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KATRIN_1 [288]
3 years ago
8

If f(x)=4x^5+5x^4+1, then what is the remainder when f(x) is divided by x-2

Mathematics
1 answer:
monitta3 years ago
4 0

Answer:

Remainder 209

Step-by-step explanation:

f(x)= 4x⁵+5x⁴+1  

x-2=0⇒x=2

Putting x=2 in the above equation

f(x)= 4(2)⁵ +5(2)⁴+1

f(x)= 4(32) +5(16) +1

f(x)= 128 +80+1

f(x)= 209

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Every rhombus is a square? true it false.​
murzikaleks [220]

Answer:

false

Step-by-step explanation:

4 0
2 years ago
HELP ME! @all smart people
larisa86 [58]
The first one is 5.3 you multiply 4x4 which is 16 then divide by 3 which then gives you the answer
6 0
3 years ago
Read 2 more answers
Applying Rate of change
olga nikolaevna [1]
\bf \begin{array}{ccllll}
minutes(x)&Mbs\ left(y)\\
-----&-----\\
2&38.4\\
8&9.6
\end{array}\\\\
-----------------------------\\\\
slope = {{ m}}= \cfrac{rise}{run} \implies 
\cfrac{{{ y_2}}-{{ y_1}}}{{{ x_2}}-{{ x_1}}}\impliedby \textit{rate of change}
\\\\\\
m=\cfrac{9.6-38.4}{8-2}\implies m=-\cfrac{24}{5}\iff m= -4.8

now, notice, the rate of change for downloading is negative, because, "y" is decreasing,  namely the Mbs to be downloaded, are less and less and less as the minutes go by, because the file is almost fully downloaded

so is -4.8

now... at 8minutes, there are 9.6Mbs to download
bear in mind that 9.6 is just 4.8 * 2
that simply means, another minute, another 4.8 Mbs, and another minute and another 4.8 Mbs and the file is done

so, the file downloaded really in 10minutes

now, we know the rate is -4.8 or -24/5,   let us nevermind the sign for now

since we know the file is downloading at 24Mbs in 5minutes, a rate of 24/5

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7 0
3 years ago
Is there any systematic tendency for part-time college faculty to hold their students to different standards than do full-time f
alexandr1967 [171]

Answer:

H0 : μ1 - μ2 = 0

H1 : μ1 - μ2 ≠ 0

-1. 34

0.1837

Step-by-step explanation:

Full time :

n1 = 125

x1 = 2.7386

s1 = 0.65342

Part time :

n2 = 88

x2 = 2.8439

s2 = 0.49241

H0 : μ1 - μ2 = 0

H1 : μ1 - μ2 ≠ 0

Test statistic :

The test statistic :

(x1 - x2) / sqrt[(s1²/n1 + s2²/n2)]

(2.7386 - 2.8439) / sqrt[(0.65342²/125 + 0.49241²/88)]

−0.1053 / sqrt(0.0034156615712 + 0.0027553)

-0.1053 /0.0785554

= - 1.34

Test statistic = - 1.34

The Pvalue :

Using df = smaller n - 1 = 88 - 1 = 87

Pvalue from test statistic score ;

Pvalue = 0.1837

Pvalue > α ; We fail to reject the null and conclude that the GPA does not differ.

At α = 0.01 ; the result is insignificant

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The answer to that one is the first box definitely 

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