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3 years ago

What is the discontinuity and the zero of the function 2x^2+5x-12/x+4?

Mathematics

1 answer:

yan [13]3 years ago

5 0

Discontinuity: x=-4
Zero: x=3/2

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Simplify 2 - 5(x-2) =

ycow [4]

Answer:

12-5x

Step-by-step explanation:

2-5(x-2)

2-5x+10

12-5x

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3 years ago

Rachel weighed her two cats, Tiger and Mittens. Tiger weighed 10.91 pounds, and Mittens weighed 178.08 ounces. Which cat is heav

Molodets [167]

Mittens is heavier. Hope this helps!

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3 years ago

Read 2 more answers

Suppose a square ABCD has a side length of 1 unit. The arcs BD and AC are circular arcs with centers at A and D, respectively. F

Likurg_2 [28]

Answer:

$1-\frac{\sqrt{3} }{4}-\frac{\pi}{6}$

Step-by-step explanation:

The missing figure is shown in the attachment

The area of the shaded region = Area of Square - (Area of sector AOB + Area of equilateral triangle BOC + Area of sector COD)

Area of Sector AOB=Area of Sector COD= $\frac{30}{360}*\pi*1^2=\frac{\pi}{12}$

Area of equilateral triangle = $\frac{1}{2}*r*\frac{\sqrt{3} }{2}*r= \frac{1}{2}*1*\frac{\sqrt{3} }{2}*=\frac{\sqrt{3} }{4}$

Area of shade region = $1^2-\frac{\sqrt{3} }{4}-\frac{\pi}{12}*2$

$1-\frac{\sqrt{3} }{4}-\frac{\pi}{6}$

6 0

3 years ago

If 10 pounds of potatos cost 6.55 how much would 1 pound cost

kap26 [50]

Divide ten by how much it cost 10/6.55=1.52

6 0

3 years ago

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An advertisement for a popular weight-loss clinic suggests that participants in its new diet program lose, on average, more than

Sauron [17]

The test statistic for the hypothesis would be 1.413.

Given that the participants in its new diet program lose, on average, more than 13 pounds and the mean weight loss of these participants as 13.8 pounds with a standard deviation of 3.1 pounds.

The objective is to text the advertisement's claim that participants in new diet program lose weight, on average, more than 13 pounds.

Hypothesis:

Null hypothesis:H₀:μ=13

Alternative hypothesis:Hₐ:μ>13

Here, μ be the mean weight loss of all participants.

n=30,

Degree of freedom n-1=30-1=29

$\bar{x}$ =13.8 and s=3.1

To test the null hypothesis H₀, the value of test static would be calculated as follows:

$\begin{aligned}t&=\frac{\bar{x}-\mu}{\frac{s}{\sqrt{n}}}\\ t_{29}&=\frac{13.8- 13}{\frac{3.1}{\sqrt{30}}}\\ &=\frac{0.8}{0.566}\\ &=1.413\end$

Hence, the value of the test static for the hypothesis with the mean weight loss of these participants as 13.8 pounds with a standard deviation of 3.1 pounds is 1.413.

Learn more about hypothesis from here brainly.com/question/14783359

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6 0

2 years ago