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Ivenika [448]
3 years ago
5

What is the volume of a 1 cm by 2 cm by 3 cm box? I got 216 cm.

Mathematics
2 answers:
Amanda [17]3 years ago
5 0

Answer:

6 cm³

Step-by-step explanation:

The volume of a box or rectangular prism is l×w×h.

length × width × height

The dimensions of the box are 1cm × 2cm × 3cm

1 × 2 × 3

Multiply the numbers.

= 6

The volume of the box is 6 cm³.

bogdanovich [222]3 years ago
3 0

Answer:

V = 6 cm^3

Step-by-step explanation:

To find the volume you multiply the dimensions of the three sides of the box, length times width times height

The length is 1 cm the width is 2 cm and the height is 3 cm

The units for volume will be cm^3 since we are multiplying cm* cm*cm

V = l*w*h

V = 1*2*3

V = 6 cm^3

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Which expression is equivalent to -16 - 7?
lakkis [162]

Step-by-step explanation:

B. -16 + (-7) ...this is the same as -16-7

3 0
2 years ago
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Archy [21]

(a) If the particle's position (measured with some unit) at time <em>t</em> is given by <em>s(t)</em>, where

s(t) = \dfrac{5t}{t^2+11}\,\mathrm{units}

then the velocity at time <em>t</em>, <em>v(t)</em>, is given by the derivative of <em>s(t)</em>,

v(t) = \dfrac{\mathrm ds}{\mathrm dt} = \dfrac{5(t^2+11)-5t(2t)}{(t^2+11)^2} = \boxed{\dfrac{-5t^2+55}{(t^2+11)^2}\,\dfrac{\rm units}{\rm s}}

(b) The velocity after 3 seconds is

v(3) = \dfrac{-5\cdot3^2+55}{(3^2+11)^2} = \dfrac{1}{40}\dfrac{\rm units}{\rm s} = \boxed{0.025\dfrac{\rm units}{\rm s}}

(c) The particle is at rest when its velocity is zero:

\dfrac{-5t^2+55}{(t^2+11)^2} = 0 \implies -5t^2+55 = 0 \implies t^2 = 11 \implies t=\pm\sqrt{11}\,\mathrm s \imples t \approx \boxed{3.317\,\mathrm s}

(d) The particle is moving in the positive direction when its position is increasing, or equivalently when its velocity is positive:

\dfrac{-5t^2+55}{(t^2+11)^2} > 0 \implies -5t^2+55>0 \implies -5t^2>-55 \implies t^2 < 11 \implies |t|

In interval notation, this happens for <em>t</em> in the interval (0, √11) or approximately (0, 3.317) s.

(e) The total distance traveled is given by the definite integral,

\displaystyle \int_0^8 |v(t)|\,\mathrm dt

By definition of absolute value, we have

|v(t)| = \begin{cases}v(t) & \text{if }v(t)\ge0 \\ -v(t) & \text{if }v(t)

In part (d), we've shown that <em>v(t)</em> > 0 when -√11 < <em>t</em> < √11, so we split up the integral at <em>t</em> = √11 as

\displaystyle \int_0^8 |v(t)|\,\mathrm dt = \int_0^{\sqrt{11}}v(t)\,\mathrm dt - \int_{\sqrt{11}}^8 v(t)\,\mathrm dt

and by the fundamental theorem of calculus, since we know <em>v(t)</em> is the derivative of <em>s(t)</em>, this reduces to

s(\sqrt{11})-s(0) - s(8) + s(\sqrt{11)) = 2s(\sqrt{11})-s(0)-s(8) = \dfrac5{\sqrt{11}}-0 - \dfrac8{15} \approx 0.974\,\mathrm{units}

7 0
2 years ago
Solve the equation 3x + 2 = 4x + 5 using algebra tiles. Which tiles need to be added to both sides to remove the smaller coeffic
Anastasy [175]

Answer:

Step-by-step explanation:

Given the expression

3x + 2 = 4x + 5

1. The smaller coefficient of x is 3

to remove the smaller coefficient

we need to add - 3x to both sides

3x +2 + (-3x) = 4x +5 (-3x)

3x + 2 - 3x = 4x + 5 - 3x

Collecting like terms we have

3x-3x+2= 4x-3x+5

2 = x+ 5

2. The constant on the right side is

5,to remove the constant from the right side of the equation we need to add - 5 to both sides

3x + 2+ (- 5) = 4x + 5 +(-5)

3x+ 2-5 =4x +5-5

3x-3 = 4x

3 0
3 years ago
The table shows a proportional relationship between the number of pounds of grapes
WINSTONCH [101]

Answer:

Pounds of grapes: 8; total cost: $5.52

Step-by-step explanation:

  1. Find the proportional relationship for 1. 1:0.69
  2. You can eliminate 3 off the top
  3. Try 11 first since it is the largest and the table is increasing in pounds. 11:7.59 (incorrect)
  4. Try 8. 8:5.52 (correct)
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3 years ago
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How do I simplify this?
chubhunter [2.5K]
You can multiply each top and bottom by sqrt 15
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