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Ghella [55]
3 years ago
5

(NO LINKS) HELP ASAP! THIS IS DUE TODAY! ILL GIVE BRAINLIEST TO DECENT ANSWER AND 20+ POINTS!

Mathematics
1 answer:
IrinaVladis [17]3 years ago
8 0

Answer:

Part A: a histogram

Step-by-step explanation:

Part A:

A histogram would be the best to use since they gave values that are x - y, or a ranged amount for the score.

Part B:

You would set up each value for the scores on the bottom. You would make the number of students on the left in increments of 1 (1,2,3,4,5 etc. )

You would make the first value ( 0-4 ) go up to 4 students.  The second value ( 5-9) would go to 5. The third value ( 10-14 ) would go to 2. The fourth and fifth values ( 15-19 and 20-24 ) would go to 3.

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Help me please thank you
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SSS - the side they're sharing is also congruent so if you just add in the other congruent sides you get SSS
4 0
3 years ago
PLEASE ANSWER ASAP! The equation of a line is y = 2x + 3. What is the equation of the line that is parallel to the first line an
Alik [6]
B. Y-y1=m(X-X1)
Y--1= 2(X-2)
Y+1= 2x-4
Sub 1
Y= 2x-5

The m is 2 cuz it's parallel to the other equation
7 0
3 years ago
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Let z=3+i, <br>then find<br> a. Z²<br>b. |Z| <br>c.<img src="https://tex.z-dn.net/?f=%5Csqrt%7BZ%7D" id="TexFormula1" title="\sq
zysi [14]

Given <em>z</em> = 3 + <em>i</em>, right away we can find

(a) square

<em>z</em> ² = (3 + <em>i </em>)² = 3² + 6<em>i</em> + <em>i</em> ² = 9 + 6<em>i</em> - 1 = 8 + 6<em>i</em>

(b) modulus

|<em>z</em>| = √(3² + 1²) = √(9 + 1) = √10

(d) polar form

First find the argument:

arg(<em>z</em>) = arctan(1/3)

Then

<em>z</em> = |<em>z</em>| exp(<em>i</em> arg(<em>z</em>))

<em>z</em> = √10 exp(<em>i</em> arctan(1/3))

or

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

(c) square root

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

√<em>z</em> = √(√10) exp(<em>i</em> arctan(1/3) / 2)

and

√<em>z</em> = √(√10) exp(<em>i</em> (arctan(1/3) + 2<em>π</em>) / 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 <em>z</em> lies in the first quadrant, so

0 < arg(<em>z</em>) = arctan(1/3) < <em>π</em>/2

which means

0 < 1/2 arctan(1/3) < <em>π</em>/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(<em>x</em> + <em>π</em>) = -cos(<em>x</em>) and sin(<em>x</em> + <em>π</em>) = -sin(<em>x</em>),

\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 <em>y</em> such that tan(<em>y</em>) = 1/3. In a right triangle satisfying this relation, we would see that cos(<em>y</em>) = 3/√10 and sin(<em>y</em>) = 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 <em>z</em> 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
Read 2 more answers
If you work for an hourly wage, your gross pay is a function of the number of hours that you work. Your hourly wage is $8.50 per
Ket [755]

Answer:

f(n) = 8.5n

Step-by-step explanation:

$8.50 * n = Gross

3 0
3 years ago
HELP ME PLZZZ!!!! URGENT!!!​
finlep [7]

Answer:

Step-by-step explanation:

Given equation is,

x² + (p + 1)x = 5 - 2p

x² + (p + 1)x - (5 - 2p) = 0

x² + (p + 1)x + (2p - 5) = 0

Properties for the roots of a quadratic equation,

1). Quadratic equation will have two real roots, discriminant will be greater than zero. [(b² - 4ac) > 0]

2). If the equation has exactly one root, discriminant will be zero [(b² - 4ac) = 0]

3). If equation has imaginary roots, discriminant will be less than zero [(b² - 4ac) < 0].

Discriminant of the given equation = (p+1)^2-4(1)(2p-5)

For real roots,

(p+1)^2-4(1)(2p-5)>0

p² + 2p + 1 - 8p + 20 > 0

p² - 6p + 21 > 0

For all real values of 'p', given equation will be greater than zero.

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3 years ago
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