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3 years ago

10

Find the zeros of polynomial function and solve polynomials equations. f(x)=81x^4-16

1 answer:

irga5000 [103]3 years ago

3 0

X = 2/3, -2/3

“Solve” : (9x^2+4)•(3x+2)•(3x-2)

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Darla is taking a survey at her school. She finds that 20 out of 100 students have fish as pets. This means that her experimenta

Arisa [49]

Answer:

The second option

Step-by-step explanation:

This is because the more samples she has, the closer her experimental probability will be to the real value.

6 0

3 years ago

joe spends $8 on lunch and $6.50 on dry cleaning. He also buys 2 shiets that cost the same amount. joe spendes a total of $52. w

ANEK [815]

X=amout the shirts cost
52=8+6.50+2x
add 8 and 6.5
52=14.5+2x
subtract 14.5 from 52
37.5=2x
divide 37.5 by 2
x=18.75
It cost $18.75 for each shirt.

4 0

2 years ago

What is the value of 6 times 7 - 3^2 times 9 + 4^3<br> show work..THX

Kitty [74]

Step-by-step explanation:

6 times7-3^2 times 9+4^3=

(6*(7-3^2))*(9+4^3) =

(6(7-9)) *(9+64) =

(6*-2) (73) =

-12*73

3 0

3 years ago

Troyanec [42]

Answer:

B is correct.

Step-by-step explanation:

Quadratic Expression (Standard Form) is:

$\displaystyle \large{a {x}^{2} + bx + c}$

To say it simply, the expression must have 2 as the highest degree.

This means that if there are any higher or lower degrees than 2 then they are not quadratic expression.

A choice is not quadratic expression because it has 1 as the highest degree.

B choice is correct because 2 is the highest degree.

C choice is wrong because 3 is the highest degree.

D choice is wrong because it is not a polynomial.

Therefore, B is correct.

7 0

2 years ago

Three consecutive odd integers have a sum of 27. Find the integers.

jek_recluse [69]

$\sf \bf {\boxed {\mathbb {GIVEN:}}}$

Sum of three consecutive odd integers = $27$

$\sf \bf {\boxed {\mathbb {TO\:FIND:}}}$

The values of the three integers.

$\sf \bf {\boxed {\mathbb {SOLUTION:}}}$

$\sf\purple{The\:three\:consecutive \:odd\:integers\:are\:7,\:9\:and\:11.}$

$\sf \bf {\boxed {\mathbb {STEP-BY-STEP\:\:EXPLANATION:}}}$

Let us assume the three consecutive odd integers to be $x$ , $(x+2)$ and $(x+4)$ .

As per the condition, we have

$Sum \: \: of \: \: the \: \: three \: \: consecutive \: \: odd \: \: integers = 27$

$➺ \: x + (x + 2) + (x + 4) = 27$

$➺ \: x + x + 2 + x + 4 = 27$

Now, collect the like terms.

$➺ \: (x + x + x) + (2 + 4) = 27$

$➺ \: 3x + 6 = 27$

$➺ \: 3x = 27 - 6$

$➺ \: 3x = 21$

$➺ \: x = \frac{21}{3} \\$

$➺ \: x = 7$

Therefore, the three consecutive odd integers whose sum is $27$ are $\boxed{ 7 }$ , $\boxed{ 9 }$ and $\boxed{ 11 }$ respectively.

$\sf \bf {\boxed {\mathbb {TO\:VERIFY :}}}$

$⇢ 7 + 9 + 11 = 27$

$⇢ 27 = 27$

⇢ L. H. S. = R. H. S.

$\sf\blue{Hence\:verified.}$

$\red{\large\qquad \qquad \underline{ \pmb{{ \mathbb{ \maltese \: \: Mystique35ヅ}}}}}$

4 0

3 years ago