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3 years ago

10

7/21 equivalent fractions

}{?21} " alt=" \frac{7}{?21} " align="absmiddle" class="latex-formula">

1 answer:

pashok25 [27]3 years ago

6 0

Answer:

1/103

Step-by-step explanation:

7/721=1/103

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A box with a square base and open top must have a volume of 296352 c m 3 . We wish to find the dimensions of the box that minimi

mestny [16]

Answer:

Base Length of 84cm
Height of 42 cm.

Step-by-step explanation:

Given a box with a square base and an open top which must have a volume of 296352 cubic centimetre. We want to minimize the amount of material used.

Step 1:

Let the side length of the base =x

Let the height of the box =h

Since the box has a square base

Volume, $V=x^2h=296352$

$h=\dfrac{296352}{x^2}$

Surface Area of the box = Base Area + Area of 4 sides

$A(x,h)=x^2+4xh\\$Substitute h=\dfrac{296352}{x^2}\\A(x)=x^2+4x\left(\dfrac{296352}{x^2}\right)\\A(x)=\dfrac{x^3+1185408}{x}$

Step 2: Find the derivative of A(x)

$If\:A(x)=\dfrac{x^3+1185408}{x}\\A'(x)=\dfrac{2x^3-1185408}{x^2}$

Step 3: Set A'(x)=0 and solve for x

$A'(x)=\dfrac{2x^3-1185408}{x^2}=0\\2x^3-1185408=0\\2x^3=1185408\\$Divide both sides by 2\\x^3=592704\\$Take the cube root of both sides\\x=\sqrt[3]{592704}\\x=84$

Step 4: Verify that x=84 is a minimum value

We use the second derivative test

$A''(x)=\dfrac{2x^3+2370816}{x^3}\\$When x=84$\\A''(x)=6$

Since the second derivative is positive at x=84, then it is a minimum point.

Recall:

$h=\dfrac{296352}{x^2}=\dfrac{296352}{84^2}=42$

Therefore, the dimensions that minimizes the box surface area are:

Base Length of 84cm
Height of 42 cm.

5 0

3 years ago

If f(x) = (1/8)(8^x), what is f(3)?

faust18 [17]

Answer:

f(3)=64

Step-by-step explanation:

f(3)=(1/8)(8 cubed)

f(3)= 8 to the negative one times 8 cubed

f(3)= 8 squared

f(3)=64

5 0

3 years ago

Read 2 more answers

Salem High School's winter sports teams WON 72% of their games last season. If the teams played 50 games, how many games did the

Dmitriy789 [7]

Answer

Find out the how many games did the teams lose .

To prove

Formula

$Percentage = \frac{Part\ value\times 100}{Total\ values}$

As given

Salem High School's winter sports teams WON 72% of their games last season.

If the teams played 50 games .

Percentage = 72%

Total value = 50

Put in the formula

$72 = \frac{Number\ of\ games\ win\times 100}{50}$

$Number\ of\ games\ win = \frac{72\times 50}{100}$

Number of games win = 36

Number of games teams loses = Total number of games - number of games win

= 50 - 36

= 14

Therefore the number of games team loses are 14 .

5 0

3 years ago

When the blue trapezoid is translated to the green one, the new coordinates of the point (0, 0) will be

creativ13 [48]

Your point (0,0) on the blue shape, when translated to the green shape, becomes (-6,-8)

6 0

3 years ago

Read 2 more answers

Week 1 2 3 4 5 Water Level (inches) − 1 1 4 1 3 8 2 1 2 − 1 5 8 − 1 3 4 Between which two weeks did the water level change the m

suter [353]

Answer:

The water level change the most in week 3 and week 4

The change = -370 inches

Step-by-step explanation:

Given - Week 1 2 3 4 5

Water Level (inches) − 1 1 4 1 3 8 2 1 2 -1 5 8 -1 3 4

To find - Between which two weeks did the water level change the most? Calculate the change .

Proof -

The formula for change in water level with respect to week be-

Rate of Change in Water Level = Difference in water level / Difference in weeks

Now,

For week 1 and week 2

Rate of change in water level = $\frac{138 - 114}{2 - 1}$ = $\frac{24}{1}$ = 24 inches

Now,

For week 1 and week 3

Rate of change in water level = $\frac{212 - 114}{3 - 1}$ = $\frac{98}{2}$ = 49 inches

Now,

For week 1 and week 4

Rate of change in water level = $\frac{-158 - 114}{4 - 1}$ = $-\frac{272}{3}$ = -90.67 inches

Now,

For week 1 and week 5

Rate of change in water level = $\frac{-134 - 114}{5 - 1}$ = $-\frac{248}{4}$ = -62 inches

Now,

For week 2 and week 3

Rate of change in water level = $\frac{212 - 138}{3 - 2}$ = $\frac{74}{1}$ = 74 inches

Now,

For week 2 and week 4

Rate of change in water level = $\frac{-158 - 138}{4 - 2}$ = $-\frac{296}{2}$ = -148 inches

Now,

For week 2 and week 5

Rate of change in water level = $\frac{-134 - 138}{5 - 2}$ = $-\frac{272}{3}$ = -90.67 inches

Now,

For week 3 and week 4

Rate of change in water level = $\frac{-158 - 212}{4 - 3}$ = $-\frac{370}{1}$ = -370 inches

Now,

For week 3 and week 5

Rate of change in water level = $\frac{-134 - 212}{5 - 3}$ = $-\frac{346}{2}$ = 173 inches

Now,

For week 4 and week 5

Rate of change in water level = $\frac{-134 + 158}{5 - 4}$ = $\frac{24}{1}$ = 24 inches

∴ we get

The water level change the most in week 3 and week 4

The change = -370 inches

Note :

The highest change means which temperature goes the most change in water level. It can be negative also.

So, We have to take the modulus and then find the highest number.

Here, 370 > 173

So, Highest change occurs in Week 3 and Week 4 but not in Week 3 and Week 5.

7 0

2 years ago