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1 year ago

How to solve a pair of equations

1 answer:

5 0

I assume what you mean by this question is an equation like this:

1 + 2 x 3 + 4
Or,
3 x 4 + 6 /div 5
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What you'd do:
1+2 x 3+4 First, add either 1+2 or 3+4 which is 3 & 7. then, youd multiply your answers as, 3 x 7 which is, 21.
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Please ask if you have any questions

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The diagram shows an empty cylindrical container. Ana puts a solid cube of side length 8 cm into the container. She then pours 1

marshall27 [118]

Answer:

<h3><u>Cylinder</u></h3>

$\textsf{Volume of a cylinder}=\sf \pi r^2 h \quad\textsf{(where r is the radius and h is the height)}$

Given:

r = 6 cm

h = 18 cm

$\begin{aligned}\implies \sf V &= \sf \pi (6)^2(18)\\& = \sf 648 \pi \: cm^3\end{aligned}$

$\textsf{Volume of a cube}=\sf x^3\quad\textsf{(where x is the side length)}$

Given:

x = 8 cm

$\begin{aligned}\implies \sf V &= 8^3\\& = \sf 512 \: cm^3\end{aligned}$

<h3><u>Volume available to be filled with water</u></h3>

Volume of cylinder - volume of cube

= 684π - 512

= 1532.75204 cm³

1 litre = 1000 cm³

⇒ 1.5 litres = 1000 × 1.5 = 1500 cm³

As 1500 < 1532.75204, the volume of water poured into the container is smaller than the empty space available in the cylinder. Therefore, the water will <u>not</u> come over the top of the container.

4 0

2 years ago

Question Down below.

expeople1 [14]

Answer:

there equal

Step-by-step explanation:

the absolute value of 3 is 3

and the absolute value of -3 is 3

hope this helps

let me know if you need help with anymore

Have a great day!

8 0

3 years ago

Backgammon is a board game for two players in which the playing pieces are moved according to the roll of two dice. players win

Ivahew [28]

The rolls of the dice are independent, i.e. the outcome of the second die doesn't depend in any way on the outcome of the first die.

In cases like this, the probability of two events happening one after the other is the multiplication of the probabilities of the two events.

So, the probability of rolling two 6s is the multiplication of the probabilities of rolling a six with the first die, and another six with the second:

$P(\text{rolling two 6s}) = P(\text{rolling a 6}) \cdot P(\text{rolling a 6}) = \dfrac{1}{6} \cdot \dfrac{1}{6} = \dfrac{1}{36}$

Similarly,

$P(\text{rolling two 3s}) = P(\text{rolling a 3}) \cdot P(\text{rolling a 3}) = \dfrac{1}{6} \cdot \dfrac{1}{6} = \dfrac{1}{36}$

Actually, you can see that the probability of rolling any ordered couple is always 1/36, since the probability of rolling any number on both dice is 1/6:

$P(\text{rolling any ordered couple}) = P(\text{rolling the first number}) \cdot P(\text{rolling the second number}) = \dfrac{1}{6} \cdot \dfrac{1}{6} = \dfrac{1}{36}$