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3 years ago

Evaluate the following function for f(-3),

Mathematics

1 answer:

TEA [102]3 years ago

8 0

Answer: −3f

Step-by-step explanation:

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What are the coordinates of the midpoint of CD¯¯¯¯¯ where C(2, −6) and D(4, 10)?

Aleks [24]

You can use the midpoint formulae as I have in the picture below

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3 years ago

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What is the missing polynomial?

german

Answer:

C. 40-4x-12x^2

Step-by-step explanation:

very simple

? – (20 – 4x – 5x2) = 20 – 7x2, so ? = 20 – 7x2 + (20 – 4x – 5x2)

? = 20 – 7x2 + (20 – 4x – 5x2)=20-7x²+20-4x-5x²=40-12x²-4x

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3 years ago

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A rectangle has a length of 20 inches. if its perimeter is 64 inches, what is the area?

zzz [600]

Answer:

$\boxed{ \bold{ \huge{ \boxed{ \sf 240 \: {inches}^{2} }}}}$

Step-by-step explanation:

Given,

Length of a rectangle = 20 inches

Perimeter of a rectangle = 64 inches

Area of a rectangle = ?

Let width of a rectangle be ' w ' .

First, finding the width of a rectangle

$\boxed{ \sf{perimeter = 2(l + w)}}$

plug the values

⇒ $\sf{64 = 2(20 + w)}$

Distribute 2 through the parentheses

⇒ $\sf{64 = 40 + 2w}$

Swap the sides of the equation

⇒ $\sf{40 + 2w = 64}$

Move 2w to right hand side and change it's sign

⇒ $\sf{2w = 64 - 40}$

Subtract 40 from 64

⇒ $\sf{2w = 24}$

Divide both sides of the equation by 2

⇒ $\sf{ \frac{2w}{2} = \frac{24}{2} }$

Calculate

⇒ $\sf{w = 12 \: inches}$

Width of a rectangle ( w ) = 12 inches

Now, finding the area of a rectangle having length of 20 inches and width of 12 inches

$\boxed{ \sf{area \: of \: rectangle = length \: \times \: \: width}}$

plug the values

⇒ $\sf{area \: of \: rectangle =20 \times 12 }$

Multiply the numbers : 20 and 12

⇒ $\sf{area \: of \: rectangle = 240 \: {inches}^{2} }$

Hence, Area of a rectangle = 240 inches²

Hope I helped !

Best regards!

7 0

3 years ago

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

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Please help really confused

marysya [2.9K]

X intercept: (- 1/7, 0)
Y intercept: (0, 1/4)

3 0

3 years ago