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3 years ago

Consider the Piece-Wise graph described by the following equation:

Mathematics

1 answer:

Scilla [17]3 years ago

8 0

Answer:

Hey, Dalilah. I hope I'm not too late, but the answer is (3,5).

Step-by-step explanation:

You have to use the equation that includes the x value of 3.

4x+1 only includes x values that are less than 3.

2x-1 includes x values that are greater than or equal to 3.

I plugged 3 into both of these equations (4x+1; x < 3 & 2x-1; x ≥ 3) in order to see which equation would've been true.

4(3)+1 = 13; 13 < 3 [False] 13 isn't less than 3 -- and 2(3)-1 = 5; 5 ≥ 3 [True] 5 is greater than or equal to 3.

Since the equation 2(3)-1; 5≥3 is true, you should use this equation. I used Demos in order to graph Y=2(3)-1. At the bottom of that I put x=3, and it showed me (3,5).

I'll post the pictures for proof and better clarification if this is confusing, love.

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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

ΔMNO is shown.

JulijaS [17]

Answer:

ABC - AAS

DEF - not enough information

GHI - not enough information

JKL - SAS

Step-by-step explanation:

SAS postulate states that if two sides and the included angle of one triangle are congruent to two sides and the included angle of another triangle, then these two triangles are congruent.

AAS postulate states that if two angles and the non-included side one triangle are congruent to two angles and the non-included side of another triangle, then these two triangles are congruent.

HL postulate states that if the hypotenuse and leg of one right triangle are congruent to the hypotenuse and leg of another right triangle, then the two triangles are congruent.

ASA postulate states that if two angles and the included side of one triangle are congruent to two angles and the included side of another triangle, then the triangles are congruent.

SSS postulate states that if three sides of one triangle are congruent to three sides of another triangle, then the triangles are congruent.

1. In triangles MNO and ABC, there are two congruent sides and non-included angle - AAS

2. In triangles MNO and DEF, there are two congruent sides - there is not enough information

3. In triangles MNO and GHI, there are three congruent angles - there is not enough information

4. In triangles MNO and JKL, there are two congruent sides and included angle - SAS

5 0

4 years ago

Pls help decide what i should type ..

Mashutka [201]

1.) It snowed like it was the North Pole!

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3 years ago

Which lengths could be the sides of a triangle?

Ugo [173]

The answer is A. 17 cm, 5 cm, 13 cm

For the triangle:
a+b > c
b+c > a
a+c > b

Check all choices:
A. 17 cm, 5 cm, 13 cm
a = 17
b = 5
c = 13
17+5 = 22
22 > 13
5+13= 18
18 > 17
17+13=30
30>5

If you check other choices, you will see they are incorrect.

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What does 3 divided by 5 equal to

LiRa [457]

0.6 is the answer.

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