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2 years ago

Question 5

Mathematics

1 answer:

Darina [25.2K]2 years ago

8 0

Answer:

the answer is 12.24 .

Step-by-step explanation:

that answer is right is because when you add of those you get 530 , then you divide by how many numbers there are which is 7 , which equals to 75.7 . when you get that number ( 75.7) you subtract that number to 65,90,85,70,70,95, and 55 . when you get the total of all of those you add those . which is 10.7, 14.3 , 9.3 , 5.7 , 5.7 , 19.3 , and 20.7 . from adding those you get 85.7. you divide by 7 and get 12.24 . hope this helps :) .

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∠1 and ∠2 are supplementary. ∠1=124°∠2=(2x+4)° Select from the drop down menu to correctly answer the question.

Alex_Xolod [135]

Remark
Supplementary angles add up to 180o.

Givens
<1 = 124
<2 = 2x + 4

I imagine you are looking for either x or <2.

Equation
<1 + <2 = 180 Substitute the givens
124 + 2x + 4 = 180 Collect like terms on the left.
128 + 2x = 180 Subtract 128 from both sides
2x = 180 - 128 Collect like terms on the right
2x = 52 divide by 2
x = 52/2
x = 26 <<<<<<<<< answer

<1 = 124
<2 = 2x + 4 = 2*26 + 4
<2 = 52 + 4
<2 = 56 <<<<<<<< answer

We need choices if you want an exact answer.

3 0

3 years ago

Consider the function g (x) = StartFraction x squared + 7 x minus 18 Over x squared + 2 x minus 8 EndFraction. Which type of dis

Scrat [10]

Answer:

D. Removeable

Step-by-step explanation:

ed2021

3 0

3 years ago

Read 2 more answers

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

$\cos\left(\dfrac12 \arctan\left(\dfrac13\right)\right) = \sqrt{\dfrac{1+\dfrac3{\sqrt{10}}}2} = \sqrt{\dfrac{10+3\sqrt{10}}{20}}$