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2 years ago

Help ASAP please help

Mathematics

1 answer:

Morgarella [4.7K]2 years ago

5 0

Answer:

Step-by-step explanation:

another isosceles triangle

SOH CAH TOA

Cos(x)=Adj/ Hyp

Hyp*Cos(45) = Adj

26 * $\sqrt{2}/2\\$ = Adj

13 $\sqrt{2}$ both legs

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MATH HELP PLZ!!!

RoseWind [281]

Answer:

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

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

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

cos 330° = 1- 2 sin² (165°)

Step-by-step explanation:

Step(i):-

By using trigonometry formulas

cos2∝ = 2 cos² ∝-1

cos∝ = 2 cos² ∝/2 -1

1+ cos∝ = 2 cos² ∝/2

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

cos2∝ = 1- 2 sin² ∝

cos∝ = 1- 2 sin² ∝/2

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

Step(i):-

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

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

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

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

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

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

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

A school is organising a competition. There are 128 students from the Maths specialist class and 80 students from the English sp

Elden [556K]

Answer:

5 english students in each team

Step-by-step explanation:

So there are multiple ways to do this, you could do trial and error or you could factor out common numbers between the two.

Through trial and error, the lowest possible number of groups that could be divided between the number of students whilst still being able to maintain a whole # was 16 groups

128 math students ÷16 groups = 8 math students/group
80 english students ÷ 16 groups = 5 english students/group

The other way to solve this is to factor out common numbers between the two:

3 0

2 years ago

Which algebraic expression represents “Nicholas tripled his workout time”?

Nata [24]

Answer:

times 3

Step-by-step explanation:

workouttime³

6 0

2 years ago

Read 2 more answers

Which expression is equivalent to 1/5m -20

muminat

There are infinite equivalent expressions. Here are some:

1/5(m-100)

20(1/100m-1)

1/5m-(4•5)

If you expand any of these or any of the terms, you will get an equivalent expression.

7 0

2 years ago

PLEASE HELPPPPPP The perimeter is 50 ft and the length is 15 ft. What's the width?

SVEN [57.7K]

Answer:

Step-by-step explanation:

I assume it's a rectangle, therefore:

P = 50

a = 15 ft

b = ?

P = 2a + 2b

50 = 2 * 15 + 2b

50 = 30 + 2b

2b = 50 - 30

2b = 20

b = 10

8 0

2 years ago