A survey about the student government program at a school finds that of 245 students:

If f(x)=5x^2-3x-1f(x)=5x 2 −3x−1, then what is the remainder when f(x)f(x) is divided by x+10x+10?

nata0808 [166]

Answer:

$5x-53+\frac{529}{x+10}$

Step-by-step explanation:

8 0

1 year ago

Find the sum

lakkis [162]

Answer:

Numerator = 2(b^2+a^2) or equivalently 2b^2+2a^2

Denominator = (b+a)^2*(b-a), or equivalently b^3+ab^2-a^2b0-a^3

Step-by-step explanation:

Let

S = 2b/(b+a)^2 + 2a/(b^2-a^2) factor denominator

= 2b/(b+a)^2 + 2a/((b+a)(b-a)) factor denominators

= 1/(b+a) ( 2b/(b+a) + 2a/(b-a)) find common denominator

= 1/(b+a) ((2b*(b-a) + 2a*(b+a))/((b+a)(b-a)) expand

= 1/(b+a)(2b^2-2ab+2ab+2a^2)/((b+a)(b-a)) simplify & factor

= 2/(b+a)(b^2+a^2)/((b+a)(b-a)) simplify & rearrange

= 2(b^2+a^2)/((b+a)^2(b-a))

Numerator = 2(b^2+a^2) or equivalently 2b^2+2a^2

Denominator = (b+a)^2*(b-a), or equivalently b^3+ab^2-a^2b0-a^3

4 0

2 years ago

Use the Pythagorean Theorem to determine which of the following give the measures of the legs and hypotenuse of a right triangle

kakasveta [241]

The Pythagorean theorem states that the square of the hypotenuse is equal to the sum of the squares of the legs:

$c^2=a^2+b^2.$

The hypotenuse is the side with the greatest length.

Check all options:

1. c=5, a=3, b=4:

$5^2=25,\\3^2+4^2=9+16=25,\\5^2=3^2+4^2$ true.

2. c=14, a=4, b=11:

$14^2=196,\\4^2+11^2=16+121=137,\\14^2\neq 4^2+11^2$ false.

3. c=17, a=9, b=14:

$17^2=289,\\9^2+14^2=81+196=277,\\17^2\neq 9^2+14^2$ false.

4. c=16, a=8, b=14:

$16^2=256,\\8^2+14^2=64+196=260,\\16^2\neq 8^2+14^2$ false.

5. c=17, a=8, b=15:

$17^2=289,\\8^2+15^2=64+225=289,\\17^2=8^2+15^2$ true.

Answer: correct options are A and E.

4 0

2 years ago

Read 2 more answers

30 yards = how many feet?

mrs_skeptik [129]

30 yards = 90 ft

3.5 pounds = 56 oz

3.5 cm = 18.8976 in

12ft and 10 inches (12.833333333333334 = 154 im

8 0

3 years ago

Read 2 more answers

Use mathematical induction to prove the statement is true for all positive integers n, or show why it is false:

kondaur [170]

$\text{Proof by induction:}$
$\text{Test that the statement holds or n = 1}$

$LHS = (3 - 2)^{2} = 1$
$RHS = \frac{6 - 4}{2} = \frac{2}{2} = 1 = LHS$
$\text{Thus, the statement holds for the base case.}$

$\text{Assume the statement holds for some arbitrary term, n= k}$
$1^{2} + 4^{2} + 7^{2} + ... + (3k - 2)^{2} = \frac{k(6k^{2} - 3k - 1)}{2}$

$\text{Prove it is true for n = k + 1}$
$RTP: 1^{2} + 4^{2} + 7^{2} + ... + [3(k + 1) - 2]^{2} = \frac{(k + 1)[6(k + 1)^{2} - 3(k + 1) - 1]}{2} = \frac{(k + 1)[6k^{2} + 9k + 2]}{2}$

$LHS = \underbrace{1^{2} + 4^{2} + 7^{2} + ... + (3k - 2)^{2}}_{\frac{k(6k^{2} - 3k - 1)}{2}} + [3(k + 1) - 2]^{2}$
$= \frac{k(6k^{2} - 3k - 1)}{2} + [3(k + 1) - 2]^{2}$
$= \frac{k(6k^{2} - 3k - 1) + 2[3(k + 1) - 2]^{2}}{2}$
$= \frac{k(6k^{2} - 3k - 1) + 2(3k + 1)^{2}}{2}$
$= \frac{k(6k^{2} - 3k - 1) + 18k^{2} + 12k + 2}{2}$
$= \frac{k(6k^{2} - 3k - 1 + 18k + 12) + 2}{2}$
$= \frac{k(6k^{2} + 15k + 11) + 2}{}$
$= \frac{(k + 1)[6k^{2} + 9k + 2]}{2}$
$= \frac{(k + 1)[6(k + 1)^{2} - 3(k + 1) - 1]}{2}$
$= RHS$

Since it is true for n = 1, n = k, and n = k + 1, by the principles of mathematical induction, it is true for all positive values of n.

3 0

2 years ago