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2 years ago

How much greater is the weight of the large box at 3/4lb than the small box at 1 1/8lbs

1 answer:

8 0

Well, is just their difference, let's first convert the mixed fraction to "improper" and then subtract.

$\bf \stackrel{mixed}{1\frac{1}{8}}\implies \cfrac{1\cdot 8+1}{8}\implies \stackrel{improper}{\cfrac{9}{8}}\\\\
-------------------------------\\\\
\cfrac{9}{8}-\cfrac{3}{4}\impliedby \textit{our LCD is then 8}\implies \cfrac{(1\cdot 9)~~-~~(2\cdot 3)}{8}
\\\\\\
\cfrac{9-6}{8}\implies \cfrac{3}{8}$

You might be interested in

2- 1/2n=3n+16<br><br> I need you to solve the whole equation step by step

Alex Ar [27]

Answer:

n = -4

Step-by-step explanation:

2 - 1/2 n = 3n + 16

Combine the constants by subtracting 2 from both sides.

-1/2n = 3n + 14

Combine the variable by subtracting 3n from both sides.

-7/2n = 14

Multiply both sides by 2 to simplify the fraction.

-7n = 28

Isolate n by dividing by -7 on both sides.

n = -4

Check.

2 - 1/2(-4) = 3(-4) + 16

2 + 2 = -12 + 16

4 = 4

Your answer is correct.

3 0

2 years ago

Read 2 more answers

Brian invests £2500 into his bank account.

klio [65]

2.5% per year multiplay by 5 equals 12.5% then multiple 2500 with 12.5 equals 312.5

5 0

2 years ago

If c is the curve given by \mathbf{r} \left( t \right = \left( 1 5 \sin t \right \mathbf{i} \left( 1 3 \sin^{2} t \right \mathbf

jonny [76]

With the curve $C$ parameterized by

$C:\mathbf r(t)=\underbrace{15\sin t}_{x(t)}\,\mathbf i+\underbrace{13\sin^2t}_{y(t)}\,\mathbf j+\underbrace{12\sin^3t}_{z(t)}\,\mathbf k$

with $0\le t\le\dfrac\pi2$ , and given the vector field

$\mathbf f(x,y,z)=x\,\mathbf i+y\,\mathbf j+z\,\mathbf k$

the work done by $\mathbf f$ on a particle moving on along $C$ is given by the line integral

$\displaystyle\int_C\mathbf f\cdot\mathrm d\mathbf r=\int\limits_{t=0}^{t=\pi/2}\mathbf f(x(t),y(t),z(t))\cdot\frac{\mathrm d\mathbf r(t)}{\mathrm dt}\,\mathrm dt$

where

$\mathrm d\mathbf r=(15\cos t\,\mathbf i+26\sin t\cos t\,\mathbf j+36\sin^2t\cos t\,\mathbf k)\,\mathrm dt$

The integral is then

$\displaystyle\int_0^{\pi/2}(15\sin t\,\mathbf i+13\sin^2t\,\mathbf j+12\sin^3t\,\mathbf k)\cdot(15\cos t\,\mathbf i+13\sin2t\,\mathbf j+18\sin t\sin2t\,\mathbf k)\,\mathrm dt$
$=\displaystyle\int_0^{\pi/2}(432\sin^5t\cos t+338\sin^3t\cos t+225\sin t\cos t$
$=269$

6 0

3 years ago

425 is decreased to 319

tatuchka [14]

425 is decreased to 319.

The above statement means that a number should be deducted from 425 to get the difference of 319.

Let x be the unknown number.

425 - x = 319
425 - 319 = x
106 = x

106 is the unknown number.

It can be phrased like this: 425 decreased by 106 is 319.

7 0

2 years ago

In ΔUVW, the measure of ∠W=90°, VW = 32 feet, and WU = 78 feet. Find the measure of ∠V to the nearest degree.

DanielleElmas [232]

Answer: $67.68^{\circ}$

Step-by-step explanation:

Given

$\angle W=90^{\circ}$

$VW=32\ ft$

$WU=78\ ft$

from the figure, we can write

$\Rightarrow \tan v=\dfrac{UW}{VW}\\\\\Rightarrow \tan v=\dfrac{78}{32}\\\\\Rightarrow \tan v=2.437\\\Rightarrow v=\tan^-1(2.437)\\\Rightarrow v=67.68^{\circ}$

7 0

2 years ago