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Virty [35]
3 years ago
11

Someone please help me :'(

Mathematics
1 answer:
eimsori [14]3 years ago
8 0
Hello,

Let's assume v=the normal bicycling speed with no wind
and t the same amount of time...

60=(v+2)*t (1)
48=(v-2)*t (2)

(1)/(2)==>60/48=(v+2)/(v-2)

==>5(v-2)=4(v+2)
==>5v-4v=8+10
==>v=18 (mi/h)
 
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Which is the best estimate for 10 drops of liquid?
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The answer is a A. I just took a short test about it.

Step-by-step explanation:

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3 years ago
By which rule are these triangles congruent?<br><br> A) AAS<br> B) ASA<br> C) SAS <br> D) SSS
Greeley [361]

Answer:

C) SAS

Step-by-step explanation:

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7 0
3 years ago
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Please help me to prove this!<br>I need is no.(c). So, please help me do it.<br>​
zloy xaker [14]

Answer:  see proof below

<u>Step-by-step explanation:</u>

Given: A + B + C = 90°                  → A + B = 90° - C

                                                     → C = 90° - (A + B)

Use the Double Angle Identity:      cos 2A = 1 - 2 sin² A

                                                       → sin² A = (1 - cos 2A)/2

Use Sum to Product Identity: cos A + cos B = 2 cos [(A + B)/2] · cos [(A - B)/2]

Use the Product to Sum Identity: cos (A - B) - cos (A + B) = 2 sin A · sin B

Use the Cofunction Identities:      cos (90° - A) = sin A

                                                       sin (90° - A) = cos A

<u>Proof LHS → RHS:</u>

LHS:                       sin² A + sin² B + sin² C

\text{Double Angle:}\qquad \dfrac{1-\cos 2A}{2}+\dfrac{1-\cos 2B}{2}+\sin^2 C\\\\\\.\qquad \qquad \qquad =\dfrac{1}{2}\bigg(2-\cos 2A-\cos 2B\bigg)+\sin^2 C\\\\\\.\qquad \qquad \qquad =1-\dfrac{1}{2}\bigg(\cos 2A+\cos 2B\bigg)+\sin^2 C

\text{Sum to Product:}\quad 1-\dfrac{1}{2}\bigg[2\cos \bigg(\dfrac{2A+2B}{2}\bigg)\cdot \cos \bigg(\dfrac{2A-2B}{2}\bigg)\bigg]+\sin^2 C\\\\\\.\qquad \qquad \qquad =1-\cos (A+B)\cdot \cos (A-B)+\sin^2 C

Given:                1 - cos (90° - C) · cos (A - B) + sin² C

Cofunction:       1 - sin C · cos (A - B) + sin² C

Factor:               1 - sin C [cos (A - B) + sin C]

Given:                1 - sin C[cos (A - B) - sin (90° - (A + B))]

Cofunction:       1 - sin C[cos (A - B) - cos (A + B)]

Sum to Product:       1 - sin C [2 sin A · sin B]

                            = 1 - 2 sin A · sin B · sin C

LHS = RHS: 1 - 2 sin A · sin B · sin C = 1 - 2 sin A · sin B · sin C   \checkmark

6 0
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4 0
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Some one help ts also says explain ur thinking
AysviL [449]

Answer:

Step-by-step explanation:

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