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2 years ago

7

You have four ice cubes in your cup of lemonade each ice cube has a length of 2 1/4 cm a width of 2 2/3 cm and a height of 2 cm

what is the total volume of all the ice cubes

1 answer:

Margarita [4]2 years ago

4 0

Answer: 6

Step-by-step explanation:

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I do not understand what number I am supposed to use for “y=4x+__. I thought it would be 6 because that’s how many hexagons are

skad [1K]

Based on the table, it should be 2.
Y=4x+2 works for all x and y values in the table.

8 0

2 years ago

Please help me I need help!!!!

Lelu [443]

The answer to your question is C.

6 0

3 years ago

Need help finding this answer

ch4aika [34]

Answer: Option 1

Step-by-step explanation:

1. You know the lenght of both rectangular prism, therefore if both are similar, the scale factor is:

$sf=\frac{12cm}{4cm}\\sf=3$

2. Then the scale factor of the volumes is:

$sf_v=3^{3}\\sf_v=27$

3. Now, you must multiply the volume of the smalller rectangular prism by the scale factor obtained, then you obtain the following result:

$V_2=(24cm^{3})(27)=648cm^{3}$

6 0

3 years ago

Read 2 more answers

Divide 21 by g. Then, subtract 6.

IrinaVladis [17]

21/g - 6 = (21-6g)/g. G

8 0

3 years ago

Read 2 more answers

You are given the parametric equations x=2cos(θ),y=sin(2θ). (a) List all of the points (x,y) where the tangent line is horizonta

vladimir1956 [14]

Answer:

The solutions listed from the smallest to the greatest are:

x: $-\sqrt{2}$ $-\sqrt{2}$ $\sqrt{2}$ $\sqrt{2}$

y: -1 1 -1 1

Step-by-step explanation:

The slope of the tangent line at a point of the curve is:

$m = \frac{\frac{dy}{dt} }{\frac{dx}{dt} }$

$m = -\frac{\cos 2\theta}{\sin \theta}$

The tangent line is horizontal when $m = 0$ . Then:

$\cos 2\theta = 0$

$2\theta = \cos^{-1}0$

$\theta = \frac{1}{2}\cdot \cos^{-1} 0$

$\theta = \frac{1}{2}\cdot \left(\frac{\pi}{2}+i\cdot \pi \right)$ , for all $i \in \mathbb{N}_{O}$

$\theta = \frac{\pi}{4} + i\cdot \frac{\pi}{2}$ , for all $i \in \mathbb{N}_{O}$

The first four solutions are:

x: $\sqrt{2}$ $-\sqrt{2}$ $-\sqrt{2}$ $\sqrt{2}$

y: 1 -1 1 -1

The solutions listed from the smallest to the greatest are:

x: $-\sqrt{2}$ $-\sqrt{2}$ $\sqrt{2}$ $\sqrt{2}$

y: -1 1 -1 1

6 0

2 years ago