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xxTIMURxx [149]
3 years ago
10

Find an equation of the line that satisfies the given conditions. through (−5, −7); perpendicular to the line passing through (−

2, 5) and (2, 3)
Mathematics
1 answer:
murzikaleks [220]3 years ago
7 0

Answer:

The equation would be y = 2x + 3

Step-by-step explanation:

In order to solve this, we first need to find the slope of the line between (-2, 5) and (2, 3). In order to do this, we use the slope formula.

m(slope) = (y2 - y1)/(x2 - x1)

m = (3 - 5)/(2 - -2)

m = -2/4

m = -1/2

Now that we have the original line with a slope of -1/2, we can tell a perpendicular line would have a slope of 2. This is because perpendicular lines have opposite and reciprocal slopes. Now we can use that slope and the given point in point-slope form to get the answer. Be sure to solve for y.

y - y1 = m(x - x1)

y + 7 = 2(x + 5)

y + 7 = 2x + 10

y = 2x + 3

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A survey is being planned to determine the mean amount of time corporation executives watch television. a pilot survey indicated
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Missing question: How many executives should be surveyed?

Solution:
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Where N = Sample population

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Therefore, 554 executives should be surveyed to yield such results.
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3 years ago
John borrowed $ from bank for 3 years at a rate of 10%
garik1379 [7]
<h3>Given:</h3>
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  • T= 3 years
<h3>Note that:</h3>
  • P= Principal amount
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  • T= Time period
<h3>To find:</h3>
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<h3>Solution:</h3>

\large\boxed{Formula: I= \frac{PRT}{100}}

First, we'll have to multiply, principal amount (12500), rate (10) and time period (3).

12500×10×3

=375000

Now, we'll have to divide the amount (375000) by 100.

\frac{375000}{100}

<em>I=$3750</em>

Now, we can find the total amount paid.

\large\boxed{A=SI+P}

Let's substitute according to the formula.

A=3750+12500

<em>A=$16250</em>

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5 0
1 year ago
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
Calculate the slope of the line between (3,7) and (2,7).
goldenfox [79]

Answer:

It’s y2-y1 over x2-x1

Y2 is 5 and y1 is 5

5-5

X2 is 7 and x1 is -2

7- negative 2

5-5 is 0

7- negative 2 is 9

Slope is 0

Step-by-step explanation:

Hope this helps:)

7 0
2 years ago
Read 2 more answers
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