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3 years ago

Simplify f(x)=(x+1)(x-3) to standard form

Mathematics

1 answer:

Ivahew [28]3 years ago

4 0

Answer:

x^2-2x-3

Explanation: Use FOIL

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which is the slope of a line that is perpendicular to the line that passes through the points (29, 24) and (3, 4)?

Nataliya [291]

Answer:

26/20

Step-by-step explanation:

y2-y1 divided by x2-x1

4-24= -20

3-29= -26

-20/-26= m

perpendicular means to switch the numbers and change the sign

a perpendicular slope to -20/-26 is 26/20.

3 0

2 years ago

If f(x) = 2x²+1 and g(x) =x²-7find (f-g)(x)

Karolina [17]

(2x^2-x^2)= x^2
(1+7) = 8

X^2 + 8

7 0

2 years ago

Can someone plz help me find the area for this composite figure plzzzz I beg u

Anarel [89]

$S_1 = \frac{a+b}{2}h = \frac{7+9}{2}*5=\frac{18*5}{2}= 9 * 5 = 45~m^2$

$S_\triangle = \frac{ab}{2} = \frac{5*9}{2} = \frac{45}{2} = 22,5~m^2$

$S = S_1 + S_\triangle = 45 + 22,5 = 67,5~m^2$

Answer: 67,5 m².

5 0

3 years ago

Read 2 more answers

In a ballistic pendulum experiment, projectile 1 results in a max height,h, of the pendulum equal to 2.6cm. A second projectile

prohojiy [21]

The second projectile was travelling 1.4 times faster than the first. I am hoping that this answer has satisfied your query and it will be able to help you in your endeavor, and if you would like, feel free to ask another question.

4 0

2 years ago

In a study of wait times at an amusement park, the most popular roller coaster has a mean wait time of 17.4 minutes with a stand

Ann [662]

Answer: A: 0.0031

Step-by-step explanation:

Given : In a study of wait times at an amusement park, the most popular roller coaster has a mean wait time of 17.4 minutes with a standard deviation of 5.2 minutes.

i.e. $\mu=17.4$ and $\sigma=5.2$

We assume that the wait times are normally distributed.

samples size : n= 30

Let x denotes the sample mean wait time.

Then, the probability that the mean wait time is greater than 20 minutes will be :

$P(x>20)=1-P(x\leq20)\\\\=1-P(\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}\leq\dfrac{20-17.4}{\dfrac{5.2}{\sqrt{30}}})\\\\=1-P(z\leq2.74)\ \ [\because\ z=\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}]\\\\=1-0.9969\ \ [\text{ By z table}]\\\\=0.0031$

Hence, the probability that the mean wait time is greater than 20 minutes.= 0.0031

Thus , the correct answer is A: 0.0031 .

5 0

2 years ago