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3 years ago

12

A frost has caused a temporary spike in the price of orange juice. The price is now $10 a gallon. Before the frost, the price wa

s $4 a gallon. How much more expensive is orange juice now? A) 0.25 times b) 1.5 times c) 2.5 times d) 5 times e) 6 times

1 answer:

alex41 [277]3 years ago

6 0

C) 2.5 times more expensive

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Determine whether each of the following functions is even, odd, or neither even nor odd.

Illusion [34]

Answer:

A.) Even.

Step-by-step explanation:

If a function is an even function, then

F(-x) = f(x)

Also, if a function is an odd function, then, f(-x) = -f(x)

You are given the below function

f(x) = 1 + 3x^2 − x^4

Let x = 2

Substitute 2 for x in the function

F(x) = 1 + 3(2)^2 - (2)^4

F(x) = 1 + 3(4) - 16

F(x) = 1 + 12 - 16

F(x) = -3

Also, Substitute -2 for x in the function

F(x) = 1 + 3(-2)^2 - (-2)^4

F(x) = 1 + 3(4) - 16

F(x) = 1 + 12 - 16

F(x) = -3

Since f(-x) = f(x), we can conclude that

F(x) = 1 + 3x^2 - x^4 is even

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3 years ago

Which number is a solution of the inequality

SashulF [63]

x(7-x)-8>8-8

x(7-x)-8>0

8 0

3 years ago

How do I find number 2?

prisoha [69]

Answer:

y = 3/4x +4

Step-by-step explanation:

From point A to point B, you show a rise of 3 units and a run of 4 units. (The rise is the difference in height of the first two squares; the run is the side length of the first square.) The ratio rise/run = 3/4 is the slope of the line you want.

The upper-left corner of the first square is the y-intercept of the line (4). So, in slope-intercept form, the equation of the dotted line is ...

y = mx + b . . . . . m = slope; b = y-intercept

y = 3/4x + 4

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4 years ago

If a is divided by 7 and has a remainder of 6, what is the remainder when 4a is divided by 7

puteri [66]

3 duh that’s so easy

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4 years ago

Find the exact value of cot(15degrees) using half angle formulas?

liubo4ka [24]

Double angle (or half angle, depending how you look at it) identities:

$\cos^2\dfrac x2=\dfrac{1+\cos x}2$

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

$\implies\cot^2\dfrac x2=\dfrac{\cos^2\frac x2}{\sin^2\frac x2}=\dfrac{1+\cos x}{1-\cos x}$

So we have

$\cot^215^\circ=\dfrac{1+\cos30^\circ}{1-\cos30^\circ}=\dfrac{1+\frac{\sqrt3}2}{1-\frac{\sqrt3}2}$

$\implies\cot^215^\circ=7+4\sqrt3$

$\implies\cot15^\circ=\sqrt{7+4\sqrt3}=2+\sqrt3$

Note that when taking the square root, we should take into account that that could yield two possible solutions, but we know $\cos15^\circ>0$ and $\sin15^\circ>0$ , so it's also the case that $\cot15^\circ>0$ .

Also, the reason we have equality in the last step can be explained like so:

$7+4\sqrt3=4+4\sqrt3+3=4+4\sqrt3+(\sqrt3)^2=(2+\sqrt3)^2$

(not unlike the process used to complete the square)

8 0

3 years ago

Read 2 more answers