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2 years ago

14

For the table, determine whether the relationship is a function. Then represent the relationship using words, an equation, graph

. X Y (0,4) (1,3) (2,2) (3,1)

1 answer:

Flura [38]2 years ago

6 0

Answer:

don't know

Step-by-step explanation:

not really sure

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How do you rewrite (5a) to the negative 2nd power into a positive exponent

timurjin [86]

1 over 5a to the 2nd power

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3 years ago

Which of the following order pairs is a solution of x+1/2y=1?<br><br> (-2,6)<br> (2,-6)<br> (-2,-6)

julsineya [31]

Y/2=1-x

y=2-2x

y(-2)=2-2(-2)=2+4=6, (-2,6)

So (-2, 6) is an ordered pair from x+y/2=1

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3 years ago

Isabella painted her bedroom. She used 2 gallons of paint and finished in 8 hours. At what rate did Isabella paint the room?

GrogVix [38]

Answer: " \frac{1}{4} gal /hr ; or, write as: " 0.25 gal/hr. " .

_______________

Step-by-step explanation:

_______________

This is a "unit rate" problem;

that is: we want to solve for a particular quantity per [single] unit.

_______________

→ in this case, per single "hour" ;

_______________

So we convert:

_______________

" 2 gal per 8 hours" ; to the correct number of:

_?_ gal per [single] hr.

_______________

{<u>Note</u>: "gal" refers to: "gallon(s)" ;

"hr" refers to: "hour(s)" .}.

_______________

→ " $\frac{2 gal}{ 8 hr} = [\frac{2}{8}]$ gal / hr. " ;

_______________

→ " { $\frac{2}{8} =$ [2÷2] / [8÷2] = 1/4 ;

_______________

So, the answer is:

→ " $\frac{1}{4} gal /hr$ " ;

_______________

or: write as decimal:

→ " $\frac{1}{4}$ = (1 ÷ 4) = 0.25 " ;

_______________

write the answer as:

_______________

→ " 0.25 gal/hr. "

_______________

Or, when we had:

→ "2/8" gal/hr ;

_______________

Use calculator:

→ " $\frac{2}{8}$ = "2 ÷ 8" ;

= " 0.25 gal/ hr.

_______________

So: The correct answer is:

_______________

" \frac{1}{4} gal /hr ; or, write as: " 0.25 gal/hr. " ;

_______________

Hope this is helpful to you!

Best wishes in your academic endeavors—

and within the "Brainly" community!

_______________

6 0

3 years ago

Help me please !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

GarryVolchara [31]

X=6 Is your answer :)

3 0

3 years ago

Precalc - trig identities. Decent explanation, please. THX!!!

LuckyWell [14K]

12. Recall the half-angle identity,

$\sin^2\dfrac y2=\dfrac{1-\cos y}2\implies\sin\dfrac y2=\sqrt{\dfrac{1-\cos y}2}$

where we take the positive square root because $y$ is an angle in a right triangle, which means $0^\circ$ , so $0^\circ$ . For such an angle, it's always the case that $\sin\dfrac y2>0$ .

Use the Pythagorean theorem to find the length of the hypotenuse:

$\sqrt{5^2+12^2}=\sqrt{169}=13$

Then

$\sin\dfrac y2=\sqrt{\dfrac{1-\frac5{13}}2}=\sqrt{\dfrac4{13}}=\dfrac2{\sqrt{13}}$

###

13. $x$ is in quadrant II, which means $90^\circ$ , so $45^\circ$ . In other words, $\dfrac x2$ is in quadrant I, so $\sin\dfrac x2>0$ . From the half-angle identity we get

$\sin\dfrac x2=\sqrt{\dfrac{1-\cos x}2}=\sqrt{\dfrac23}$

###

14. Simplification follows from the definitions of each function:

$\sec x=\dfrac1{\cos x}$

$\tan x=\dfrac{\sin x}{\cos x}$

$\csc x=\dfrac1{\sin x}$

So we have

$\sec x\tan x\cos x\csc x=\dfrac{\sin x\cos x}{\cos^2x\sin x}=\dfrac1{\cos x}$

###

15. Use the Pythagorean identity:

$\cos^2x+\sin^2x=1\implies\dfrac{\cos^2x}{\cos^2x}+\dfrac{\sin^2x}{\cos^2x}=\dfrac1{\cos^2x}\implies1+\tan^2x=\sec^2x$

Then

$\tan^3x+\tan x=\tan x(\tan^2x+1)=\tan x\sec^2x$

6 0

3 years ago