If AZ is a midsegment of WXY, find XZ
A.7
B.9
C.12
D.14

Answer:

-37260

Step-by-step explanation:

did it in my head

Answer:

Option A is correct

the expression is equal to [tex]\frac{5}{ \sqrt{11}}[/tex] is [tex]\frac{5\sqrt{11}} {11}[/tex]

Explanation:

Given expression is, [tex]\frac{5}{ \sqrt{11}}[/tex]

Multiply and divide by the denominator by [tex]\sqrt{11}[/tex] in the given expression, we have,

[tex]\frac{5}{\sqrt{11}} \times \frac{\sqrt{11}} { \sqrt{11}}[/tex]

or

[tex]\frac{5 \cdot \sqrt{11}} {\sqrt{11} \cdot \sqrt{11}}[/tex]

use : [tex]\sqrt{a}\cdot\sqrt{a}=(\sqrt{a} )^2 = a[/tex]

then;

[tex]\frac{5\sqrt{11}} { (\sqrt{11})^2} =\frac{5\sqrt{11}} {11}[/tex]

Therefore, the expression is equal to [tex]\frac{5}{ \sqrt{11}}[/tex] is, [tex]\frac{5\sqrt{11}} {11}[/tex]

Answer:

Answer:

16 miles per gallon

Step-by-step explanation:

Final answer:

The number of commercial orbital launches is 18, and the number of noncommercial orbital launches is 55.

Explanation:

Let's assume that the number of commercial orbital launches is represented by x. According to the problem statement, the number of noncommercial orbital launches is more than three times the number of commercial orbital launches. So, the number of noncommercial orbital launches would be 3x + 1.

Given that the total number of commercial and noncommercial orbital launches is 73, we can set up the equation:

x + (3x + 1) = 73

Combining like terms, we get:

4x + 1 = 73

Subtracting 1 from both sides, we get:

4x = 72

Dividing both sides by 4, we get:

x = 18

Therefore, there were 18 commercial orbital launches and 55 noncommercial orbital launches.

The number of commercial orbital launches is 24, and the number of noncommercial orbital launches is 49.

Let's assume that the number of commercial orbital launches is represented by 'x'. According to the question, the number of noncommercial orbital launches is one more than three times the number of commercial orbital launches. So, the number of noncommercial orbital launches can be represented by '3x + 1'.

Given that the total number of commercial and noncommercial orbital launches is 73, we can set up the equation 'x + (3x + 1) = 73' to represent the total launches. Solving this equation will give us the values of 'x' and '3x + 1', which represent the number of commercial and noncommercial orbital launches, respectively.

By solving the equation, we find that 'x = 24' and '3x + 1 = 73 - 24 = 49'.

Final answer:

The inequality |2x + 4| > 8 is resolved by considering the inside expression being greater than 8 and less than -8, which leads to x > 2 and x < -6. The proper representation on a number line is with open circles on -6 and 2, shading away from these points, indicating the range of solutions.

Explanation:

To solve the absolute value inequality |2x + 4| > 8, we must consider the two scenarios that could make the inequality true: when (2x + 4) is greater than 8 and when -(2x + 4) is less than -8. Absolute values denote the distance from zero, so they are never negative. Since the inequality is strict (indicated by '>'), we will use open circles in the graph.

First, let's consider when the expression inside the absolute value is positive:

2x + 4 > 8

2x > 4

x > 2
This scenario corresponds to all x-values greater than 2.

Now for the negative scenario:

-(2x + 4) > 8

-2x - 4 > 8

-2x > 12

x < -6
This scenario corresponds to all x-values less than -6.

The correct graph of the solution set on a number line would have open circles on -6 and 2, with shading going in opposite directions away from these points, indicating all the numbers greater than 2 and all the numbers less than -6 satisfy the original inequality.

Based on this, the third choice is correct: A number line with open circles on negative 6 and 2, shading going in the opposite directions.

The cone of dimension 4.5 inches height and 1.5 inches radius can hold 9.425 inches^3 of ice cream.

A sugar cone can be seen as a three-dimensional object, specifically a cone, which has a known formula for calculating its volume.

Volume of a cone: (1/3) x π x radius2 x height

Substituting the given values: (1/3) x π x 1.52 x 4.5 = 9.425 inches^3.

Answer: x=6 and y=3

Step-by-step explanation:

4x - 7y = 3 -------(1)

x - 7y = -15 ------(2)

equation(1) - equation (2)

3x = 18

Divide bothside by 3

x = 18/3

x= 6

also, multiply equation (2) by 4

4x - 28y = -60--------------(3)

subtract equation(3) from equation(1)

21y = 63

divide bothside by 21

y = 63/21

y = 3

2(b−4)=6(b) 2b - 8 = 6b 2b - 6b = 8 -4b = 8 Now you can solve.....

Answer:

Two rational solutions.  This is the right answer

Step-by-step explanation:

The task is to determine the coordinates of two 2-cm holes in a quarter-circle steel plate, but there is insufficient information to provide a precise location without additional details or constraints.

The question requires us to find the coordinates for the centers of two 2-cm holes to be drilled in a steel plate, which is shaped like a quarter of a circle with a radius of 49 cm. Assuming the plate is positioned with its corner at the origin (0,0), with the circular edge encompassed by the positive x-axis and positive y-axis, the holes should be placed in such a way that their centers within the quarter circle are equidistant from the axes and from the plate's circular edge.

However, the question does not provide sufficient information or constraints about the location of the holes within the quarter-circle plate. To determine the exact coordinates, we need more specifics, such as how far from the edges or any other centering constraints for the holes. Without this additional information, there are infinitely many possible coordinates for the holes that meet the given condition of being within the quarter-circle plate.

If the holes are to be evenly distributed and placed symmetrically, a possible choice could be to align the centers of the holes along the line y = x, which bisects the quarter-circle. Yet, without precise instructions, we cannot provide a definitive answer to the coordinates of the holes.

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The fifth root of -32 is -2. To find it, think of a number that, when raised to the power of 5, gives the original number.

The fifth root of -32 is -2. To find the fifth root of a number, we can think of it as finding a number that, when raised to the power of 5, gives us the original number. In this case, -2 * -2 * -2 * -2 * -2 = -32.

Answer:

Option 4th is correct

[tex]2 \cdot (x-4)(x+4)[/tex]

Step-by-step explanation:

Using difference of square:

[tex]a^2-b^2=(a-b)(a+b)[/tex]

GCF is the largest number that divide the given polynomial or expression.

Given the expression:

[tex]2x^2-32[/tex]

We have to completely factor the expression.

GCF of [tex]2x^2[/tex] and 32 is, 2

then;

[tex]2 \cdot (x^2-16)[/tex]

Using difference of square

[tex]2 \cdot (x-4)(x+4)[/tex]

therefore, the completely factor of [tex]2x^2-32[/tex] is, [tex]2 \cdot (x-4)(x+4)[/tex]

Answer:

D.  2(x + 4)(x − 4)

Step-by-step explanation:

Final answer:

The greatest common factor of the monomials 16x and 36x is 4x, obtained by finding the common prime factors of 16 and 36 and including the variable x, which is present in both monomials.

Explanation:

To find the greatest common factor of the monomials 16x and 36x, we need to break each monomial into its prime factors and compare them.

Prime factors of 16 are 2 x 2 x 2 x 2.

Prime factors of 36 are 2 x 2 x 3 x 3.

Both 16 and 36 have two 2's in common.

Therefore, the GCF of 16 and 36 is 2 x 2 = 4. Since x is present in both monomials, x is also part of their GCF.

The greatest common factor of the monomials 16x and 36x is therefore 4x.