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4 years ago

What is 5 3/8- 3 7/8

Mathematics

2 answers:

grigory [225]4 years ago

6 0

Conversion: 5 3/8 = 5 · 8 + 3

= 43

Conversion: 3 7/8 = 3 · 8 + 7

= 31

Subtract: 43

- 31

= 43 - 31

= 12

= 3

irina1246 [14]4 years ago

5 0

The answer is 1.5

Hope this helps

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HEELP I'll give brainliest

andrey2020 [161]

Answer:

28.3

Step-by-step explanation:

12/20 = 17/x

cross multiply

12x=340

divide each side by 12

x=28.3333333333.......

round it to nearest tenth

28.3

7 0

3 years ago

Evaluate the integral using the indicated trigonometric substitution. (use c for the constant of integration.) x^3 / sqrt x^2 +

slava [35]

$\displaystyle\int\frac{x^3}{\sqrt{x^2+49}}\,\mathrm dx$

Taking $x=7\tan\theta$ gives $\mathrm dx=7\sec^2\theta\,\mathrm d\theta$ , so that the integral becomes

$\displaystyle\int\frac{(7\tan\theta)^3}{\sqrt{(7\tan\theta)^2+49}}(7\sec^2\theta)\,\mathrm d\theta$
$=\displaystyle7^4\int\frac{\tan^3\theta\sec^3\theta}{\sqrt{49\tan^2\theta+49}}\,\mathrm d\theta$
$=\displaystyle7^3\int\frac{\tan^3\theta\sec^3\theta}{\sqrt{\tan^2\theta+1}}\,\mathrm d\theta$
$=\displaystyle7^3\int\frac{\tan^3\theta\sec^3\theta}{\sqrt{\sec^2\theta}}\,\mathrm d\theta$
$=\displaystyle7^3\int\frac{\tan^3\theta\sec^3\theta}{|\sec\theta|}\,\mathrm d\theta$

When $\sec\theta>0$ , we have

$=\displaystyle7^3\int\frac{\tan^3\theta\sec^3\theta}{\sec\theta}\,\mathrm d\theta$
$=\displaystyle7^3\int\tan^3\theta\sec^2\theta\,\mathrm d\theta$

and from here we can substitute $u=\tan\theta$ to proceed from here.

Quick note: When we set $x=7\tan\theta$ , we are implicitly enforcing $-\dfrac\pi2$ just so that the substitution can be undone later via $\theta=\tan^{-1}\dfrac x7$ . But note that over this domain, we automatically guarantee that $\sec\theta>0$ , so the absolute value bars can be dropped immediately.

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3 years ago

Katrina buys the two vases shown. How do the volumes of the vases compare ?

White raven [17]

They are the same one is just curved.
Hope this helped!!

8 0

3 years ago

The gardening club at school is growing vegetables. The club has 300 square feet of planting beds. Cucumber plants require 6 squ

nadya68 [22]

The question doesn't ask anything in particular, I will show the set of inequalities defined in the problem.

Answer:

System of inequalities:

$6c+4t\leq 300$

$c>0\\t > 0\\c+t\geq 60$

Step-by-step explanation:

Inequalities

The express relations between expressions with a sign other than the equal sign. Common relationals are 'less than', 'greater than', 'not equal to', and many others.

The gardening club at school has 300 square feet of planting beds to plant cucumber and tomato. Each cucumber plant requires 6 square feet of growing space and each tomato plant requires 4 square feet of growing space. We know the total area cannot exceed 300 square feet, so

$6c+4t\leq 300$

Being c and t the number of cucumber and tomato plants respectively.

We also know the students want to plant some of each type of plant and have at least 60 plants. This lead us to more conditions

$c>0\\t > 0\\c+t\geq 60$

Note: The set of inequalities shown is not enough to uniquely solve the problem. We need something to maximize or minimize to optimize c and t

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3 years ago

A 8-foot tall man is standing beside a 32-foot flagpole. The flagpole is

kipiarov [429]

Answer:

The flagpole's shadow is 16.875 feet longer than the man's shadow

Step-by-step explanation:

The total length of the shadow is expressed by taking its actual length by a factor that depends on the position of the sun which is constant for the man too. The expression is as follows;

Height of the shadow=actual height of the flagpole×factor

where;

length of the flagpole's shadow=22.5 feet

actual height of the flagpole=32 feet

factor=f

replacing;

22.5=32×f

32 f=22.5

f=22.5/32

f=0.703125

Using this factor in the expression below;

Length of man's shadow=actual height of man×factor

where;

length of man's shadow=m

actual height of man=8 feet

factor=0.703125

replacing;

length of man's shadow=8×0.703125=5.625 feet

Determine how much longer the flagpole's shadow is as follows;

flagpoles shadow-man's shadow=22.5-5.625=16.875 feet

The flagpole's shadow is 16.875 feet longer than the man's shadow

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3 years ago