-4+x)>56$

applying the distributive property left side

$16-4x>56$

Subtract 16 both sides

$-4x>56-16$

$-4x>40$

Multiply by -1 both sides

$4x$

Divide by 4 both sides

$x$

the solution is the interval ------> (-∞,-10)

The solution is the shaded area to the left of the dashed line at number -10 (open circle).

see the attached figure

Part 5) $-b-2>8$

adds 2 both sides

$-b>8+2$

$-b>10$

Multiply by -1 both sides

$b$

the solution is the interval ------> (-∞,-10)

The solution is the shaded area to the left of the dashed line at number -10 (open circle).

Part 6) $-4(3+n)>-32$

applying the distributive property left side

$-12-4n>-32$

adds 12 both sides

$-4n>-32+12$

$-4n>-20$

multiply by -1 both sides

$4n$

divide by 4 both sides

$n$

the solution is the interval ------> (-∞,5)

The solution is the shaded area to the left of the dashed line at number 5 (open circle).

see the attached figure

Part 7) $4+\frac{n}{3}$

Subtract 4 both sides

$\frac{n}{3}$

Multiply by 3 both sides

$n$

the solution is the interval ------> (-∞,6)

The solution is the shaded area to the left of the dashed line at number 6 (open circle).

see the attached figure

Part 8) $-3(r-4)\geq 0$

applying the distributive property left side

$-3r+12\geq 0$

subtract 12 both sides

$-3r\geq -12$

divide by -1 both sides

$3r\leq 12$

divide by 3 both sides

$r\leq 4$

the solution is the interval ------> (-∞,4]

The solution is the shaded area to the left of the solid line at number 4 (closed circle).

see the attached figure

Part 9) $-7x-7\leq -56$

Adds 7 both sides

$-7x\leq -56+7$

$-7x\leq -49$

Multiply by -1 both sides

$7x\geq 49$

Divide by 7 both sides

$x\geq 7$

the solution is the interval ------> [7,∞)

The solution is the shaded area to the right of the solid line at number 7 (closed circle).

see the attached figure

Part 10) $-3(p-7)\geq 21$

applying the distributive property left side

$-3p+21\geq 21$

subtract 21 both sides

$-3p\geq 21-21$

$-3p\geq 0$

Multiply by -1 both sides

$3p\leq 0$

$p\leq 0$

the solution is the interval ------> (-∞,0]

The solution is the shaded area to the left of the solid line at number 0 (closed circle).

see the attached figure

Part 11) $-11x-4> -15$

Adds 4 both sides

$-11x> -15+4$

$-11x> -11$

Multiply by -1 both sides

$11x$

Divide by 11 both sides

$x$

the solution is the interval ------> (-∞,1)

The solution is the shaded area to the left of the dashed line at number 1 (open circle).

see the attached figure

Part 12) $\frac{-9+a}{15}>1$

Multiply by 15 both sides

$-9+a > 15$

Adds 9 both sides

$a > 15+9$

$a > 24$

the solution is the interval ------> (24,∞)

The solution is the shaded area to the right of the dashed line at number 24 (open circle).

see the attached figure

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