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3 years ago

1. (1 pt) Which is a solution to the equation y=3x+1 ?A. (3, 10) B. (2, 6) C. (1, 5) D. (0, 3)

Mathematics

1 answer:

frozen [14]3 years ago

7 0

Answer:

the easy way to figure this out is to just plug in answers.

start with a, (10)=3(3)+1

10=9+1

10=10

this makes this a true statement

feel free to try the other options but this is your answer

A. (3,10)

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The width of a rectangle is 22 inches and the perimeter is at least 165 inches. what inequality could be used to find the minimu

Lera25 [3.4K]

The answer is D.

We know that a rectangle has two widths that are equal and two lengths that are equal. One width is 22, so the other one is also 22.

If you wanted to find the lengths, you would add both widths together (same as multiplying a width by two) and add that to the two lengths equaled to the perimeter.

So, 22 * 2 + 2x = perimeter of rectangle. We added all four sides together.

We know that the perimeter is at least 165, so 22 * 2 + 2x = 165. Here's the twist. They want the most minimum possible length. So, what answer choice gives you 165 or less for the most minimum or smallest length while still getting to 165?
That is D.
22 * 2 + 2x < = 165.
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4 years ago

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Find f if f′(x)=2x−3/x^4, x>0 and f(1)=2.

slavikrds [6]

$f'(x)=2x-\dfrac3{x^4}$
$\displaystyle\int f'(x)\,\mathrm dx=\int\left(2x-\frac3{x^4}\right)\,\mathrm dx$
$f(x)=x^2+\dfrac1{x^3}+C$

Given that $f(1)=2$ , we get

$2=1^2+\dfrac1{1^3}+C$
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$\implies C=2$

so that

$f(x)=x^2+\dfrac1{x^3}$

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3 years ago

Let z=3+i, then find a. Z² b. |Z| c.<img src="https://tex.z-dn.net/?f=%5Csqrt%7BZ%7D" id="TexFormula1" title="\sq

zysi [14]

Given z = 3 + i, right away we can find

(a) square

z ² = (3 + i )² = 3² + 6i + i ² = 9 + 6i - 1 = 8 + 6i

(b) modulus

|z| = √(3² + 1²) = √(9 + 1) = √10

(d) polar form

First find the argument:

arg(z) = arctan(1/3)

Then

z = |z| exp(i arg(z))

z = √10 exp(i arctan(1/3))

z = √10 (cos(arctan(1/3)) + i sin(arctan(1/3))

Any complex number has 2 square roots. Using the polar form from part (d), we have

√z = √(√10) exp(i arctan(1/3) / 2)

and

√z = √(√10) exp(i (arctan(1/3) + 2π) / 2)

Then in standard rectangular form, we have

$\sqrt z = \sqrt[4]{10} \left(\cos\left(\dfrac12 \arctan\left(\dfrac13\right)\right) + i \sin\left(\dfrac12 \arctan\left(\dfrac13\right)\right)\right)$

and

$\sqrt z = \sqrt[4]{10} \left(\cos\left(\dfrac12 \arctan\left(\dfrac13\right) + \pi\right) + i \sin\left(\dfrac12 \arctan\left(\dfrac13\right) + \pi\right)\right)$

We can simplify this further. We know that z lies in the first quadrant, so

0 < arg(z) = arctan(1/3) < π/2

which means

0 < 1/2 arctan(1/3) < π/4

Then both cos(1/2 arctan(1/3)) and sin(1/2 arctan(1/3)) are positive. Using the half-angle identity, we then have

$\cos\left(\dfrac12 \arctan\left(\dfrac13\right)\right) = \sqrt{\dfrac{1+\cos\left(\arctan\left(\dfrac13\right)\right)}2}$

$\sin\left(\dfrac12 \arctan\left(\dfrac13\right)\right) = \sqrt{\dfrac{1-\cos\left(\arctan\left(\dfrac13\right)\right)}2}$

and since cos(x + π) = -cos(x) and sin(x + π) = -sin(x),

$\cos\left(\dfrac12 \arctan\left(\dfrac13\right)+\pi\right) = -\sqrt{\dfrac{1+\cos\left(\arctan\left(\dfrac13\right)\right)}2}$

$\sin\left(\dfrac12 \arctan\left(\dfrac13\right)+\pi\right) = -\sqrt{\dfrac{1-\cos\left(\arctan\left(\dfrac13\right)\right)}2}$

Now, arctan(1/3) is an angle y such that tan(y) = 1/3. In a right triangle satisfying this relation, we would see that cos(y) = 3/√10 and sin(y) = 1/√10. Then

$\cos\left(\dfrac12 \arctan\left(\dfrac13\right)\right) = \sqrt{\dfrac{1+\dfrac3{\sqrt{10}}}2} = \sqrt{\dfrac{10+3\sqrt{10}}{20}}$

$\sin\left(\dfrac12 \arctan\left(\dfrac13\right)\right) = \sqrt{\dfrac{1-\dfrac3{\sqrt{10}}}2} = \sqrt{\dfrac{10-3\sqrt{10}}{20}}$

$\cos\left(\dfrac12 \arctan\left(\dfrac13\right)+\pi\right) = -\sqrt{\dfrac{10-3\sqrt{10}}{20}}$

$\sin\left(\dfrac12 \arctan\left(\dfrac13\right)+\pi\right) = -\sqrt{\dfrac{10-3\sqrt{10}}{20}}$

So the two square roots of z are

$\boxed{\sqrt z = \sqrt[4]{10} \left(\sqrt{\dfrac{10+3\sqrt{10}}{20}} + i \sqrt{\dfrac{10-3\sqrt{10}}{20}}\right)}$

and

$\boxed{\sqrt z = -\sqrt[4]{10} \left(\sqrt{\dfrac{10+3\sqrt{10}}{20}} + i \sqrt{\dfrac{10-3\sqrt{10}}{20}}\right)}$

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I'm so bad at math please anyone help :(

kondaur [170]

It should be d. , Because if you take the equation 3x+y=9, and make it slope intercept form, it would be y= -3x+9. -3 is the slope, and 9 is the y intercept. D is the answer.

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The squares of the first five whole numbers

antiseptic1488 [7]

Answer:

Square each number: 1 , 2 , 3 , 4 , 5:

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2² = 2 * 2 = 4

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5² = 5 * 5 = 25

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