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damaskus [11]
3 years ago
7

(radical) √450 estimated to the nearest hundredth

Mathematics
2 answers:
borishaifa [10]3 years ago
7 0
The correct answer would be 21.21
MAVERICK [17]3 years ago
7 0
EZ brainliest for the other guy lol
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Choose the right classification of 5x + 3x^4 - 7x^3 + 10 by number terms and he degree
jeka57 [31]

Answer:

2. 4th degree polynomial

Step-by-step explanation:

The degree is the highest exponent in the expression. Since the highest exponent is 4, this is a 4th degree polynomial.

8 0
3 years ago
Solve for the value of e.<br> pls help
Zinaida [17]

Answer:

44

Step-by-step explanation:

the steps are in the pic above

4 0
3 years ago
An architect designs a rectangular flower garden such that the width is exactly​ two-thirds of the length. If 390 feet of antiqu
Ksenya-84 [330]

Length (L): L

width (w): (2/3)L

Perimeter (P) = 2L + 2w

               390 = 2(L) + 2(2/3)(L)

               1170 = 6L + 4L

               1170 = 10L

                  117 = L

width (w): (2/3)L   = (2/3)(117) = 2(39)   = 78

Answer: width = 78 ft, length = 117 ft


3 0
3 years ago
Chase has 3 gallons of a solution that is 30% antifeeze that he wants to use to winterize his car. How much pure antifreeze shou
makvit [3.9K]

Let  x gallons be the amount of pure antifreeze that should be added to the 30% solution to produce a solution that is 65% antifreeze. Then the total amount of antifreeze solution will be x+3 gallons.

There are 30% of pure antifreeze in 3 gallons of solution, then

3 gallons - 100%,

a gallons - 30%,

where a gallons is the amount of pure antifreeze in given solution.

Mathematically,

\dfrac{3}{a}=\dfrac{100}{30},\\ \\a=\dfrac{3\cdot 30}{100}=0.9\ gallons.

Now in new solution there will be x+0.9 gallons of pure antefreeze.

x+3 gallons - 100%,

x+0.9 - 65%

or

\dfrac{x+3}{x+0.9}=\dfrac{100}{65},\\ \\65(x+3)=100(x+0.9),\\ \\65x+195=100x+90,\\ \\35x=105,\\ \\x=3\ gallons.

Answer: he should add 3 gallons of pure antifreeze.

8 0
3 years ago
Can somebody explain how these would be done? The selected answer is incorrect, and I was told "Nice try...express the product b
trapecia [35]

Answer:

Solution ( Second Attachment ) : - 2.017 + 0.656i

Solution ( First Attachment ) : 16.140 - 5.244i

Step-by-step explanation:

Second Attachment : The quotient of the two expressions would be the following,

6\left[\cos \left(\frac{2\pi }{5}\right)+i\sin \left(\frac{2\pi \:}{5}\right)\right] ÷ 2\sqrt{2}\left[\cos \left(\frac{-\pi }{2}\right)+i\sin \left(\frac{-\pi \:}{2}\right)\right]

So if we want to determine this expression in standard complex form, we can first convert it into trigonometric form, then apply trivial identities. Either that, or we can straight away apply the following identities and substitute,

( 1 ) cos(x) = sin(π / 2 - x)

( 2 ) sin(x) = cos(π / 2 - x)

If cos(x) = sin(π / 2 - x), then cos(2π / 5) = sin(π / 2 - 2π / 5) = sin(π / 10). Respectively sin(2π / 5) = cos(π / 2 - 2π / 5) = cos(π / 10). Let's simplify sin(π / 10) and cos(π / 10) with two more identities,

( 1 ) \cos \left(\frac{x}{2}\right)=\sqrt{\frac{1+\cos \left(x\right)}{2}}

( 2 ) \sin \left(\frac{x}{2}\right)=\sqrt{\frac{1-\cos \left(x\right)}{2}}

These two identities makes sin(π / 10) = \frac{\sqrt{2}\sqrt{3-\sqrt{5}}}{4}, and cos(π / 10) = \frac{\sqrt{2}\sqrt{5+\sqrt{5}}}{4}.

Therefore cos(2π / 5) = \frac{\sqrt{2}\sqrt{3-\sqrt{5}}}{4}, and sin(2π / 5) = \frac{\sqrt{2}\sqrt{5+\sqrt{5}}}{4}. Substitute,

6\left[ \left\frac{\sqrt{2}\sqrt{3-\sqrt{5}}}{4}+i\left\frac{\sqrt{2}\sqrt{5+\sqrt{5}}}{4}\right] ÷ 2\sqrt{2}\left[\cos \left(\frac{-\pi }{2}\right)+i\sin \left(\frac{-\pi \:}{2}\right)\right]

Remember that cos(- π / 2) = 0, and sin(- π / 2) = - 1. Substituting those values,

6\left[ \left\frac{\sqrt{2}\sqrt{3-\sqrt{5}}}{4}+i\left\frac{\sqrt{2}\sqrt{5+\sqrt{5}}}{4}\right] ÷ 2\sqrt{2}\left[0-i\right]

And now simplify this expression to receive our answer,

6\left[ \left\frac{\sqrt{2}\sqrt{3-\sqrt{5}}}{4}+i\left\frac{\sqrt{2}\sqrt{5+\sqrt{5}}}{4}\right] ÷ 2\sqrt{2}\left[0-i\right] = -\frac{3\sqrt{5+\sqrt{5}}}{4}+\frac{3\sqrt{3-\sqrt{5}}}{4}i,

-\frac{3\sqrt{5+\sqrt{5}}}{4} = -2.01749\dots and \:\frac{3\sqrt{3-\sqrt{5}}}{4} = 0.65552\dots

= -2.01749+0.65552i

As you can see our solution is option c. - 2.01749 was rounded to - 2.017, and 0.65552 was rounded to 0.656.

________________________________________

First Attachment : We know from the previous problem that cos(2π / 5) = \frac{\sqrt{2}\sqrt{3-\sqrt{5}}}{4}, sin(2π / 5) = \frac{\sqrt{2}\sqrt{5+\sqrt{5}}}{4}, cos(- π / 2) = 0, and sin(- π / 2) = - 1. Substituting we receive a simplified expression,

6\sqrt{5+\sqrt{5}}-6i\sqrt{3-\sqrt{5}}

We know that 6\sqrt{5+\sqrt{5}} = 16.13996\dots and -\:6\sqrt{3-\sqrt{5}} = -5.24419\dots . Therefore,

Solution : 16.13996 - 5.24419i

Which rounds to about option b.

7 0
3 years ago
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