All categories

3 years ago

Use the Remainder Theorem to evaluate f(10) for: f(x) = 2x^3 – 14x^2 - 56x - 40

Mathematics

1 answer:

MAXImum [283]3 years ago

7 0

Step-by-step explanation:

Since we are told that 10 is a zero,

(x−10) must be a factor:

F(x)=2x3−14x2−56x−40

F(x)=(x−10) (2x2+6x+4)

F(x)=2(x−10)(x2+3x+2)

F(x)=2(x−10)(x+2)(x+1)

So the other two zeros are

x = -2 and x= −1

You might be interested in

If you remove the last digit (one’s place) from a 4-digit whole number, the resulting number is a factor of the 4-digit number.

shutvik [7]

Answer:

900

Step-by-step explanation:

We assume that your 4-digit number must be in the range 1000 to 9999. Clearly, any number ending in zero will meet your requirement:

1000/100 = 10

3890/389 = 10

However, the requirement cannot be met when the 1s digit is other than zero.

For some 3-digit number N and some 1s digit x, the 4-digit number will be

4-digit number: 10N+x

Dividing this by N will give ...

(10N+x)/N = 10 remainder x

N will only be a factor of 10N+x when x=0.

So, there are 900 4-digit numbers that meet your requirement. They range from 1000 to 9990.

6 0

3 years ago

On a finance exam, 15 accounting majors had an average grade of 90. On the same exam, 7 marketing majors averaged 85, and 10 fin

lilavasa [31]

Answer:

89.84375

Step-by-step explanation:

Given that on a finance exam, 15 accounting majors had an average grade of 90. On the same exam, 7 marketing majors averaged 85, and 10 finance majors averaged 93.

To find weighted mean.

Here weights can be considered as number of majors in each subject

$Marks x 90 85 93\\Weights 15 7 10\\Product 1350 595 930\\$

Weighted total = 1350+595+930 = 2875

Weighted mean = weighted total/total students

$=\frac{2875}{32} =89.84375$

6 0

3 years ago

2(y+4)−8=24<br> what is y?

Law Incorporation [45]

I got the answer of 12

4 0

2 years ago

Read 2 more answers

Lifetime of $1 Bills The average lifetime of circulated $1 bills is 18 months. A researcher believes that the average lifetime i

OLEGan [10]

<h2>Answer with explanation:</h2>

Let $\mu$ be the population mean lifetime of circulated $1 bills.

By considering the given information , we have :-

$H_0:\mu=18\\\\H_a:\mu\neq18$

Since the alternative hypotheses is two tailed so the test is a two tailed test.

We assume that the lifetime of circulated $1 bills is normally distributed.

Given : Sample size : n=50 , which is greater than 30 .

It means the sample is large so we use z-test.

Sample mean : $\overline{x}=18.8$

Standard deviation : $\sigma=2.8$

Test statistic for population mean :-

$z=\dfrac{\overline{x}-\mu}{\dfrac{\sigma}{\sqrt{n}}}$

$\Rightarrow\ z=\dfrac{18.8-18}{\dfrac{2.8}{\sqrt{50}}}\approx2.02$

The p-value= $2P(z>2.02)=0.0433834$

Since the p-value (0.0433834) is greater than the significance level (0.02) , so we do not reject the null hypothesis.

Hence, we conclude that we do not have enough evidence to support the alternative hypothesis that the average lifetime of a circulated $1 bill differs from 18 months.

6 0

2 years ago

Two out of every 5 muffins are blueberry in one muffin variety pack. how many muffins are blueberry muffins if there are a total

love history [14]

Since the ratio is 2/5 then for 25 (which is five 5s) there will be 2 × 5 which is 10
Answer: there are 10 bluemuffins

3 0

2 years ago