Please help

Sequence of Transformation

weeeeeb [17]

Answer:

F (4, 2) G (2, 2) H (0, -4)

Step-by-step explanation:

8 0

4 years ago

Read 2 more answers

NEED HELP ASAP!!!!!!!

sertanlavr [38]

ANSWER: x > -6
1. subtract 12 from each side
2. divide by -3 which will make the sign flip

8 0

3 years ago

Help me solve this please

shutvik [7]

Answer:

D'G' = 52.5 units

Step-by-step explanation:

Since the dilatation is centred at the origin then multiply the original coordinates by the scale factor 3.5

D' = (1 × 3.5, 7 × 3.5 ) = (3.5, 24.5 )

G' = (- 8 × 3.5, - 5 × 3.5 ) = (- 28, - 17.5 )

Calculate D'G' using the distance formula

d = $\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2 }$

with (x₁, y₁ ) = D' (3.5, 24.5) and (x₂, y₂ ) = G' (- 28, - 17.5)

D'G' = $\sqrt{(-28-3.5)^2+(-17.5-24.5)^2}$

= $\sqrt{(-31.5)^2+42^2}$

= $\sqrt{992.25+1764}$

= $\sqrt{2756.25}$

= 52.5 units

7 0

3 years ago

Maggie is standing on a platform that is 8 feet above the ground. She throws a ball in the air that hits the ground after 2 seco

Paha777 [63]

The equation that offers the best approximation to this result is: $h = -16\cdot t^{2}+28\cdot t + 8$ . (Choice D)

<h3>How to find the free fall formula for a given scenario</h3>

An object experiments a free fall when it is solely accelerated by gravity on the assumption of an <em>uniform</em> acceleration. The formula is described below:

$h =h_{o} + v_{o}\cdot t + \frac{1}{2}\cdot g\cdot t^{2}$ (1)

Where:

$h_{o}$ - Initial height, in feet.
$v_{o}$ - Initial speed, in feet per second.
$t$ - Time, in seconds.
$g$ - Gravitational acceleration, in feet per square second.

If we know that $t = 2\,s$ , $h_{o} = 8\,ft$ , $h = 0\,ft$ , $g = -32.174\,\frac{ft}{s^{2}}$ , then the height formula is:

$0 = 8 +2\cdot v_{o} - \frac{1}{2}\cdot (32.174)\cdot (2)^{2}$

$0 = -56.348+2\cdot v_{o}$

$v_{o} = 28.174\,\frac{ft}{s}$

The equation that offers the best approximation to this result is: $h = -16\cdot t^{2}+28\cdot t + 8$ . (Choice D)

To learn more on free fall, we kindly invite to check this verified question: brainly.com/question/13796105

5 0

3 years ago

From a large number of actuarial exam scores, a random sample of scores is selected, and it is found that of these are passing s

Mnenie [13.5K]

<u>Supposing 60 out of 100 scores are passing scores</u>, the 95% confidence interval for the proportion of all scores that are passing is (0.5, 0.7).

The lower limit is 0.5.
The upper limit is 0.7.

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the z-score that has a p-value of $\frac{1+\alpha}{2}$ .

60 out of 100 scores are passing scores, hence $n = 100, \pi = \frac{60}{100} = 0.6$

95% confidence level

So $\alpha = 0.95$ , z is the value of Z that has a p-value of $\frac{1+0.95}{2} = 0.975$ , so $z = 1.96$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.6 - 1.96\sqrt{\frac{0.6(0.4)}{100}} = 0.5$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.6 + 1.96\sqrt{\frac{0.6(0.4)}{100}} = 0.7$

The 95% confidence interval for the proportion of all scores that are passing is (0.5, 0.7).

The lower limit is 0.5.
The upper limit is 0.7.

A similar problem is given at brainly.com/question/16807970

5 0

3 years ago