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Nastasia [14]
3 years ago
5

Which line is a graph of the equation x = 6? Identify the intercept.

Mathematics
2 answers:
Step2247 [10]3 years ago
3 0
Correct answer is D.

Line x = a is parallel with y-axis and passes through the point (a,0).

So our line is parallel with y-axis and passes through the point (6,0).
Umnica [9.8K]3 years ago
3 0
It would be d ...
..........
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URGENT! A ball gets hit upwards from a height of 5ft and an initial velocity of 200f/s. When does the ball hit the ground to the
Vinil7 [7]

Answer:

The equation for the object's height s at time t seconds after launch is s(t) = –4.9t2 + 19.6t + 58.8, where s is in meters. When does the object strike the ground.

Step-by-step explanation:

5 0
4 years ago
If i drive 45 miles per hour how many minutes to drive 1 mile
Galina-37 [17]
45 miles / 1 hour = 45 miles / 60 minutes
Set up a ratio:
45mi/60m = x mi/1m
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5 0
4 years ago
Find the cube roots of 27(cos 330° + i sin 330°)
Aleksandr-060686 [28]

Answer:

See below for all the cube roots

Step-by-step explanation:

<u>DeMoivre's Theorem</u>

Let z=r(cos\theta+isin\theta) be a complex number in polar form, where n is an integer and n\geq1. If z^n=r^n(cos\theta+isin\theta)^n, then z^n=r^n(cos(n\theta)+isin(n\theta)).

<u>Nth Root of a Complex Number</u>

If n is any positive integer, the nth roots of z=rcis\theta are given by \sqrt[n]{rcis\theta}=(rcis\theta)^{\frac{1}{n}} where the nth roots are found with the formulas:

  • \sqrt[n]{r}\biggr[cis(\frac{\theta+360^\circ k}{n})\biggr] for degrees (the one applicable to this problem)
  • \sqrt[n]{r}\biggr[cis(\frac{\theta+2\pi k}{n})\biggr] for radians

for  k=0,1,2,...\:,n-1

<u>Calculation</u>

<u />z=27(cos330^\circ+isin330^\circ)\\\\\sqrt[3]{z} =\sqrt[3]{27(cos330^\circ+isin330^\circ)}\\\\z^{\frac{1}{3}} =(27(cos330^\circ+isin330^\circ))^{\frac{1}{3}}\\\\z^{\frac{1}{3}} =27^{\frac{1}{3}}(cos(\frac{1}{3}\cdot330^\circ)+isin(\frac{1}{3}\cdot330^\circ))\\\\z^{\frac{1}{3}} =3(cos110^\circ+isin110^\circ)

<u>First cube root where k=2</u>

<u />\sqrt[3]{27}\biggr[cis(\frac{330^\circ+360^\circ(2)}{3})\biggr]\\3\biggr[cis(\frac{330^\circ+720^\circ}{3})\biggr]\\3\biggr[cis(\frac{1050^\circ}{3})\biggr]\\3\biggr[cis(350^\circ)\biggr]\\3\biggr[cos(350^\circ)+isin(350^\circ)\biggr]

<u>Second cube root where k=1</u>

\sqrt[3]{27}\biggr[cis(\frac{330^\circ+360^\circ(1)}{3})\biggr]\\3\biggr[cis(\frac{330^\circ+360^\circ}{3})\biggr]\\3\biggr[cis(\frac{690^\circ}{3})\biggr]\\3\biggr[cis(230^\circ)\biggr]\\3\biggr[cos(230^\circ)+isin(230^\circ)\biggr]

<u>Third cube root where k=0</u>

<u />\sqrt[3]{27}\biggr[cis(\frac{330^\circ+360^\circ(0)}{3})\biggr]\\3\biggr[cis(\frac{330^\circ}{3})\biggr]\\3\biggr[cis(110^\circ)\biggr]\\3\biggr[cos(110^\circ)+isin(110^\circ)\biggr]

4 0
2 years ago
In kite ABCD, AC = 12, and CD = 16, what is ED?
anygoal [31]

Answer:

Step-by-step explanation:

EC = 12÷2 = 6

ED = √(CD² - EC²) = √(16² - 6²) = √220 = 2√55 ≅ 14.8 units

6 0
3 years ago
A triangle has a base of x 1/2 m and a height of x 3/4 m. If the area of te triangle is 16m to the power of 2, what are the base
Olenka [21]

Answer:

The base of triangle is  \frac{8}{\sqrt{3}} \ m and the height of triangle is  \frac{12}{\sqrt{3}} \ m.

Step-by-step explanation:

Given:

A triangle has a base of x 1/2 m and a height of x 3/4 m. If the area of the triangle is 16m to the power of 2.

Now, to find the base and height of the triangle.

The base of triangle = x\times\frac{1}{2} =\frac{x}{2} \ m.

The height of triangle = x\times \frac{3}{4} =\frac{3x}{4}\ m.

The area of triangle = 16\ m^2.

Now, we put the formula of area to solve:

Area=\frac{1}{2} \times base\times height

16=\frac{1}{2} \times \frac{x}{2} \times \frac{3x}{4}

16=\frac{3x^2}{16}

<em>Multiplying both sides by 16 we get:</em>

<em />256=3x^2<em />

<em>Dividing both sides by 3 we get:</em>

<em />\frac{256}{3} =x^2<em />

<em>Using square root on both sides we get:</em>

\frac{16}{\sqrt{3}}=x

x=\frac{16}{\sqrt{3}}

Now, by substituting the value of x to get the base and height:

Base=\frac{x}{2}\\\\Base=\frac{\frac{16}{\sqrt{3}}}{2} \\\\Base=\frac{8}{\sqrt{3}} \ m.

<em>So, the base of triangle = </em>\frac{8}{\sqrt{3}} \ m.<em />

Height=x\times\frac{3}{4} \\\\Height=\frac{16}{\sqrt{3}}\times \frac{3}{4} \\\\Height=\frac{12}{\sqrt{3}} \ m.

<em>Thus, the height of triangle =  </em>\frac{12}{\sqrt{3}} \ m.<em />

Therefore, the base of triangle is  \frac{8}{\sqrt{3}} \ m and the height of triangle is  \frac{12}{\sqrt{3}} \ m.

7 0
4 years ago
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