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3 years ago

Someone please explain I am so confused

Mathematics

1 answer:

Arisa [49]3 years ago

5 0

Answer:

0, 6

Step-by-step explanation:

No matter whatever the number x is, (4/5)^x will not be below 0. You can put (4/5)^x>0. since g(x) has a +6, the first will be 0 and the second would be 6

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5 0

4 years ago

Please help , thanks

Bogdan [553]

Answer:

u want to find volume?

if so then, it is = 150cm³

Step-by-step explanation:

1/3×5×9×10=150cm³

8 0

3 years ago

Find the slope given the graph below:

RoseWind [281]

Answer:

5/3

Step-by-step explanation:

rise over run

(y2-y1)/(x2-x1)

(-2-3)=-5

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6 0

3 years ago

Find the domain and range!!! 10 points!! Help needed

larisa86 [58]

ANSWER

B. Domain is (-∞,∞) and Range is (-∞,∞).

The given function is

$f(x) = \sqrt[3]{x - 6} + 5$

This is function is obtained by shifting the base cubic root function 6 units to the right and 5 units up.

This function is defined for all real values of x.

Therefore the domain is all real numbers.

The range is all real numbers.

The domain for this function becomes the range for the inverse function and the range becomes the domain.

Hence the domain is (-∞,∞) and the range is (-∞,∞).

4 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