What is answer to -14r-19=303

Find the roots of the equation x^2+2x+5=0

stiv31 [10]

Answer:

x = - 1 ± 2i

Step-by-step explanation:

we can use the discriminant b² - 4ac to determine the nature of the roots

• If b² - 4ac > , roots are real and distinct

• If b² - 4ac = 0, roots are real and equal

• If b² - ac < 0, roots are not real

for x² + 2x + 5 = 0

with a = 1, b = 2 and c = 5, then

b² - 4ac = 2² - (4 × 1 × 5 ) = 4 - 20 = - 16

since b² - 4ac < 0 there are 2 complex roots

using the quadratic formula to calculate the roots

x = ( - 2 ± $\sqrt{-16}$ ) / 2

= (- 2 ± 4i ) / 2 = - 1 ± 2i

7 0

4 years ago

Consider two boxes, one containing one black and one white marble, the other, two black and one white marble. A box is selected

valina [46]

Answer:

$\frac{7}{12}$

Step-by-step explanation:

Probability refers to chance of happening of some event.

Conditional probability is the probability of an event A, given that another event B has already occurred.

$B_1,B_2$ denote the two boxes.

In box $B_1$ :

No. of black balls = 1

No. of white balls = 1

In box $B_2$ :

No. of black balls = 2

No. of white balls = 1

Let B, W denote black and white marble.

So, probability that either of the boxes $B_1,B_2$ is chosen is $\frac{1}{2}$

Probability that a black ball is chosen from box $B_1$ = $\frac{1}{2}$

Probability that a black ball is chosen from box $B_2=\frac{2}{3}$

To find:probability that the marble is black

Solution:

Probability that the marble is black = $\frac{1}{2}(\frac{1}{2} )+\frac{1}{2}(\frac{2}{3})=\frac{1}{4}+\frac{1}{3}=\frac{7}{12}$

6 0

3 years ago

Can someone give me some 4th grade math problems

stich3 [128]

1. What is 32×25?

2. What is 856×4?

3. What is 567×40?

Show your work. Sorry if these are too hard.

8 0

4 years ago

Read 2 more answers

Which expression can be used to find the value of expression below -(7/10) + 2/8 + 1/10 - (4/8)

maks197457 [2]

Answer:

(-17)/20

Step-by-step explanation:

Simplify the following:

-7/10 + 2/8 + 1/10 - 4/8

The gcd of 2 and 8 is 2, so 2/8 = (2×1)/(2×4) = 2/2×1/4 = 1/4:

-7/10 + 1/4 + 1/10 - 4/8

The gcd of -4 and 8 is 4, so (-4)/8 = (4 (-1))/(4×2) = 4/4×(-1)/2 = (-1)/2:

-7/10 + 1/4 + 1/10 + -1/2

Put -7/10 + 1/4 + 1/10 - 1/2 over the common denominator 20. -7/10 + 1/4 + 1/10 - 1/2 = (2 (-7))/20 + 5/20 + 2/20 - 10/20:

(2 (-7))/20 + 5/20 + 2/20 - 10/20

2 (-7) = -14:

(-14)/20 + 5/20 + 2/20 - 10/20

-14/20 + 5/20 + 2/20 - 10/20 = (-14 + 5 + 2 - 10)/20:

(-14 + 5 + 2 - 10)/20

-14 + 5 + 2 - 10 = (5 + 2) - (14 + 10):

((5 + 2) - (14 + 10))/20

5 + 2 = 7:

(7 - (14 + 10))/20

| 1 | 4

+ | 1 | 0

| 2 | 4:

(7 - 24)/20

7 - 24 = -(24 - 7):

(-(24 - 7))/20

| 1 | 14

| 2 | 4

- | | 7

| 1 | 7:

Answer: (-17)/20

4 0

3 years ago

Match each interval with its corresponding average rate of change for q(x) = (x + 3)2. 1. -6 ≤ x ≤ -4 1 2. -3 ≤ x ≤ 0 -4 3. -6 ≤

MrMuchimi

The average rate of change of a function f(x) in an interval, a < x < b is given by
$\frac{f(b) - f(a)}{b - a}$

Given q(x) = (x + 3)^2

1.) The average rate of change of q(x) in the interval -6 ≤ x ≤ -4 is given by $\frac{q(-4)-q(-6)}{-4-(-6)} = \frac{(-4+3)^2-(-6+3)^2}{-4+6} = \frac{1-9}{2} = \frac{-8}{2} =-4$

2.) The average rate of change of q(x) in the interval -3 ≤ x ≤ 0 is given by $\frac{q(0)-q(-3)}{0-(-3)} = \frac{(0+3)^2-(-3+3)^2}{0+3} = \frac{9-0}{3} = \frac{9}{3} =3$

3.) The average rate of change of q(x) in the interval -6 ≤ x ≤ -3 is given by $\frac{q(-3)-q(-6)}{-3-(-6)} = \frac{(-3+3)^2-(-6+3)^2}{-3+6} = \frac{0-9}{3} = \frac{-9}{3} =-3$

4.) The average rate of change of q(x) in the interval -3 ≤ x ≤ -2 is given by $\frac{q(-2)-q(-3)}{-2-(-3)} = \frac{(-2+3)^2-(-3+3)^2}{-2+3} = \frac{1-0}{1} = \frac{1}{1} =1$

5.) The average rate of change of q(x) in the interval -4 ≤ x ≤ -3 is given by $\frac{q(-3)-q(-4)}{-3-(-4)} = \frac{(-3+3)^2-(-4+3)^2}{-3+4} = \frac{0-1}{1} = \frac{-1}{1} =-1$

6.) The average rate of change of q(x) in the interval -6 ≤ x ≤ 0 is given by $\frac{q(0)-q(-6)}{0-(-6)} = \frac{(0+3)^2-(-6+3)^2}{0+6} = \frac{9-9}{6} = \frac{0}{6} =0$

3 0

3 years ago

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