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prisoha [69]
3 years ago
10

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

Mathematics
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
HURRYYYYY
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
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