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2 years ago

True or false? "If two angles form a linear pair, they must be supplementary?"

Mathematics

2 answers:

castortr0y [4]2 years ago

6 0

I think it is True

vodka [1.7K]2 years ago

4 0

Answer:

true?

Step-by-step explanation:

im guessing.

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Inequalities Which is the graph of the system x + 3y > -3 and y < 1/2x +1?

OLga [1]

Answer:

the 4th graph.

Step-by-step explanation:

by solving for Y in the first equation, we get the y-intercept of -1 and the second equation has a y-intercept of 1. 4th graph is the only graph with those 2 intercepts

6 0

1 year ago

can someone please help me solve these problems I did accept it on another paper I'm trying to see if I got them all correct I n

Hunter-Best [27]

3.
$3 \div 30 = 10$
sorry all I can help with at 10:15 at night

7 0

2 years ago

For what values of a and m does f(x) have a horizontal asymptote at y = 2 and a vertical asymptote at x = 1? f(x)=2x^m/x+a {a =

iVinArrow [24]

Answer:

See below

Step-by-step explanation:

$f(x)=\frac{2x^{m} }{x+a}$

$f(x)=\frac{2x^{1} }{x-1}$

If x=1, then 1-1=0, which implies a vertical asymptote at x=1 since dividing by 0 is undefined.

3 0

2 years ago

A candidate running for office is putting on a gala as a fundraiser, with hopes of raising more than $7,900 in political contrib

baherus [9]

The simplest interpretation would go a little something like this:

We know that we want the total donation amount to be more than $7,900, so we can set up this inequality to begin with

$D \ \textgreater \ 7,900$

Where D is the total donations raised (in dollars). How do we find D? Well, we just add up the total number of table reservations sold and the total number of single tickets sold. If we let r stand for the number of reservation tickets and s stand for the number of single tickets, then we have

$D=540r+34s$

So, the inequality representing this situation would be

$540r+34s\ \textgreater \ 7,900$

And that would probably be fine for this problem.

Footnote: Of course, if this were a real-life scenario, we'd need to take some additional details into account: How many tables do we have? How many people can be seated at each table?

3 0

2 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

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

3 0

3 years ago

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