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Nookie1986 [14]
3 years ago
6

Factor and solve the problems in the photo ……….. pleaseeeee help i haven’t done algebra in 2 years

Mathematics
1 answer:
vladimir1956 [14]3 years ago
5 0

Answer:

30. 2x²-x-1=0

or, 2x²-2x+x-1=0

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

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

so, x-1 = 0 or, x = 1

and 2x+1 = 0 or, 2x = -1 or, x = -1/2

so, x = 1 and 1/2

31. 3x²-14x=5

or, 3x²-14x-5=0

or, 3x²-15x+x-5=0

or, 3x(x-5)+1(x-5)=0

or, (x-5)(3x+1)=0

so, x-5=0 or, x=5

and 3x+1=0 or, 3x=-1 or, x=-1/3

so, x = 5 and -1/3

Answered by GAUTHMATH

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4. If DF = 9x - 39, find EF.<br><br> Help I’m so confused!!!!
topjm [15]

Answer:

EF =58

Step-by-step explanation:

from the illustration,

EF =DF - DE

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<em> </em><em> </em>

<em>3</em><em>x</em><em> </em><em>+</em><em>1</em><em>0</em><em>=</em><em>9</em><em>x</em><em>-</em><em>3</em><em>9</em><em> </em><em>-</em><em> </em><em>4</em><em>7</em>

<em>sol</em><em>ving</em><em> </em><em>for</em><em> </em><em>x</em>

<em>3</em><em>x</em><em> </em><em>+</em><em>1</em><em>0</em><em> </em><em>=</em><em>9</em><em>x</em><em> </em><em>-</em><em>8</em><em>6</em>

<em>grou</em><em>ping</em><em> like</em><em> </em><em>ter</em><em>ms</em>

<em>3</em><em>x</em><em> </em><em>-</em><em> </em><em>9x</em><em> </em><em>=</em><em>-</em><em>8</em><em>6</em><em> </em><em>-</em><em>1</em><em>0</em>

<em>-6x</em><em>=</em><em>-</em><em>9</em><em>6</em>

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<em>\frac{ - 6x}{ - 6}  =  \frac{ - 96}{ - 6}</em>

<em>x  =  \frac{96}{6}</em>

<em>x = 16</em>

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substitute x=16 into it to get the EF

EF= 3(16)+10

EF=48+10

EF=58

6 0
3 years ago
Read 2 more answers
Please answer correctly !!!!! Will mark Brianliest !!!!!!!!!!!!!!
Sedbober [7]

Answer:

24/25

Step-by-step explanation:

Please see the attached image.

The cosine of an angle is the adjacent side relative to that angle over the hypotenuse.

Answer: 24/25

8 0
2 years ago
Express sin P as a fraction in simplest terms.​
d1i1m1o1n [39]

Answer:

1/15

Step-by-step explanation:

sin P = ON / PN = 2/30 = 1/15

8 0
3 years ago
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A group of researchers are interested in the possible effects of distracting stimuli during eating, such as an increase or decre
Dmitry [639]

Using the t-distribution, it is found that since the p-value of the test is of 0.0302, which is <u>less than the standard significance level of 0.05</u>, the data provides convincing evidence that the average food intake is different for the patients in the treatment group.

At the null hypothesis, it is <u>tested if the consumption is not different</u>, that is, if the subtraction of the means is 0, hence:

H_0: \mu_1 - \mu_2 = 0

At the alternative hypothesis, it is <u>tested if the consumption is different</u>, that is, if the subtraction of the means is not 0, hence:

H_1: \mu_1 - \mu_2 \neq 0

Two groups of 22 patients, hence, the standard errors are:

s_1 = \frac{45.1}{\sqrt{22}} = 9.6154

s_2 = \frac{26.4}{\sqrt{22}} = 5.6285

The distribution of the differences is has:

\overline{x} = \mu_1 - \mu_2 = 52.1 - 27.1 = 25

s = \sqrt{s_1^2 + s_2^2} = \sqrt{9.6154^2 + 5.6285^2} = 11.14

The test statistic is given by:

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

In which \mu = 0 is the value tested at the null hypothesis.

Hence:

t = \frac{25 - 0}{11.14}

t = 2.2438

The p-value of the test is found using a <u>two-tailed test</u>, as we are testing if the mean is different of a value, with <u>t = 2.2438</u> and 22 + 22 - 2 = <u>42 df.</u>

  • Using a t-distribution calculator, this p-value is of 0.0302.

Since the p-value of the test is of 0.0302, which is <u>less than the standard significance level of 0.05</u>, the data provides convincing evidence that the average food intake is different for the patients in the treatment group.

A similar problem is given at brainly.com/question/25600813

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3 years ago
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Answer:

$167.03

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ok

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1 year ago
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