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olga nikolaevna [1]

3 years ago

12

janelle came to the bat 262 times in 91 games. at this rate, how many times should she expect to have at bat in full season of 1

62 games?

1 answer:

adell [148]3 years ago

7 0

Answer:

465 times

Step-by-step explanation:

$r=\frac{262}{91} = 2.87$

In full season of 162 games

$162r=162(2.87)$ = 465

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Draw a square that has a circumference of 12 cm.

Rasek [7]

a square has four equal sides so 12 ÷ 4 = 3, so each side is 3 then, 3+3+3+3 = perimeter.

check the picture below.

6 0

3 years ago

A right triangle has side lengths a, b, and c as shown below.

Musya8 [376]

Answer:

$\bold{\sin x = \frac{b}{c}}\bold{\\\cos x = \frac{a}{c}}\bold{\\\tan x = \frac{b}{a}}$

Step-by-step explanation:

Given that

A right triangle has side lengths a, b, and c

Because you did not attached photo of the right triangle so I will assume that:

Side a is the adjacent (A)
Side b is the opposite (O)
Side c is the hypotenuse (H)

(Please have a look at the attached photo)

To solve for the trigonometric functions of x, we need to recall the ratios they represent as shown below.

$\sin x = \frac{O}{H}\\\cos x = \frac{A}{H}\\\tan x = \frac{O}{A}$

EX: the sine of x is equal to the side opposite of angle x over the hypotenuse. Hence, we have the expressions of the trigonometric functions as shown below:

$\bold{\sin x = \frac{b}{c}}\bold{\\\cos x = \frac{a}{c}}\bold{\\\tan x = \frac{b}{a}}$

Hope it will find you well

4 0

3 years ago

Which integer in the set has the smallest value (-25, -5, -13, 3)? PLZ I NEED A ANWER!! 20 points

garik1379 [7]

Answer:

D -25

Have a good day!

Brainliest?

4 0

2 years ago

Read 2 more answers

At a distance of 10 cm from a presumably isotropic, radioactive source, a pair of students measure 55 cps (cps = counts per seco

professor190 [17]

With an isotropic source, you can think on the radiation like a expanding sphere, where you will measure the same cps in any point while you keep the same distance to the source, ence the radius of the sphere remains constant. So, you know that the relation depends on the radius, also, when the sphere grows, the radiation in it expands in more surface, ence the density of particles will decrease, particularly, like $\frac{1}{r^{2} }$

then, in 15cm you will have 55cps = A/(10*10) where A is the "intensity of the source" whit this you can know the value of A = 55cps*100 $cm^{2}$ .

then, for 22cm, Y = 55cps*100/(22*22) = 20.66 cps.

7 0

3 years ago

MATH HELP PLZ!!!

RoseWind [281]

Answer:

a) $tan (157.5) = \frac{1-cos 315}{sin315}$

b)

$sin (165) =\sqrt{ \frac{1-cos (330) }{2}}$

c)

$sin^{2} (157.5) = \frac{1-cos (315) }{2}$

d)

cos 330° = 1- 2 sin² (165°)

Step-by-step explanation:

<u><em>Step(i):-</em></u>

By using trigonometry formulas

a)

cos2∝ = 2 cos² ∝-1

cos∝ = 2 cos² ∝/2 -1

1+ cos∝ = 2 cos² ∝/2

$cos^{2} (\frac{\alpha }{2}) = \frac{1+cos\alpha }{2}$

b)

cos2∝ = 1- 2 sin² ∝

cos∝ = 1- 2 sin² ∝/2

$sin^{2} (\frac{\alpha }{2}) = \frac{1-cos\alpha }{2}$

<u><em>Step(i):-</em></u>

Given

$tan\alpha = \frac{sin\alpha }{cos\alpha }$

we know that trigonometry formulas

$sin\alpha = 2sin(\frac{\alpha }{2} )cos(\frac{\alpha }{2} )$

1- cos∝ = 2 sin² ∝/2

Given

$tan(\frac{\alpha }{2} ) = \frac{sin(\frac{\alpha }{2} )}{cos(\frac{\alpha }{2}) }$

put ∝ = 315

$tan(\frac{315}{2} ) = \frac{sin(\frac{315 }{2} )}{cos(\frac{315 }{2}) }$

multiply with ' 2 sin (∝/2) both numerator and denominator

$tan (\frac{315}{2} )= \frac{2sin^{2}(\frac{315)}{2} }{2sin(\frac{315}{2} cos(\frac{315}{2}) }$

Apply formulas

$sin\alpha = 2sin(\frac{\alpha }{2} )cos(\frac{\alpha }{2} )$

1- cos∝ = 2 sin² ∝/2

now we get

$tan (157.5) = \frac{1-cos 315}{sin315}$

b)

$sin^{2} (\frac{\alpha }{2}) = \frac{1-cos\alpha }{2}$

put ∝ = 330° above formula

$sin^{2} (\frac{330 }{2}) = \frac{1-cos (330) }{2}$

$sin^{2} (165) = \frac{1-cos (330) }{2}$

$sin (165) =\sqrt{ \frac{1-cos (330) }{2}}$

c )

$sin^{2} (\frac{\alpha }{2}) = \frac{1-cos\alpha }{2}$

put ∝ = 315° above formula

$sin^{2} (\frac{315 }{2}) = \frac{1-cos (315) }{2}$

$sin^{2} (157.5) = \frac{1-cos (315) }{2}$

d)

cos∝ = 1- 2 sin² ∝/2

put ∝ = 330°

$cos 330 = 1 - 2sin^{2} (\frac{330}{2} )$

cos 330° = 1- 2 sin² (165°)

3 0

3 years ago