Simplify the expression show work<br> 15-3[2+6(-3)]

(I) x²-8x = -16<br>(ii) x²+2x+1 = 0<br>(iii) 4x²-24x-36=0<br>Solve please...........

algol13

(I)

Given equation is x²-8x = -16

⇛ x²-8x+16 = 0

⇛ x²-4x-4x+16 = 0

⇛ x(x-4)-4(x-4) = 0

⇛ (x-4)(x-4) = 0

⇛ x-4 = 0 or x-4 = 0

⇛ x = 4 or x = 4

(ii)

Given equation is x²+2x+1 = 0

⇛ x²+x+x+1 = 0

⇛ x(x+1)+1(x+1) = 0

⇛ (x+1)(x+1) = 0

⇛ x+1 = 0 or x+1 = 0

⇛ x = -1 or x = -1

(iii)

Given equation is 4x²-24x-36 = 0

⇛ 4(x²-6x-9) = 0

⇛ x²-6x-9 = 0

⇛ x²-2(x)(3) = 9

On adding 3² both sides then

⇛ x²-2(3)x)+3² = 9+9

⇛ (x-3)² = 18

⇛ (x-3) = ±√18

⇛ x-3 = ±√(2×9)

⇛ x-3 = ±3√2

⇛ x = 3±3√2

⇛ x = 3+3√2 or 3-3√2

5 0

3 years ago

A force of 3 pounds is required to hold a spring stretched 0.6 feet beyond its natural length. how much work (in foot-pounds) is

ioda

The work done (in foot-pounds) in stretching the spring from its natural length to 0.7 feet beyond its natural length is 1.23 foot-pound

<h3>Data obtained from the question</h3>

From the question given above, the following data were obtained:

Force (F) = 3 pounds
Extension (e) = 0.6 feet
Work done (Wd) =?

<h3>How to determine the spring constant</h3>

Force (F) = 3 pounds
Extension (e) = 0.6 feet
Spring constant (K) =?

F = Ke

Divide both sides by e

K = F/ e

K = 3 / 0.6

K = 5 pound/foot

Thus, the spring constant of the spring is 5 pound/foot

<h3>How to determine the work done</h3>

Spring constant (K) = 5 pound/foot
Extention (e) = 0.7 feet
Work done (Wd) =?

Wd = ½Ke²

Wd = ½ × 5 × 0.7²

Wd = 2.5 × 0.49

Wd = 1.23 foot-pound

Therefore, the work done in stretching the spring 0.7 feet is 1.23 foot-pound

Learn more about spring constant:

brainly.com/question/9199238

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5 0

1 year ago

Several programs attempt to address the shortage of qualified teachers by placing uncertified instructors in schools with acute

ss7ja [257]

Answer:

We conclude that the mean scores with uncertified teachers is higher or equal as compared to certified teachers.

Step-by-step explanation:

We are given that reading scores of the students of certified teachers averaged 35.62 points with standard deviation 9.31. The scores of students instructed by uncertified teachers had mean 32.48 points with standard deviation 9.43 points on the same test.

There were 44 students in each group.

Let $\mu_1$ = <em><u>mean scores with uncertified teachers.</u></em>

$\mu_2$ = <em><u>mean scores with certified teachers.</u></em>

So, Null Hypothesis, $H_0$ : $\mu_1\geq \mu_2$ {means that the mean scores with uncertified teachers is higher or equal as compared to certified teachers}

Alternate Hypothesis, $H_A$ : $\mu_1$ {means that the mean scores with uncertified teachers is lower as compared to certified teachers}

The test statistics that would be used here <u>Two-sample t test statistics</u> as we don't know about the population standard deviations;

T.S. = $\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ ~ $t__n__1-_n__2-2$

where, $\bar X_1$ = sample mean scores of students instructed by uncertified teachers = 32.48 points

$\bar X_2$ = sample mean scores of students instructed by certified teachers = 35.62 points

$s_1$ = sample standard deviation of scores of students instructed by uncertified teachers = 9.43 points

$s_2$ = sample standard deviation of scores of students instructed by certified teachers = 9.31 points

$n_1$ = sample of students under uncertified teachers = 44

$n_2$ = sample of students under certified teachers = 44

Also, $s_p=\sqrt{\frac{(n_1-1)s_1^{2} +(n_2-1)s_2^{2} }{n_1+n_2-2} }$ = $\sqrt{\frac{(44-1)\times 9.43^{2} +(44-1)\times 9.31^{2} }{44+44-2} }$ = 9.37

So, <u><em>the test statistics</em></u> = $\frac{(32.48-35.62)-(0)}{9.37 \times \sqrt{\frac{1}{44} +\frac{1}{44} } }$ ~ $t_8_6$

= -1.572

The value of t test statistics is -1.572.

Since, in the question we are not given the level of significance so we assume it to be 5%. <u>Now, at 5% significance level the t table gives critical values of -1.665 at 86 degree of freedom for left-tailed test.</u>

Since our test statistic is more than the critical values of t as -1.572 > -1.665, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which <u>we fail to reject our null hypothesis</u>.

Therefore, we conclude that the mean scores with uncertified teachers is higher or equal as compared to certified teachers.

7 0

3 years ago

Distribute and Combine Like Terms<br><br> 11(2t - 4) + 6t - 6

Nutka1998 [239]

It’s actually 28t - 10

11(2t-4)+6t-6
22t-4+6t-6 add 22t and 6t
28t-4-6 combine -4 & -6
28t-10

5 0

3 years ago

5.<br> Enter the number of pound equal to 448 ounces.<br> 16 ounces = 1 pound

lara31 [8.8K]

Answer:

27.875

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

Simplify the expression show work 15-3[2+6(-3)]