All categories

3 years ago

(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.

You might be interested in

Help me please thank you

iris [78.8K]

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

Read 2 more answers

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

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.

5 0

3 years ago