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3 years ago

If f(x) = |x - 4| + 5, find f(1)

Mathematics

2 answers:

tia_tia [17]3 years ago

6 0

I believe 8 is the answer

Gemiola [76]3 years ago

3 0

F(1) = |1 - 4| + 5 = !-3) + 5 = 3 + 5 = 8 (answer)

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If the functions y=9/2x^2 Was placed in the form y=ax^b , where a and b are real numbers, then which of the following is the val

exis [7]

Answer:

$\frac{13}{2}$

Step-by-step explanation:

We have that:

$a=\frac{9}{2} \\b=2$

Now we can find a+b

$\frac{9}{2}+2=\frac{9+4}{2}=\frac{13}{2}$

5 0

3 years ago

Solve the following equation: 8 - 9x = 27.

bearhunter [10]

The given equation is

$8-9x=27$

To solve for x, we have to subtract 8 to both sides

$8-9x-9 = 27-8 \\ -9x = 19$

Now we need to get rid of -9 from the left side, and for that we do division

$\frac{-9x}{-9} = \frac{19}{-9}$

$x = - \frac{19}{9}$

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3 years ago

Chuck mallory makes a deposit at an atm and walks away with $50.00 in cash and the receipt for the $538.00 total deposit he made

Delicious77 [7]

I think this answer is $176.00

6 0

3 years ago

How is this solved using trig identities (sum/difference)?

GenaCL600 [577]

FIRST PART
We need to find sin α, cos α, and cos β, tan β
α and β is located on third quadrant, sin α, cos α, and sin β, cos β are negative

Determine ratio of ∠α
Use the help of right triangle figure to find the ratio
tan α = 5/12
side in front of the angle/ side adjacent to the angle = 5/12
Draw the figure, see image attached

Using pythagorean theorem, we find the length of the hypotenuse is 13
sin α = side in front of the angle / hypotenuse
sin α = -12/13

cos α = side adjacent to the angle / hypotenuse
cos α = -5/13

Determine ratio of ∠β
sin β = -1/2
sin β = sin 210° (third quadrant)
β = 210°

$cos \beta = -\frac{1}{2} \sqrt{3}$

$tan \beta= \frac{1}{3} \sqrt{3}$

SECOND PART
Solve the questions
Find sin (α + β)
sin (α + β) = sin α cos β + cos α sin β
$sin( \alpha + \beta )=(- \frac{12}{13} )( -\frac{1}{2} \sqrt{3})+( -\frac{5}{13} )( -\frac{1}{2} )$
$sin( \alpha + \beta )=(\frac{12}{26}\sqrt{3})+( \frac{5}{26} )$
$sin( \alpha + \beta )=(\frac{5+12\sqrt{3}}{26})$

Find cos (α - β)
cos (α - β) = cos α cos β + sin α sin β
$cos( \alpha + \beta )=(- \frac{5}{13} )( -\frac{1}{2} \sqrt{3})+( -\frac{12}{13} )( -\frac{1}{2} )$
$cos( \alpha + \beta )=(\frac{5}{26} \sqrt{3})+( \frac{12}{26} )$
$cos( \alpha + \beta )=(\frac{5\sqrt{3}+12}{26} )$

Find tan (α - β)
$tan( \alpha - \beta )= \frac{ tan \alpha-tan \beta }{1+tan \alpha tan \beta }$
$tan( \alpha - \beta )= \frac{ \frac{5}{12} - \frac{1}{2} \sqrt{3} }{1+(\frac{5}{12}) ( \frac{1}{2} \sqrt{3})}$

Simplify the denominator
$tan( \alpha - \beta )= \frac{ \frac{5}{12} - \frac{1}{2} \sqrt{3} }{1+(\frac{5\sqrt{3}}{24})}$
$tan( \alpha - \beta )= \frac{ \frac{5}{12} - \frac{1}{2} \sqrt{3} }{ \frac{24+5\sqrt{3}}{24} }$

Simplify the numerator
$tan( \alpha - \beta )= \frac{ \frac{5}{12} - \frac{6}{12} \sqrt{3} }{ \frac{24+5\sqrt{3}}{24} }$
$tan( \alpha - \beta )= \frac{ \frac{5-6\sqrt{3}}{12} }{ \frac{24+5\sqrt{3}}{24} }$

Simplify the fraction
$tan( \alpha - \beta )= (\frac{5-6\sqrt{3}}{12} })({ \frac{24}{24+5\sqrt{3}})$
$tan( \alpha - \beta )= \frac{10-12\sqrt{3} }{ 24+5\sqrt{3}}$

7 0

3 years ago

What Is The Surface Area Of The Rectangular Prism Below 9, 12, 15

Sholpan [36]

There isn't a rectangular prism. Guess 12.

7 0

3 years ago

Read 2 more answers