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Stella [2.4K]
3 years ago
12

Whats the reason for 8x-5-2x=2x+1-2x

Mathematics
1 answer:
Irina-Kira [14]3 years ago
4 0
Your answer is X= -1
you have to move terms first to get this
8x-5=2x+1
then collect like terms and calculate sum
8x-2x=1+5
lastly you divide both sides by 6
X=1
HOPE THIS HELPS! :)
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Please help. Which polynomial represents the quotient for the division problem shown?
NARA [144]

Answer:

B

Step-by-step explanation:

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8 0
3 years ago
Read 2 more answers
Please answer correctly !!!!!!! Will mark brainliest !!!!!!!!!!!!!!
Mamont248 [21]

Answer:

y= -x^2+5

Step-by-step explanation:

It normally expontionally goes up, so across the x axis is down so it is negative, and is is all scales up, so the start is 5 up from the origin on the y axis, so it would be y=-x+5

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7 0
2 years ago
You plant a rectangular rose garden along the side of your garage. You enclose 3 sides of the garden with 40 feet of fencing. Th
solmaris [256]

Let's assume

length of rectangle =L

width of rectangle =W

You enclose 3 sides of the garden with 40 feet of fencing

so, we get

L+2W=40

now, we can solve for L

L=40-2W

we know that

area of rectangle is

A=L*W

100=L*W

now, we can plug

100=(40-2W)*W

now, we can solve for W

-2W^2+40W-100=0

we can use quadratic formula

W=\frac{-40+\sqrt{40^2-4\left(-2\right)\left(-100\right)}}{2\left(-2\right)}

W=\frac{-40-\sqrt{40^2-4\left(-2\right)\left(-100\right)}}{2\left(-2\right)}

we can take anyone value ..because both are giving positive value

first dimensions:

W=2.929

now, we can find L

L=40-2*2.929

L=34.142

so, length is 34.142feet

width is 2.929 feet

Second dimensions:

W=17.071

now, we can find L

L=40-2*17.071

L=5.858

so, length is 5.858feet

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6 0
3 years ago
Data is collected to compare two different types of batteries. We measure the time to failure of 12 batteries of each type. Batt
-Dominant- [34]

Using the t-distribution, it is found that since the <u>test statistic is greater than the critical value for the right-tailed test</u>, it is found that there is enough evidence to conclude that Battery B outlasts Battery A by more than 2 hours.

At the null hypothesis, it is <u>tested if it does not outlast by more than 2 hours</u>, that is, the subtraction is not more than 2:

H_0: \mu_B - mu_A \leq 2

At the alternative hypothesis, it is <u>tested if it outlasts by more than 2 hours</u>, that is:

H_1: \mu_B - \mu_A > 2

  • The sample means are: \mu_A = 8.65, \mu_B = 11.23
  • The standard deviations for the samples are s_A = s_B = 0.67

Hence, the standard errors are:

s_{Ea} = S_{Eb} = \frac{0.67}{\sqrt{12}} = 0.1934

The distribution of the difference has <u>mean and standard deviation</u> given by:

\overline{x} = \mu_B - \mu_A = 11.23 - 8.65 = 2.58

s = \sqrt{s_{Ea}^2 + s_{Eb}^2} = \sqrt{0.1934^2 + 0.1934^2} = 0.2735

The test statistic is given by:

t = \frac{\overline{x} - \mu}{s}

In which \mu = 2 is the value tested at the hypothesis.

Hence:

t = \frac{\overline{x} - \mu}{s}

t = \frac{2.58 - 2}{0.2735}

t = 2.12

The critical value for a <u>right-tailed test</u>, as we are testing if the subtraction is greater than a value, with a <u>0.05 significance level</u> and 12 + 12 - 2 = <u>22 df</u> is given by t^{\ast} = 1.71

Since the <u>test statistic is greater than the critical value for the right-tailed test</u>, it is found that there is enough evidence to conclude that Battery B outlasts Battery A by more than 2 hours.

A similar problem is given at brainly.com/question/13873630

7 0
2 years ago
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