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2 years ago

The figure shows a kite inside a rectangle. Which expression represents the area of the shaded region?

Mathematics

2 answers:

Leto [7]2 years ago

5 0

Answer:

$4x^{2}\ units^{2}$

Step-by-step explanation:

we know that

The area of the shaded region is equal to the area of the rectangle minus the area of a kite

Step 1

Find the area of the rectangle

The area of the rectangle is equal to

$A=bh$

In this problem we have

$b=(3x+x)=4x\ units$

$h=(x+x)=2x\ units$

substitute

$A=(4x)(2x)=8x^{2}\ units^{2}$

Step 2

Find the area of a kite

The area of a kite is equal to

$A=\frac{1}{2}D1D2$

where D1 and D2 are the diagonals of a kite

In this problem we have

$D1=(3x+x)=4x\ units$

$D2=(x+x)=2x\ units$

substitute

$A=\frac{1}{2}(4x)(2x)=4x^{2}\ units^{2}$

Step 3

Find the area of the shaded region

$8x^{2}\ units^{2}-4x^{2}\ units^{2}=4x^{2}\ units^{2}$

Nuetrik [128]2 years ago

4 0

1/2 of 4x+2x which is 4x

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Sam is observing the velocity of a car at different times. After two hours, the velocity of the car is 50 km/h. After six hours,

Eva8 [605]

Let:
x = hours of travel
y = velocity

slope= rise/run slope=(y2-y1)/(x2-x1)
(x1,y1) = (2,50) (x2,y2) = (6,54)
sub values back into the equation m = (54-50)/(6-2) m = 1
POINT SLOPE FORMy-y1 = m(x-x1) y-50= 1(x-2) y = x -2 +50
y = x + 48

B)

the graph within the first seven hours can be obtained at point B

x = 7

y = 7+48 = 55
B(7,55)

4 0

1 year ago

Type two statements that use nextint() to print 2 random integers between (and including) 100 and 149. end with a newline. ex: 1

solong [7]

NB- Solution is emboldened

import java.util.Scanner;

import java.util.Random;

public class RandomGenerateNumbers {

public static void main (String [] args) {

Random randGen = new Random();

int seedVal = 0;

seedVal = 4;

randGen.setSeed(seedVal);

System.out.println(randGen.nextInt(50) + 100);

return;

}

4 0

1 year ago

Read 2 more answers

Question 1: Classify the system of equations and identify the number of solutions. 3 = 4x + y 2y = 6 − 8x Answers: inconsistent;

Mariana [72]

Answer:

1st: consistent and dependent

2nd: (-2,0). Simply put the called and check.

7 0

1 year ago

Employment data at a large company reveal that 74% of the workers are married, 42% are college graduates, and that 56% are marri

ValentinkaMS [17]

Answer:

(E) None of these above are true.

Step-by-step explanation:

Married = 74% or 0.74

College graduates = 42% or 0.42

pr(married | college graduates) = 0.56

(A) These events are pairwise disjoint. This is false. Pairwise disjoint are also known as mutually exclusive events. Here we can see that both events are occurring at same time.

(B) These events are independent events. This is also false.

(D) A worker is either married or a college graduate always. False

Here Probability(A or B) shall be 1

= Pr(A) + Pr(B) - Pr( A and B) = 0.74 + 0.42 - 0.56 * 0.42 = 0.9248

This is not equal to 1.

(E) None of these above are true. This is true.

8 0

2 years ago

Determine the area (in units2) of the region between the two curves by integrating over the x-axis. y = x2 − 24 and y = 1

astra-53 [7]

Answer:

The area of the region between the two curves by integration over the x-axis is 9.9 square units.

Step-by-step explanation:

This case represents a definite integral, in which lower and upper limits are needed, which corresponds to the points where both intersect each other. That is:

$x^{2} - 24 = 1$

Given that resulting expression is a second order polynomial of the form $x^{2} - a^{2}$ , there are two real and distinct solutions. Roots of the expression are:

$x_{1} = -5$ and $x_{2} = 5$ .

Now, it is also required to determine which part of the interval $(x_{1}, x_{2})$ is equal to a number greater than zero (positive). That is:

$x^{2} - 24 > 0$

$x^{2} > 24$

$x < -4.899$ and $x > 4.899$ .

Therefore, exists two sub-intervals: $[-5, -4.899]$ and $\left[4.899,5\right]$ . Besides, $x^{2} - 24 > y = 1$ in each sub-interval. The definite integral of the region between the two curves over the x-axis is:

$A = \int\limits^{-4.899}_{-5} [{1 - (x^{2}-24)]} \, dx + \int\limits^{4.899}_{-4.899} \, dx + \int\limits^{5}_{4.899} [{1 - (x^{2}-24)]} \, dx$

$A = \int\limits^{-4.899}_{-5} {25-x^{2}} \, dx + \int\limits^{4.899}_{-4.899} \, dx + \int\limits^{5}_{4.899} {25-x^{2}} \, dx$

$A = 25\cdot x \right \left|\limits_{-5}^{-4.899} -\frac{1}{3}\cdot x^{3}\left|\limits_{-5}^{-4.899} + x\left|\limits_{-4.899}^{4.899} + 25\cdot x \right \left|\limits_{4.899}^{5} -\frac{1}{3}\cdot x^{3}\left|\limits_{4.899}^{5}$

$A = 2.525 -2.474+9.798 + 2.525 - 2.474$

$A = 9.9\,units^{2}$

The area of the region between the two curves by integration over the x-axis is 9.9 square units.

4 0

1 year ago