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3 years ago

What is the remainder of (x 3 + 5x 2 − 32x − 7) ÷ (x − 4)?

1 answer:

8 0

(x^3+5x^2-32x-7)/(x-4)

x^2 remainder 9x^2-32x-7

9x remainder 4x-7

4 remainder 9

(x-4)(x^2+9x+4)+9/(x-4)

The remainder is 9/(x-4)

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2/9 x s = 1/2 I don’t know

WITCHER [35]

Answer:

$\frac{9}{4}$

Step-by-step explanation:

$\frac{2}{9} \times s = \frac{1}{2} \\ \\ \frac{2s}{9} = \frac{1}{2} \\ cross \: multiply \\ 9 \times 1 = 2s \times 2 \\ 9 = 4s \\ divivde \: both \: sides \: by \: 4 \\ \frac{9}{4} = \frac{4s}{4} \\ s = \frac{9}{4}$

hope you understand

3 0

2 years ago

Please help me now plz I will mark you brainliest

krok68 [10]

Answer:

y = -5/4x -4

8 0

2 years ago

The sum of a number and 3 times its reciprocal is -4.Find the number

Butoxors [25]

Answer:

x = -1 or x = -3

Step-by-step explanation:

x + 3/x = -4

x² + 3 = -4x

x² + 4x + 3 = 0

(x + 1) (x + 3) = 0

x = -1 or x = -3

check: (-1) + (3/-1) = -1 - 3 = -4

(-3) + (3/(-3)) = -3 -1 = 14

6 0

3 years ago

100 points link down below! no off sight links please no I don't knows just a thing that state all the answers Ill give you anot

dolphi86 [110]

Answer:

All of the questions?

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

Assume the readings on thermometers are normally distributed with a mean of 0 degrees C and a standard deviation of 1.00 degrees

lozanna [386]

Answer: 0.0035

Step-by-step explanation:

Given : The readings on thermometers are normally distributed with a mean of 0 degrees C and a standard deviation of 1.00 degrees C.

i.e. $\mu=0$ and $\sigma= 1$

Let x denotes the readings on thermometers.

Then, the probability that a randomly selected thermometer reads greater than 2.17 will be :_

$P(X>2.7)=1-P(\xleq2.7)\\\\=1-P(\dfrac{x-\mu}{\sigma}\leq\dfrac{2.7-0}{1})\\\\=1-P(z\leq2.7)\ \ [\because\ z=\dfrac{x-\mu}{\sigma}]\\\\=1-0.9965\ \ [\text{By z-table}]\ \\\\=0.0035$

Hence, the probability that a randomly selected thermometer reads greater than 2.17 = 0.0035

The required region is attached below .

8 0

3 years ago