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USPshnik [31]
3 years ago
8

The cafeteria surveyed 50 sixth grade students about their

Mathematics
1 answer:
liubo4ka [24]3 years ago
6 0
Can you attach the image of the table ?
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What is the solution of 5sin2θ=3 for 0≤θ≤2π? Please help.
grigory [225]

Answer:

C) \theta\approx\{0.3218,1.249,3.463,4.3908\}

Step-by-step explanation:

5\sin2\theta=3;\: 0\leq\theta\leq 2\pi\\\\\sin2\theta=\frac{3}{5}\\ \\2\theta\approx\{0.6436,2.498,6.926,8.7816\}\\\\\theta\approx\{0.3218,1.249,3.463,4.3908\}

7 0
2 years ago
Use trigonometric identities to solve each equation within the given domain.
katrin [286]

Recall that the tangent function is defined by

tan(<em>x</em>) = sin(<em>x</em>)/cos(<em>x</em>)

Also recall the double angle identity for sine,

sin(2<em>x</em>) = 2 sin(<em>x</em>) cos(<em>x</em>)

Then the equation is the same as

3 sin(<em>x</em>)/cos(<em>x</em>) = 4 sin(<em>x</em>) cos(<em>x</em>)

Move everything to one side to prepare to factorize:

3 sin(<em>x</em>)/cos(<em>x</em>) - 4 sin(<em>x</em>) cos(<em>x</em>) = 0

sin(<em>x</em>)/cos(<em>x</em>) (3 - 4 cos²(<em>x</em>)) = 0

As long as cos(<em>x</em>) ≠ 0, we can omit the term in the denominator, so we're left with

sin(<em>x</em>) (3 - 4 cos²(<em>x</em>)) = 0

and so

sin(<em>x</em>) = 0   <u>or</u>   3 - 4 cos²(<em>x</em>) = 0

sin(<em>x</em>) = 0   <u>or</u>   cos²(<em>x</em>) = 3/4

sin(<em>x</em>) = 0   <u>or</u>   cos(<em>x</em>) = ±√3/2

On the interval [0, 2<em>π</em>),

• sin(<em>x</em>) = 0 for <em>x</em> = 0 and <em>x</em> = <em>π</em>

• cos(<em>x</em>) = √3/2 for <em>x</em> = <em>π</em>/6 and <em>x</em> = 11<em>π</em>/6

• cos(<em>x</em>) = -√3/2 for <em>x</em> = 5<em>π</em>/6 and <em>x</em> = 7<em>π</em>/6

(None of these <em>x</em> make cos(<em>x</em>) = 0, so we don't have to omit any extraneous solutions.)

6 0
3 years ago
Solve for m GHI if m GHJ = 59° and m JHI = 39º.
Setler79 [48]

Answer:

m GHI = 82°

Step-by-step explanation:

59+39=98

180-98=82 <==== answer

180 is the total amount of degree

7 0
3 years ago
The simplified expression
Romashka-Z-Leto [24]

Answer:

5x^2 y^2

Step-by-step explanation:

We need to use the properties shown below to solve this:

1. \sqrt[n]{x^a} =x^{\frac{a}{n}}

2. \sqrt{x}\sqrt{x}  =x

3.  \sqrt{x} \sqrt{y}=\sqrt{x*y}

Area of a triangle is given by  1/2 * base * height, so we do that and simplify:

A=\frac{1}{2}(\sqrt{5x^3} )(2\sqrt{5xy^4} )\\A=\frac{1}{2}(5x^3)^{\frac{1}{2}}*2*(5xy^4)^{\frac{1}{2}}\\A=\sqrt{5}x^{\frac{3}{2}}*\sqrt{5}\sqrt{x} }  y^2\\A=\sqrt{5} \sqrt{5}x^{\frac{3}{2}} x^{\frac{1}{2}}y^2\\A=5*x^2y^2\\A=5x^2 y^2

6 0
3 years ago
What is the equivlalent expression of 10 plus 10 plus 10​
Nuetrik [128]

Answer:

10 + 10 + 10 = 10 x 3

                   = 30

6 0
3 years ago
Read 2 more answers
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