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2 years ago

The table below shows all of the possible outcomes for rolling two six-sided number cubes.

Mathematics

2 answers:

Ede4ka [16]2 years ago

6 0

Answer: your answer would be 36

Masteriza [31]2 years ago

6 0

The answer to this question is D

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PLEASE HELP QUICKLY AS POSSIBLE THANK YOU :)

Eduardwww [97]

Answer:

the answers are true, false, true.

Step-by-step explanation:

4 0

2 years ago

Read 2 more answers

If W(-10, 4), X(-3, -1), and Y(-5, 11) classify ΔWXY by its sides. Show all work to justify your answer.

solniwko [45]

Given:

The vertices of ΔWXY are W(-10, 4), X(-3, -1), and Y(-5, 11).

To find:

Which type of triangle is ΔWXY by its sides.

Solution:

Distance formula:

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

Using distance formula, we get

$WX=\sqrt{(-3-(-10))^2+(-1-4)^2}$

$WX=\sqrt{(-3+10)^2+(-5)^2}$

$WX=\sqrt{(7)^2+(-5)^2}$

$WX=\sqrt49+25}$

$WX=\sqrt{74}$

Similarly,

$XY=\sqrt{\left(-5-\left(-3\right)\right)^2+\left(11-\left(-1\right)\right)^2}=2\sqrt{37}$

$WY=\sqrt{\left(-5-\left(-10\right)\right)^2+\left(11-4\right)^2}=\sqrt{74}$

Now,

$WX=WY$

So, triangle is an isosceles triangles.

and,

$WX^2+WY^2=(\sqrt{74})^2+(\sqrt{74})^2$

$WX^2+WY^2=74+74$

$WX^2+WY^2=148$

$WX^2+WY^2=(2\sqrt{37})^2$

$WX^2+WY^2=WY^2$

So, triangle is right angled triangle.

Therefore, the ΔWXY is an isosceles right angle triangle.

3 0

2 years ago

Help me with this exercise

Svetradugi [14.3K]

$\left(\frac{-3}4\right)^3 = \frac{-3}4 \times \frac{-3}4 \times \frac{-3}4 = \boxed{-\frac{27}{64}}$

3 0

2 years ago

What is the mean of this? 48, 12, 35, 57, 42, 15, 74, 29

malfutka [58]

$\begin{gathered} \text{Mean = }\frac{\sum ^{}_{}X}{n} \\ =\text{ }\frac{48\text{ +12+35+57+42+15+74+29}}{8} \\ =\frac{312}{8} \\ =39 \end{gathered}$

5 0

8 months ago

PLS HELP!!!

stira [4]

Answer: Option 4

Step-by-step explanation:

$\mathbf{u}=\langle 70 \cos 230^{\circ}, 70 \sin 230^{\circ} \rangle\\\\\mathbf{v}=\langle 60\cos 140^{\circ}, 60 \sin 140^{\circ} \rangle\\\\\mathbf{w}=\langle 80 \cos 30^{\circ}, 80 \sin 30^{\circ} \rangle\\\\\\\therefore \mathbf{u}+\mathbf{v}+\mathbf{w}=\langle 70 \cos 230^{\circ}+60 \cos 140^{\circ}+80 \cos 30^{\circ}, 70 \sin 230^{\circ}+60 \sin 140^{\circ}+80 \sin 30^{\circ} \rangle\\\\ \approx \langle -21.675, 24.944 \rangle$

$\implies |\mathbf{u}+\mathbf{v}+\mathbf{w}| \approx \sqrt{(-21.675)^2 + (24.944)^2} \approx 33.046$

Direction: The vector is in the second quadrant, so the direction is about $180^{\circ}-\arctan \left(\frac{24.944}{-21.675} \right) \approx 131^{\circ}$

8 0

1 year ago