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2 years ago

Here are five number cards.

Mathematics

1 answer:

Luden [163]2 years ago

5 0

Answer:

Card:: 17

Step-by-step explanation:

Mean is nothing but all of the numbers added together then multiplied by the amount of numbers that there are. In this case, the mean for all of them are 8.2. When you take away the 17, you would be let with 2+5+8+9/4 this results in 24/4 which equals 6.

HOPE THIS HELPS!!!

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Given the set of vertices, determine whether parallelogram ABCD is a rhombus, rectangle or square. List all that apply. A(7,-4),

Sloan [31]

Given:

Vertices of a parallelogram ABCD are A(7,-4), B(-1,-4), C(-1,-12), D(7, -12).

To find:

Whether the parallelogram ABCD is a rhombus, rectangle or square.

Solution:

Distance formula:

$D=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Using distance formula, we get

$AB=\sqrt{(-4-(-4))^2+(-1-7)^2}$

$AB=\sqrt{(-4+4)^2+(-8)^2}$

$AB=\sqrt{0+64}$

$AB=8$

Similarly,

$BC=\sqrt{(-1-(-1))^2+(12-(-4))^2}=8$

$CD=\sqrt{(7-(-1))^2+(-12-(-12))^2}=8$

$AD=\sqrt{(7-7)^2+(-12-(-4))^2}=8$

All sides of parallelogram are equal.

$AC=\sqrt{(-1-7)^2+(-12-(-4))^2}=8\sqrt{2}$

$BD=\sqrt{(7-(-1))^2+(-12-(-4))^2}=8\sqrt{2}$

Both diagonals are equal.

Since, all sides are equal and both diagonals are equal, therefore, the parallelogram ABCD is a square.

We know that, a square is special case of rectangles and rhombus.

So, parallelogram ABCD is a rhombus, rectangle or square. Therefore, the correct option is c.

7 0

2 years ago

Here is a photo of the question it self. I hope it makes it easier to understand

iren [92.7K]

Answer:

a. R>S

b. It means that R is to the right of S

c. City R is warmer

Step-by-step explanation:

3 0

2 years ago

23 girls have a mean height of 153cm

DanielleElmas [232]

Answer:

Mean height of 159

Step-by-step explanation:

Add 153 and 165, then divide by 2 to find the average of 159.

6 0

2 years ago

Plssssssssssssssssss slove I don't know how to solve this

Reika [66]

Answer:

I think the second blank is 5 and the third blank is 2.

Input 1: Output 5

Input 5: Output 2

4 0

2 years ago

Please help with geometry. 20 points, and another question worth 98 on my profile!

vovikov84 [41]

Answer:

Part 1) m∠1 =(1/2)[arc SP+arc QR]

Part 2) $PR^{2} =PS*PT$

Part 3) PQ=PR

Part 4) m∠QPT=(1/2)[arc QT-arc QS]

Step-by-step explanation:

Part 1)

we know that

The measure of the inner angle is the semi-sum of the arcs comprising it and its opposite.

we have

m∠1 -----> is the inner angle

The arcs that comprise it and its opposite are arc SP and arc QR

m∠1 =(1/2)[arc SP+arc QR]

Part 2)

we know that

The Intersecting Secant-Tangent Theorem, states that the square of the measure of the tangent segment is equal to the product of the measures of the secant segment and its external secant segment.

In this problem we have that

$PR^{2} =PS*PT$

Part 3)

we know that

The Tangent-Tangent Theorem states that if from one external point, two tangents are drawn to a circle then they have equal tangent segments

In this problem

PQ=PR

Part 4)

we know that

The measurement of the outer angle is the semi-difference of the arcs it encompasses.

In this problem

m∠QPT -----> is the outer angle

The arcs that it encompasses are arc QT and arc QS

therefore

m∠QPT=(1/2)[arc QT-arc QS]

4 0

2 years ago