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Naddika [18.5K]

3 years ago

7

What is the value of k if 2x^3-3x^2+kx-4 is divided by x-3 and gives a remainder of -7?

1 answer:

Alik [6]3 years ago

4 0

By the polynomial remainder theorem, the remainder upon dividing a polynomial $p(x)$ by $x-c$ is the same as the value of $p(c)$ .

Here $p(x)=2x^3-3x^2+kx-4$ and $c=3$ . We have

$p(3)=23+3k=-7\implies k=-10$

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A rectangle is drawn on a coordinate plane. Three vertices of the rectangle are points A(−7,2) , B(3,2) , and D(−7,−2) . Point C

Alexandra [31]

the distance from point b to c is 4 down the y-axis

8 0

3 years ago

Suppose on a certain planet that a rare substance known as Raritanium can be found in some of the rocks. A raritanium-detector i

aleksandrvk [35]

Answer:

0.967 = 96.7% probability the rock sample actually contains raritanium

Step-by-step explanation:

Conditional Probability

We use the conditional probability formula to solve this question. It is

$P(B|A) = \frac{P(A \cap B)}{P(A)}$

In which

P(B|A) is the probability of event B happening, given that A happened.

$P(A \cap B)$ is the probability of both A and B happening.

P(A) is the probability of A happening.

In this question:

Event A: Positive reading

Event B: Contains raritanium

Probability of a positive reading:

98% of 13%(positive when there is raritanium).

0.5% of 100-13 = 87%(false positive, positive when there is no raritanium). So

$P(A) = 0.98*0.13 + 0.005*0.87 = 0.13175$

Positive when there is raritanium:

98% of 13%

$P(A) = 0.98*0.13 = 0.1274$

What is the probability the rock sample actually contains raritanium?

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.1274}{0.13175} = 0.967$

0.967 = 96.7% probability the rock sample actually contains raritanium

7 0

3 years ago

I Need Help Quickly Please And Thank You

forsale [732]

Step 2: it should be P-2x, not +.

Hope this helped :)

7 0

2 years ago

Read 2 more answers

Find the solution set of this inequality|10x+20| ≤10

Pavlova-9 [17]

Answer:

solution is

[-3,-1]

Step-by-step explanation:

we are given

$|10x+20|\leq 10$

Firstly, we will find critical values

so, let's assume it is equal

$|10x+20|= 10$

now, we can break absolute sign

For $|10x+20|= -(10x+20)$ :

$-(10x+20)= 10$

we can solve for x

$-10x-20= 10$

Add both sides by 20

$-10x-20+20= 10+20$

$-10x= 30$

Divide both sides by -10

and we get

$x=-3$

For $|10x+20|= (10x+20)$ :

$(10x+20)= 10$

we can solve for x

$10x+20= 10$

Subtract both sides by 20

$10x+20-20= 10-20$

$10x= -10$

Divide both sides by 10

and we get

$x=-1$

so, critical values are

$x=-3$

$x=-1$

now, we can draw a number line and locate these values

and then we can check inequality on each intervals

For $(-\infty,-3)$ :

We can select any random value from this interval and plug that in inequality

and we get

we can plug x=-5

$|10\times -5+20|\leq 10$

$|-50+20|\leq 10$

$30\leq 10$

so, this is FALSE

For $[-3,-1]$ :

We can select any random value from this interval and plug that in inequality

and we get

we can plug x=-2

$|10\times -2+20|\leq 10$

$|-20+20|\leq 10$

$0\leq 10$

so, this is TRUE

For $(-1,\infty)$ :

We can select any random value from this interval and plug that in inequality

and we get

we can plug x=0

$|10\times 0+20|\leq 10$

$|0+20|\leq 10$

$20\leq 10$

so, this is FALSE

so, solution is

[-3,-1]

7 0

3 years ago

If K is the midpoint of JL, JK=9x-1 and KL=2x+27, find JL.

ratelena [41]

Answer:

The answer is B. 70

Step-by-step explanation:

First, since K is the midpoint of line segment JL, JK must equal KL.

So, set 9x-1 = 2x+27 and solve.

7x=28 so x=4

Then, since they are asking for JL, using x=4, add together JK and KL.

So, 9x-1+2x+27 = 11x+26 = 11(4) + 26 = 44+26 = 70

3 0

3 years ago