Order -12,|-25|,-10,18,<br> |29|,and 5 from greatest to least <br> It's a little more difficult

Find the coordinates of the intersection of the diagonals of the parallelogram with the vertices W(−2, 5), X(2, 5), Y(4, 0),

emmasim [6.3K]

Using the midpoint formula, the coordinates of the intersection of the diagonals of the parallelogram is: (1, 2.5).

<h3>What are Diagonals of a Parallelogram?</h3>

The diagonals of a parallelogram bisect each other, therefore, the coordinates of their intersection can be determined using the midpoint formula, which is: $M(\frac{x_2 + x_1}{2}, \frac{y_2 + y_1}{2})$ .

A diagonal is XZ.

X(2, 5) = (x1, y1)

Z(0, 0) = (x2, y2)

Plug in the values

$M(\frac{0 + 2}{2}, \frac{0 + 5}{2})$

= (1, 2.5).

Learn more about the diagonals of a parallelogram on:

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7 0

2 years ago

A firework is launched at the rate of 10 feet per second from a point on the ground 50 feet from an observer. to 2 decimal place

Kazeer [188]

The rate of change of the angle of elevation when the firework is 40 feet above the ground is 0.12 radians/second.

First we will draw a right angle triangle ΔABC, where ∠B = 90°

Lets, assume the height(AB) = h and base(BC)= x

If the angle of elevation, ∠ACB = α, then

tan(α) = $\frac{AB}{BC} = \frac{h}{x}$

Taking inverse trigonometric function, α = tan⁻¹ ( $\frac{h}{x}$ ) .............(1)

As we need to find the rate of change of the angle of elevation, so we will differentiate both sides of equation (1) with respect to time (t) :

$\frac{d\alpha}{dt}=[\frac{1}{1+ \frac{h^2}{x^2}}]*(\frac{1}{x})\frac{dh}{dt}$

Here, the firework is launched from point B at the rate of 10 feet/second and when it is 40 feet above the ground it reaches point A,

that means h = 40 feet and $\frac{dh}{dt}$ = 10 feet/second.

C is the observer's position which is 50 feet away from the point B, so x = 50 feet.

$\frac{d\alpha}{dt}= [\frac{1}{1+ \frac{40^2}{50^2}}] *\frac{1}{50} *10\\ \\ \frac{d\alpha}{dt} = [\frac{1}{1+\frac{16}{25}}] *\frac{1}{5}\\ \\ \frac{d\alpha}{dt} = [\frac{25}{41}] *\frac{1}{5}\\ \\ \frac{d\alpha}{dt}= \frac{5}{41} =0.1219512$

= 0.12 (Rounding up to two decimal places)

So, the rate of change of the angle of elevation is 0.12 radians/second.

5 0

3 years ago

Express the confidence interval 0.555 less than 0.777 in the form Modifying above p with caret plus or minus Upper E.

Elina [12.6K]

Complete Question

Express the confidence interval 0.555 less than p less than 0.777 in the form Modifying above p with caret plus or minus Upper E.

Answer:

The modified representation is $\r p \pm E = 0.666 \pm 0.111$

Step-by-step explanation:

From the question we are told that

The confidence interval interval is $0.555 < p < 0.777$

Now looking at the values that make up the up confidence interval we see that this is a symmetric confidence interval(This because the interval covers 95% of the area under the normal curve which mean that the probability of a value falling outside the interval is 0.05 which is divided into two , the first half on the left -tail and the second half on the right tail as shown on the figure in the first uploaded image(reference - Yale University ) ) which means

Now since the confidence interval is symmetric , we can obtain the sample proportion as follows

$\r p = \frac{0.555 + 0.777}{2}$