All categories

2 years ago

16x2 + 10x – 27 = -6x + 5 What are the solutions to this equation ? X x=

Mathematics

1 answer:

bonufazy [111]2 years ago

5 0

Answer:

x=-2 x=1

Step-by-step explanation:

16x2 + 10x – 27 = -6x + 5

Add 6x to each side

16x^2+6x + 10x – 27 = -6x+6x + 5

16x^2 +16x -27 = 5

Subtract 5 from each side

16x^2 + 16x – 27-5 = 5 - 5

16x^2 +16x -32 = 0

Factor out 16

16 (x^2 +x-2)=0

Factor

16 (x+2) (x-1) =0

Using the zero product property

(x+2) =0 x-1=0

x=-2 x=1

You might be interested in

What is the slope of the line through (-3,0) and (-4,-2)?

rodikova [14]

AnswerY=-10/3x+11:

Step-by-step explanation:

M=0--2/-3-4= -10/3

Y-1=-10/3(x-3)

-10/3x+10+1

Y=-10/3x+11

4 0

2 years ago

A skydiver in a computer game free-falls from a height of 3000 meters at a

mylen [45]

H= -55t + 3000. Hope this helps!

8 0

1 year ago

Make a table of ordered pairs for the equation. y= 1 2 x−3<br><br> PLEASE HELP !!

Hunter-Best [27]

All you need to do is plug in some x values and find the correspoding y value
when x=0, y=-3
when x=1, y=9
when x=2, y=21
when x=-1, y=-15

6 0

3 years ago

Average precipitation for the first 7 months of the year, the average precipitation in toledo, ohio, is 19.32 inches. if the ave

Colt1911 [192]

Part A:

The probability that a normally distributed data with a mean, μ and standard deviation, σ is greater than a given value, a is given by:

$P(x\ \textgreater \ a)=1-P(x\ \textless \ a)=1-P\left(z\ \textless \ \frac{a-\mu}{\sigma}\right)$

Given that the average precipitation in Toledo, Ohio for the past 7 months is 19.32 inches with a standard deviation of 2.44 inches, the probability that <span>a randomly selected year will have precipitation greater than 18 inches for the first 7 months is given by:

$P(x\ \textgreater \ 18)=1-P(x\ \textless \ 18) \\ \\ =1-P\left(z\ \textless \ \frac{18-19.32}{2.44}\right) \\ \\ =1-P(z\ \textless \ -0.5410) \\ \\ =1-0.29426=\bold{0.7057}$

Part B:

</span>The probability that an n randomly selected samples of a normally distributed data with a mean, μ and standard deviation, σ is greater than a given value, a is given by:

$P(x\ \textgreater \ a)=1-P(x\ \textless \ a)=1-P\left(z\ \textless \ \frac{a-\mu}{\frac{\sigma}{\sqrt{n}}}\right)$

Given that the average precipitation in Toledo, Ohio for the past 7 months is 19.32 inches with a standard deviation of 2.44 inches, the probability that <span>5 randomly selected years will have precipitation greater than 18 inches for the first 7 months is given by:

</span> $P(x\ \textgreater \ 18)=1-P(x\ \textless \ 18) \\ \\ =1-P\left(z\ \textless \ \frac{18-19.32}{\frac{2.44}{\sqrt{5}}}\right) \\ \\ =1-P(z\ \textless \ -1.210) \\ \\ =1-0.1132=\bold{0.8868}$

7 0

2 years ago

The number of bagels sold daily for two bakeries is shown in the table.

bonufazy [111]

The correct statement based on the data is: A. Mean, because there are no outliers to affect the data.

<h3>What are Outliers?</h3>

Outliers are data point that appears extreme from other data values of a data distribution. When there is an outlier in a data distribution, the median is a better measure of center, while the mean is best when outliers are absent.

From the data distribution for both bakeries, no data seem extreme from the rest of the data, therefore, the mean is a better measure of center.

The answer is: A. Mean, because there are no outliers to affect the data.

Learn more about outliers on:

brainly.com/question/4520746

#SPJ1

4 0

1 year ago