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Rashid [163]
3 years ago
7

Michael wants to rent a boat and spend less than $53 the boat cost six dollars per Michael wants to rent a boat and spend less t

han $53 the boat cost six dollars per hour and Michael to discount on for seven dollars off what are the possible in the number of hours Vigo could rent the boat
Mathematics
1 answer:
ASHA 777 [7]3 years ago
5 0
Hi hope I helped, ok so first 6×7=42 and its less than 53 so ur answer is 42
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A rectangular fence is 27.8 meters by 34.2 meters.what is its the perimeter answer
storchak [24]

The perimeter of a rectangle is calculated as

Perimeter= 2(length + breadth )

So, Perimeter = 2*length + 2*breadth

Now we are given the length as 34.2 meters and breadth as 27.8 meters

So the perimeter will be

2(34.2+27.8) meters

(2*34.2) + (2*27.8)

which equals 2(62) meters

Hence Perimeter = 124 meters

5 0
3 years ago
The point slope form of the equation of the line that passes through (-5,-1) and (10,-7) is y+7= -2/5(x-10) what is the standard
Alex787 [66]
Standard form for y+7=-2/5(x-10) is y=-7-2/5(x-10)
3 0
2 years 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
Please Help!!! DUE NOW!
Alexxandr [17]

<em>Answer</em><em>:</em><em> </em><em>3</em><em>7</em>

<em>Step</em><em> </em><em>by</em><em> </em><em>step</em><em> </em><em>explanation</em><em>:</em>

<em>y</em><em>+</em><em>2</em><em>9</em><em>+</em><em>4</em><em>0</em><em>+</em><em>2</em><em>y</em><em>=</em><em>1</em><em>8</em><em>0</em><em>°</em><em>(</em><em> </em><em>sum</em><em> </em><em>of</em><em> </em><em>angle</em><em> </em><em>in</em><em> </em><em>stra</em><em>ight</em><em> </em><em>line</em><em>)</em>

<em>or</em><em>,</em><em> </em><em>y</em><em>+</em><em>2</em><em>y</em><em>+</em><em>2</em><em>9</em><em>+</em><em>4</em><em>0</em><em>=</em><em>1</em><em>8</em><em>0</em><em>°</em>

<em>or</em><em>,</em><em>3</em><em>y</em><em>+</em><em>6</em><em>9</em><em>=</em><em>1</em><em>8</em><em>0</em>

<em>or</em><em>,</em><em>3</em><em>y</em><em>=</em><em>1</em><em>8</em><em>0</em><em>-</em><em>6</em><em>9</em>

<em>or</em><em>,</em><em>3</em><em>y</em><em>=</em><em>1</em><em>1</em><em>1</em>

<em>or</em><em>,</em><em>y</em><em>=</em><em>1</em><em>1</em><em>1</em><em>/</em><em>3</em>

<em>y</em><em>=</em><em>3</em><em>7</em>

<em>hope</em><em> </em><em>it</em><em> </em><em>helps</em>

<em>Good</em><em> </em><em>luck</em><em> on</em><em> your</em><em> assignment</em>

5 0
3 years ago
elect answers from the drop-down menus to correctly complete the statements. A scale factor of was applied to the first triangle
snow_tiger [21]
The scale factor is what u multiply the first image by to get the second image. So by dividing the second image by the first image will give u the scale factor.

6 / 1.5 = 4....so there is a scale factor of 4 because 1.5 * 4 = 6
7 0
3 years ago
Read 2 more answers
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