A rectangular tank is 100cm, 30cm wide and 12cm deep. The volume of liquid it will hold is?

1 answer:

5 0

Assuming that you mean 100cm in length, I think I know the answer, but message me if I’m wrong so that I can correct myself!!!

Volume is L • W • H where L is length, W is width, and H is height.

To find the volume you have to multiply all the dimensions, so 100 * 30 * 12 = 36,000, and since its 3 dimensional we use cm^3 which is centimetres cubed.

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Find the missing side

LiRa [457]

Answer:

-2.44

Step-by-step explanation:

Is a right angle triangle

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CAH is cos=Adjacent divided by Hypotenuse

TOA is tan=Opposite divided by Adjacent

so in this case,cos is the right thing to use

Therefore,x=adjacent

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angle=52°

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3 years ago

The probability that an event will occur is 1. Which of the following best describes the likelihood of the event occurring? (1 p

Semenov [28]

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4 years ago

Read 2 more answers

URGENTE, PERGUNTAS DE MARCAR:

Norma-Jean [14]

Answer:

2a: (c)

5o: (1, 3) and (1,1)

3a: (b)

1a: (d)

4o: (b)

Step-by-step explanation:

2a: the equation of a circle circumference needs to be transformable to the form $(x-x_C)^2 + (y - y_C)^2 = r^2$ where $C(x_C; y_C)$ is the center and r is the radius. (a) and (d) can’t be it because they contain non-zero factors on xy. (b) isn’t an equation.

5o: just put the given (x, y) into the equations and see if it holds. (2, 3) isn’t on the circumference of (1) because $2^2+3^2-2\cdot 2 -4\cdot 3 + 4 = 1 \neq 0$ , (3, 1) isn’t on it either because $3^2 + 1^2 - 2\cdot 3 - 4\cdot 1 + 4 = 4 \neq 0$ .

3a: calculate the value of the left-hand side term of the equation using (x, y) from the given point M. That’s the difference of square distance to the center to the square radius $r^2$ . Thus it’s 0 if the point is on the circumference, negative if inside and positive if outside. You get $3^2 + 4^2 - 6\cdot 3 -2\cdot 4 + 8 = 7$ , positive, so it’s outside the circle.

1a: see definition from 2a. Here, $r^2 = 9$ .

4o: insert the y from the straight line equation (r) (which can be equivalently transformed to $y = x+1$ ) into the circumference equation. If it yields no solution, that’s outside, it there’s exactly one solution, that’s a tangent and if there are two solutions, it’s a secant. $x^2+(x+1)^2+2x - 4(x+1) = x^2 + x^2 + 2x + 1 + 2x - 4x + 4 = 2x^2 + 5 = 0 \iff x^2 = 2.5 \iff x \in \{-\sqrt{2.5}, \sqrt{2.5}\}$ There are two solutions, so it’s a secant.