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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?

Mathematics
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
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