2/3(6b-12)<br> Please answer

Factor the following using the difference of squares method

Bond [772]

Answer:

a)

${x}^{2} - 16 \\ = {x}^{2} - {4}^{2} \\ = (x + 4)(x - 4)$

b)

${9y}^{2} - 25 \\ = {(3y)}^{2} - {5}^{2} \\ = (3y + 5)(3y - 5)$

here is the answer hope that will help you

6 0

3 years ago

If two angles form a linear pair and the measure of the first angle is one-fifth the measure of the second angle, what is the me

kari74 [83]

Well, one angle has one share of the linear angle, and the other one has five shares. All together, they add up to six shares of the linear angle.

BUT ... a linear angle adds up to 180°. So each share must be (180°/6) = 30°.

The smaller one (one share) is 30°, and the larger one (5 shares) is 150°.

5 0

3 years ago

On a coordinate plane, the segment with endpoints (10, 40) and (70, 120) is

OlgaM077 [116]

Answer:

The length of the resulting segment is 500.

Step-by-step explanation:

Vectorially speaking, the dilation is defined by following operation:

$P'(x,y) = O(x,y) + k\cdot [P(x,y)-O(x,y)]$ (1)

Where:

$O(x,y)$ - Center of dilation.

$P(x,y)$ - Original point.

$k$ - Scale factor.

$P'(x,y)$ - Dilated point.

First, we proceed to determine the coordinates of the dilated segment:

( $P(x,y) = (10, 40)$ , $Q(x,y) = (70, 120)$ , $O(x,y) = (0,0)$ , $k = 5$ )

$P'(x,y) = O(x,y) + k\cdot [P(x,y)-O(x,y)]$

$P(x,y) = (0,0) +5\cdot [(10,40)-(0,0)]$

$P'(x,y) = (50,200)$

$Q'(x,y) = O(x,y) + k\cdot [Q(x,y)-O(x,y)]$

$Q' (x,y) = (0,0) +5\cdot [(70,120)-(0,0)]$

$Q'(x,y) = (350, 600)$

Then, the length of the resulting segment is determined by following Pythagorean identity:

$l_{P'Q'} = \sqrt{(350-50)^{2}+(600-200)^{2}}$

$l_{P'Q'} = 500$

The length of the resulting segment is 500.

3 0

3 years ago

Each person in random samples of 207 male and 253 female working adults living in a certain town in Canada was asked how long, i

Sergeu [11.5K]

Answer:

The null hypothesis failed to be rejected.

There is not enough evidence to support the claim that the mean daily conmute time for males is significantly bigger than the mean daily conmute time for females (P-value=0.155).

Step-by-step explanation:

The data is:

Males n1=207 M1=31.6 s1=24.0

Females n2=253 M2=29.3 s2=24.3

This is a hypothesis test for the difference between populations means.

The claim is that the mean daily conmute time for males is significantly bigger than the mean daily conmute time for females.

Then, the null and alternative hypothesis are:

$H_0: \mu_m-\mu_f=0\\\\H_a:\mu_m-\mu_f> 0$

The significance level is 0.05.

The sample 1 (males), of size n1=207 has a mean of 31.6 and a standard deviation of 24.

The sample 2 (females), of size n2=253 has a mean of 29.3 and a standard deviation of 24.3.

The difference between sample means is Md=2.3.

$M_d=M_m-M_f=31.6-29.3=2.3$

The estimated standard error of the difference between means is computed using the formula:

$s_{M_d}=\sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma_2^2}{n_2}}=\sqrt{\dfrac{24^2}{207}+\dfrac{24.3^2}{253}}\\\\\\s_{M_d}=\sqrt{2.783+2.334}=\sqrt{5.117}=2.262$

Then, we can calculate the t-statistic as:

$t=\dfrac{M_d-(\mu_m-\mu_f)}{s_{M_d}}=\dfrac{2.3-0}{2.262}=\dfrac{2.3}{2.262}=1.017$

The degrees of freedom for this test are:

$df=n_1+n_2-1=207+253-2=458$

This test is a right-tailed test, with 458 degrees of freedom and t=1.017, so the P-value for this test is calculated as (using a t-table):

$P-value=P(t>1.017)=0.155$

As the P-value (0.155) is bigger than the significance level (0.05), the effect is not significant.

The null hypothesis failed to be rejected.

There is not enough evidence to support the claim that the mean daily conmute time for males is significantly bigger than the mean daily conmute time for females.

7 0

3 years ago

Enter the unknown value that makes this statement true. 25% of is 35

Eva8 [605]

Answer:

25% of 140 is 35

Step-by-step explanation:

If 35 is 25% then we can multiply it by 4 to get 100%.

35 x 4=140

4 0

3 years ago

2/3(6b-12) Please answer