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2 years ago

Factor the expression x^2-16x+63

Mathematics

1 answer:

Airida [17]2 years ago

7 0

The factors should add to the b value and multiply to the c value.

Final answer: (x-9)(x-7)

Check:
-9*-7=63
-9-7=-16

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Two sides of an obtuse triangle measure 9 inches and 14 inches. The length of longest side is unknown.

shepuryov [24]

The smallest possible whole-number length of the unknown side is 17 inches.

<h3>What is the Pythagoras theorem?</h3>

The Pythagoras theorem states that the square of the longest side must be equal to the sum of the square of the other two sides in a right-angle triangle.

From the information given, the sides of an obtuse triangle measure 9 inches and 14 inches.

Therefore, the third side will be:

c² = 9² + 14²

c² = 81 + 196

c² = 277

c = ✓277

c = 16.64

c = 17

Hence, the smallest possible whole-number length of the unknown side is 17 inches.

Learn more about triangles on:

brainly.com/question/17335144

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1 year ago

Which statement is true

EastWind [94]

Answer:

Step-by-step explanation:

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2 years ago

What is the area of a circle with a Radius of 4 and a Diameter of 2?<br><br> (Its on Khan)

algol13

Answer:4^2≈50.26548

Step-by-step explanation:

πr2.

Where r is the radius and π≈3.14 , the ratio of a circle's circumference to its diameter.

Plugging in 4 from the radius, we get.

42π

⇒16π inches.

This is our exact answer. Alternatively, we can plug in 3.14 for π to get.

50.26 inches.

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1 year ago

Can someone please help me solve the problem for my math homework

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2 years ago

Read 2 more answers

A manufacturer uses production method to produce steel rods. A random sample of 17 steel rods resulted in lengths with a standar

Arisa [49]

Answer:

$\chi^2 =\frac{17-1}{12.15} 22.09 =15.87$

Traditional method

We can find a critical value in the chi square distribution with df =16 who accumulates $\alpha/2 =0.05$ of the area on each tail and we got:

$\chi^2_{crit}= 7.962$

$\chi^2_{crit}=26.296$

Since the calculated value is between the two critical values we don't have enough evidence to conclude that the true deviation is NOT significantly different from 3.5 cm

Confidence level method

We can find the p value like this

$p_v =2*P(\chi^2 >15.87)=0.92$

Since the p value is higher then the significance level we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is NOT significantly different from 3.5 cm

Step-by-step explanation:

Data provided

$n=17$ represent the sample selected

$\alpha=0.1$ represent the significance

$s^2 =4.7^2 =22.09$ represent the sample variance

$\sigma^2_0 =3.5^2 =12.15$ represent the value to check

Null and alternative hypothesis

We want to verify if the new production method has lengths with a standard deviation different from 3.5 cm, so the system of hypothesis would be:

Null Hypothesis: $\sigma^2 = 12.15$

Alternative hypothesis: $\sigma^2 \neq 12.15$

The statistic for this case is given by:

$\chi^2 =\frac{n-1}{\sigma^2_0} s^2$

The degrees of freedom are:

$df =n-1= 17-1=16$

$\chi^2 =\frac{17-1}{12.15} 22.09 =15.87$

Traditional method

We can find a critical value in the chi square distribution with df =16 who accumulates $\alpha/2 =0.05$ of the area on each tail and we got:

$\chi^2_{crit}= 7.962$

$\chi^2_{crit}=26.296$

Since the calculated value is between the two critical values we don't have enough evidence to conclude that the true deviation is significantly different from 3.5 cm

Confidence level method

We can find the p value like this

$p_v =2*P(\chi^2 >15.87)=0.92$

Since the p value is higher then the significance level we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is significantly different from 3.5 cm

7 0

2 years ago