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3 years ago

28/32 that equals 7/8

1 answer:

8 0

Divide it by 4/4. You should get 7/8. 7/8 is the same as 28/32. 28/4=7, 32/4=8. Therefore, you can divide it by 4/4.

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A plumber charges $30 for a service call and an additional $25 per hour. Write an equation in slope-intercept form to represent

murzikaleks [220]

Answer:

The equation is c=25h+30 and the answer is 205

Step-by-step explanation:

Plug in 7 for h.

3 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

Read 2 more answers

Test worth 50 points; test score of 78%?

nataly862011 [7]

The person would have scored 39 out of 50. Hope this helps

7 0

3 years ago

What are the values of m and n in the matrix addition below?

Aleksandr [31]

ANSWER

The correct answer is $m=45,n=12$ .

EXPLANATION

We were given the matrix equation;

$\left[\begin{array}{cc}n-1&6\\-19&m+3\end{array}\right] +\left[\begin{array}{cc}-1&0\\16&-8\end{array}\right] =\left[\begin{array}{cc}10&6\\-3&40\end{array}\right]$ .

We must first simplify the Left Hand Side of the equation by adding corresponding entries.

$\left[\begin{array}{cc}n-1+-1&6+0\\-19+16&m+3-8\end{array}\right]=\left[\begin{array}{cc}10&6\\-3&40\end{array}\right]$ .

That is;

$\left[\begin{array}{cc}n-2&6\\-3&m-5\end{array}\right]=\left[\begin{array}{cc}10&6\\-3&40\end{array}\right]$ .

Since the two matrices are equal, their corresponding entries are also equal. we equate corresponding entries and solve for m and n.

This implies that;

$n-2=10$

We got this equation from row one-column one entry of both matrices.

$n=12$

Also, the row three-column three entries of both matrices will give us the equation;

$m-5=40$

$m=45$

Hence the correct answer is $m=45,n=12$ .

The correct option is option 2

6 0

3 years ago

How do you know if two line segments are perpendicular?

Brut [27]

Perpendicular lines, when intersecting, will always create a right angle at the intersection point. A right angle is an angle that is exactly 90 degrees.

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3 years ago

Read 2 more answers