Find the area of the shape
find the area of the shape - 1

Answers

Answer 1
Answer:

Answer:

The area is 91 cm²

Step-by-step explanation:

The shape is a kite.

area of a kite = ½(p*q)

Where, p and q are the diagonals of the kite.

p = 13 cm

q = 7 + 7 = 14 cm

The area of the kite = ½(13 * 14)

= ½(182)

Area = 91 cm²


Related Questions

Find an equation of the plane. the plane through the point (3, 0, 7) and perpendicular to the line x = 5t, y = 2 − t, z = 8 + 4t
Classify the angle as acute, right, obtuse, or straight.
9) Solve: a^2+ b^2= c^2 for b
Compute the area of the region D bounded by xy=1, xy=16, xy2=1, xy2=36 in the first quadrant of the xy-plane. Using the non-linear change of variables u=xy and v=xy2, find x and y as functions of u and v.x=x(u,v)= ?y=y(u,v)=?Find the determinant of the Jacobian for this change of variables.∣∣∣∂(x,y)/∂(u,v)∣∣∣=det=?Using the change of variables, set up a double integral for calculating the area of the region D.∫∫Ddxdy=?Evaluate the double integral and compute the area of the region D.Area =
Your office is raising money for charity by collecting aluminium cans for recycling. They get 36p per kilo of cans. A kilo is approximately 72 cans. They have collected 8645 cans How much money have they raised?

A and B and vertical angles. If A=(2x-10) and B= (x+8), find the measure of A

Answers

Answer:

18

Step-by-step explanation:

2x-10=x+8

2x-x=8+10

x=18

Answer is 18 hoped this help

I need help solving this equatioion.
(2xy)(3y)

Answers

Step-by-step explanation:

So first we will multiply

(2xy)(3y)

=

6x {y}^(2)

1. (5 points) At 6pm, ghost A is 5 kilometers due west of ghost B. Ghost A is flying westat 15 km/hr and ghost B is flying north at 20 km/hr. How fast is the distance between
the ghosts changing at 10pm?

Answers

Answer:

The distance between the ghost changes at 10 pm approximately at a rate of 24.981 kilometers per hour.

Step-by-step explanation:

At first we assume that north and east directions both represent positive quantities. Let suppose that \vec r_(A,o) = (0\,km,0\,km) and \vec r_(B,o) = (5\,km, 0\,km). If both ghosts moves at constant velocity such that \vec v_(A) = \left(-15\,(km)/(h), 0\,(km)/(h) \right) and \vec v_(B) = \left(0\,(km)/(h),20\,(km)/(h)  \right), then the final positions of both ghosts are, respectively:

Ghost A

\vec r_(A) = \vec r_(A,o)+t\cdot \vec v_(A)(Eq. 1)

Ghost B

\vec r_(B) = \vec r_(B,o)+t\cdot \vec v_(B)(Eq. 2)

Where t is the time, measured in hours.

Then, the equations of motion of each ghost are, respectively:

Ghost A

\vec r_(A) = (0\,km,0\,km)+t\cdot \left(-15\,(km)/(h), 0\,(km)/(h)  \right)

\vec r_(A) = \left(-15\cdot t, 0)\,\,\,\left[km \right]

Ghost B

\vec r_(B) = (5\,km, 0\,km)+t\cdot \left(0\,(km)/(h), 20\,(km)/(h)  \right)

\vec r_(B) = (5, 20\cdot t)\,\,\,\left[km\right]

Then, the distance between both ghosts is:

\vec r_(B/A) = (5,20\cdot t)-(-15\cdot t, 0)\,\,\,[km]

\vec r_(B/A) =(5+15\cdot t, 20\cdot t)\,\,\,[km](Eq. 3)

The magnitude of the relative is represented by the following Pythagorean identity:

r^(2)_(B/A) = (5+15\cdot t)^(2)+(20\cdot t)^(2)

Then, we find the rate of change of the relative distance (\dot r_(B/A)), measured in kilometers per hour, by implicit differentiation:

2\cdot r_(B/A)\cdot \dot r_(B/A) = 2\cdot (5+15\cdot t)\cdot 15+2\cdot (20\cdot t)\cdot 20

r_(B/A)\cdot \dot r_(B/A) = 15\cdot (5+15\cdot t)+20\cdot (20\cdot t)

\dot r_(B/A) = (75+625\cdot t)/(r_(B/A))

\dot r_(B/A) = \frac{75+625\cdot t}{\sqrt{(5+15\cdot t)^(2)+(20\cdot t)^(2)}}(Eq. 4)

If we know that t = 4\,h, then the rate of change of the relative distance at 10 PM is:

\dot r_(B/A) = \frac{75+625\cdot (4)}{\sqrt{[5+15\cdot (4)]^(2)+[20\cdot (4)]^(2)}}

\dot r_(B/A) \approx 24.981\,(km)/(h)

The distance between the ghost changes at 10 pm approximately at a rate of 24.981 kilometers per hour.

What is the prime factorization of 315?

Answers

Positive Integer factors of 315 = 3, 9, 5, 45, 7, 315 divided by 3, 3, 5, 7, gives no remainder. They are integers and prime numbers of 315, they are also called composite number.

In art class students are mixing blue and red paint to make purple paint. Cai mixes 5 cups of blue paint and 6 cups of red paint. Casho mixes 2 cups of blue paint and 3 cups of red paint. Use Cai and Casho’s percent of blue paint to determine whose purple paint will be bluer.

Answers

Answer:

Cai's Would Be Bluer

Step-by-step explanation:

Since 2/5 is 40% and 5/11 is 45% Cai's would be bluer!

Final answer:

Cai's paint will be bluer because she used a higher percentage of blue paint (45.45%) in her mix compared to Casho who used 40% blue paint.

Explanation:

In order to determine whose paint mix will be bluer, we must compare the ratio of blue to red paint in each student's mix. Cai mixes 5 cups of blue and 6 cups of red paint. Casho mixes 2 cups of blue and 3 cups of red paint. We find the percentage of blue paint by dividing the amount of blue paint by the total amount of paint and then multiplying by 100. For Cai, it would be (5 / (5 + 6)) * 100 = 45.45% blue. For Casho, it would be (2 / (2 + 3)) * 100 = 40% blue. Thus, Cai's purple paint will be bluer because it has a higher percentage of blue paint.

Learn more about Percentage here:

brainly.com/question/30697911

#SPJ2

Geometry Question Number 16

Answers

Answer:

No, the friend is not correct.

Step-by-step explanation:

The friend is not correct because let's call the three lines line A, line B, and line C. The line intersection says that if two lines intersect, then there will be one point of intersection. Therefore, we have to count all pairs of lines between line A, B, and C. Lines A and B can intersect, lines B and C can intersect, and lines A and C can intersect. Therefore there will be 3 lines of intersection, not 2.

Other Questions
Consider a random sample of ten children selected from a population of infants receiving antacids that contain aluminum, in order to treat peptic or digestive disorders. The distribution of plasma aluminum levels is known to be approximately normal; however its mean u and standard deviation o are not known. The mean aluminum level for the sample of n = 10 infants is found to be X = 37.20 ug/l and the sample standard deviation is s = 7.13 ug/1. Furthermore, the mean plasma aluminum level for the population of infants not receiving antacids is known to be only 4.13 ug/1.(a) Formulate the null hypothesis and complementary alternative hypothesis, for a two-sided test of whether the mean plasma aluminum level of the population of infants receiving antacids is equal to the mean plasma aluminum level of the population of infants not receiving antacids.(b) Construct a 95% confidence interval for the true mean plasma aluminum level of the population of infants receiving antacids.(c) Calculate the p-value of this sample (as best as possible), at the a=.05 significance level.(d) Based on your answers in parts (b) and (c), is the null hypothesis rejected in favor of the alternative hypothesis, at the a = .05 significance level? Interpret your conclusion: What exactly has been demonstrated, based on the empirical evidence?(e) With the knowledge that significantly elevated plasma aluminum levels are toxic to human beings, reformulate the null hypothesis and complementary alternative hypothesis, for the appropriate one-sided test of the mean plasma aluminum levels. With the same sample data as above, how does the new p-value compare with that found in part (c), and what is the resulting conclusion and interpretation?