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3 years ago

Write the equation of the graph shown below in factored form.

Mathematics

1 answer:

maksim [4K]3 years ago

6 0

Answer:

where is the graph?

Step-by-step explanation:

I could help.

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Solve the formula do you equals RT for T for the time in hours that it would take van Dyck to travel 26.2 miles of his average s

artcher [175]

Answer:

1.45 hours

Step-by-step explanation:

Total miles to travel by Dyck= 26.2 miles

Average Speed = 18 mph

we have to find the time taken to cover the distance

Total time taken to cover distance= $\frac{Total distance}{Total Time taken}$

Total time taken to cover distance= $\frac{26.2}{18}$

Total time taken= 1.45 hours

Hence, the correct answer is 1.45 hours

7 0

4 years ago

Estimate the quotient of 3.24÷82=

Jobisdone [24]

I rounded it to about 0.040... I havent done this in a while tho so

5 0

4 years ago

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

What is the area of x+2 length and x+5 width

aniked [119]

Answer:

$x^{2}$ + 7 $cm^{2}$

Step-by-step explanation:

x + 2 × x + 5 = $x^{2}$ + 7 $cm^{2}$

Add whatever unit is being used I'm just putting $cm^{2}$

5 0

3 years ago

Fgh is a right triangle true or false

Rudik [331]

Need more info bro lol

4 0

4 years ago

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