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3 years ago

What is the value of sin(t) if Cos(t)=4/5 and t is in quadrant 1

Mathematics

2 answers:

lara [203]3 years ago

8 0

3.5 is the answer for sure

Phantasy [73]3 years ago

4 0

Answer:

0.6 or 3/5

Step-by-step explanation:

If cos(t) = 4/5, then, taking the arccos of both sides, t ≈ 36.86989765. Taking the sin of that gives you 0.6, or 3/5 in fraction form.

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Please help mee!! For which value of x does f(x) = 2?

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At a concession stand, a pickle and two bags of chips costs a total of $3.25. Three pickles and four bags of chips costs a total

kolbaska11 [484]

Answer:

A bag of chips costs $1

A pickle costs $1.25

Step-by-step explanation:

P + 2c = 3.25 Start with these two equations

3p + 4c = 7.25

p = -2c + 3.25 Solve for one variable

3(-2c +3.25) + 4c = 7.25 Substitute

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4 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

1. The Senior Class of Great Ridge High School and Ann Arbor High School are planning a trip to Six Flags Theme Park. Great Ridg

Zanzabum

Answer:

A)14v + 16b = 1152

10v + 13b = 925

B)Number of students in a van = 8

Number of students in each bus = 65

C) Elimination method was easier.

Step-by-step explanation:

Let v be the number of students in a van

Let b be number of students in a bus

A) We are told the first school can fill 14 vans and 16 buses with 1152 students

Thus;

14v + 16b = 1152 - - - - - (eq 1)

Also,we are told that the second school can fill 10 vans and 13 buses with 925 students.

Thus;

10v + 13b = 925 - - - - (eq 2)

B) To solve this we will use elimination method.

Multiply eq (1) by 5 and multiply eq (2) by 7 to give;

70v + 80b = 5760 - - - - eq(3)

70v + 91b = 6475 - - - - - eq(4)

Subtract eq(3) from eq(4) to give;

91b - 80b = 6475 - 5760

11b = 715

b = 715/11

b = 65

Put 65 for b in eq(3) to give;

70v + 80(65) = 5760

70v = 5760 - 80(65)

70v = 560

v = 560/70

v = 8

Number of students in a van = 8

Number of students in each bus = 65

C) Elimination method was used because in both equations, we had different digits attached and so it would have been more tedious to use substitution method or graphical method. Thus, elimination method was more easier.

7 0

3 years ago