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3 years ago

Use the quadratic formula to solve 2x^2 = 5x + 6

Mathematics

2 answers:

Vikentia [17]3 years ago

8 0

16 answer .............

aleksklad [387]3 years ago

8 0

Answer:

Step-by-step explanation:

Let's go:

$2x^2 = 5x + 6 \\\ 2x^2 - 5x - 6 = 0 \\\ \Delta = (- 5)^2 - 4.2.(- 6) \\\ \Delta = 25 + 48 \\\ \Delta = 73 \\\ x = \frac{5 \pm \sqrt{73}}{2.2} \\\\\ x = \frac{5 \pm \sqrt{73}}{4}$

I hope I helped you.

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Answer:

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Step-by-step explanation:

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3 years ago

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3 years ago

Graph the line using a point and a slope. Write the equation of each line: a line that contains point (0, −3) and perpendicular

Sav [38]

Answer:y = 2x − -3 find the y and x and then identify if its zero slope or undefined.

Step-by-step explanation:

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3 years ago

Use the fact that the mean of a geometric distribution is μ= 1 p and the variance is σ2= q p2. A daily number lottery chooses th

butalik [34]

Answer:

a). The mean = 1000

The variance = 999,000

The standard deviation = 999.4999

b). 1000 times , loss

Step-by-step explanation:

The mean of geometric distribution is given as , $\mu = \frac{1}{p}$

And the variance is given by, $\sigma ^2=\frac{q}{p^2}$

Given : $p=\frac{1}{1000}$

= 0.001

The formulae of mean and variance are :

$\mu = \frac{1}{p}$

$\sigma ^2=\frac{q}{p^2}$

$\sigma ^2=\frac{1-p}{p^2}$

a). Mean = $\mu = \frac{1}{p}$

= $\mu = \frac{1}{0.001}$

= 1000

Variance = $\sigma ^2=\frac{1-p}{p^2}$

= $\sigma ^2=\frac{1-0.001}{0.001^2}$

= 999,000

The standard deviation is determined by the root of the variance.

$\sigma = \sqrt{\sigma^2}$

= $\sqrt{999,000}$ = 999.4999

b). We expect to have play lottery 1000 times to win, because the mean in part (a) is 1000.

When we win the profit is 500 - 1 = 499

When we lose, the profit is -1

Expected value of the mean μ is the summation of a product of each of the possibility x with the probability P(x).

$\mu=\Sigma\ x\ P(x)= 499 \times 0.001+(-1) \times (1-0.001)$

= $ 0.50

Since the answer is negative, we are expected to make a loss.

4 0

3 years ago

If a ball is thrown straight up into the air with an initial velocity of 70 ft/s, its height in feet after t seconds is given by

Vitek1552 [10]

Answer:

a) Average velocity at 0.1 s is 696 ft/s.

b) Average velocity at 0.01 s is 7536 ft/s.

c) Average velocity at 0.001 s is 75936 ft/s.

Step-by-step explanation:

Given : If a ball is thrown straight up into the air with an initial velocity of 70 ft/s, its height in feet after t seconds is given by $y = 70t-16t^2$ .

To find : The average velocity for the time period beginning when t = 2 and lasting. a. 0.1 s. , b. 0.01 s. , c. 0.001 s.

Solution :

a) The average velocity for the time period beginning when t = 2 and lasting 0.1 s.

$(\text{Average velocity})_{0.1\ s}=\frac{\text{Change in height}}{0.1}$

$(\text{Average velocity})_{0.1\ s}=\frac{h_{2.1}-h_2}{0.1}$

$(\text{Average velocity})_{0.1\ s}=\frac{(70(2.1)-16(2.1)^2)-(70(0.1)-16(0.1)^2)}{0.1}$

$(\text{Average velocity})_{0.1\ s}=\frac{69.6}{0.1}$

$(\text{Average velocity})_{0.1\ s}=696\ ft/s$

b) The average velocity for the time period beginning when t = 2 and lasting 0.01 s.

$(\text{Average velocity})_{0.01\ s}=\frac{\text{Change in height}}{0.01}$

$(\text{Average velocity})_{0.01\ s}=\frac{h_{2.01}-h_2}{0.01}$

$(\text{Average velocity})_{0.01\ s}=\frac{(70(2.01)-16(2.01)^2)-(70(0.01)-16(0.01)^2)}{0.01}$

$(\text{Average velocity})_{0.01\ s}=\frac{75.36}{0.01}$

$(\text{Average velocity})_{0.01\ s}=7536\ ft/s$

c) The average velocity for the time period beginning when t = 2 and lasting 0.001 s.

$(\text{Average velocity})_{0.001\ s}=\frac{\text{Change in height}}{0.001}$

$(\text{Average velocity})_{0.001\ s}=\frac{h_{2.001}-h_2}{0.001}$

$(\text{Average velocity})_{0.001\ s}=\frac{(70(2.001)-16(2.001)^2)-(70(0.001)-16(0.001)^2)}{0.001}$

$(\text{Average velocity})_{0.001\ s}=\frac{75.936}{0.001}$

$(\text{Average velocity})_{0.001\ s}=75936\ ft/s$

5 0

4 years ago