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3 years ago

Which choice is equivalent to 2^7 divided by 2^-3

Mathematics

1 answer:

harkovskaia [24]3 years ago

6 0

Answer:

2/3 = 2×4 / 3×4 = 8/12 which is an equivalent fraction of 2/3. Similarly, if we divide the numerator and denominator of 12/18 by 6 we get. 12/18 = 12÷6 / 18÷6 = 2/3. So 12/18 is an equivalent fraction of 2/3 as well.

Step-by-step explanation:

how you like them apples

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PLEASE HELP 100 POINTS

Olin [163]

Answer:

see below

Step-by-step explanation:

$f(x)=2/x\\\\g(x)=2/x\\\\f(g(x))=f(2/x)\\\\f(2/x)=2/(2/x)\\\\f(g(x))=2x/2\\\\f(g(x))=x\\\\$

and since f(x)=g(x) the same process applies to g(f(x))

4 0

2 years ago

Miguel made $286 for 13 hours of work.

Neporo4naja [7]

Answer:

Step-by-step explanation:

The key here is to find out how much he makes per hour. We can find that out using the info given.

286 = 13x so

x = 22 dollars per hour. For 7 hours of work then,

7(22) = $154

8 0

2 years ago

Read 2 more answers

(b) For those values of k, verify that every member of the family of functions y = A sin(kt) + B cos(kt) is also a solution. y =

frutty [35]

Answer:

Check attachment for complete question

Step-by-step explanation:

Given that,

y=Coskt

We are looking for value of k, that satisfies 4y''=-25y

Let find y' and y''

y=Coskt

y'=-kSinkt

y''=-k²Coskt

Then, applying this 4y'"=-25y

4(-k²Coskt)=-25Coskt

-4k²Coskt=-25Coskt

Divide through by Coskt and we assume Coskt is not equal to zero

-4k²=-25

k²=-25/-4

k²=25/4

Then, k=√(25/4)

k= ± 5/2

b. Let assume we want to use this

y=ASinkt+BCoskt

Since k= ± 5/2

y=A•Sin(±5/2t)+ B •Cos(±5/2t)

y'=±5/2ACos(±5/2t)-±5/2BSin(±5/2t)

y''=-25/4ASin(±5/2t)-25/4BCos(±5/2t

Then, inserting this to our equation given to check if it a solution to y=ASinkt+BCoskt

4y''=-25y

For 4y''

4(-25/4ASin(±5/2t)-25/4BCos(±5/2t))

-25A•Sin(±5/2t)-25B•Cos(±5/2t).

Then,

-25y

-25(A•Sin(±5/2t)+ B •Cos(±5/2t))

-25A•Sin(±5/2t) - 25B •Cos(±5/2t)

Then, we notice that, 4y'' is equal to -25y, then we can say that y=Coskt is a solution to y=ASinkt+BCoskt

4 0

2 years ago

Help! How would I solve this trig identity?

NeTakaya

Using simpler trigonometric identities, the given identity was proven below.

<h3>How to solve the trigonometric identity?</h3>

Remember that:

$sec(x) = \frac{1}{cos(x)} \\\\tan(x) = \frac{sin(x)}{cos(x)}$

Then the identity can be rewritten as:

$sec^4(x) - sen^2(x) = tan^4(x) + tan^2(x)\\\\\frac{1}{cos^4(x)} - \frac{1}{cos^2(x)} = \frac{sin^4(x)}{cos^4(x)} + \frac{sin^2(x)}{cos^2(x)} \\\\$

Now we can multiply both sides by cos⁴(x) to get:

$\frac{1}{cos^4(x)} - \frac{1}{cos^2(x)} = \frac{sin^4(x)}{cos^4(x)} + \frac{sin^2(x)}{cos^2(x)} \\\\\\\\cos^4(x)*(\frac{1}{cos^4(x)} - \frac{1}{cos^2(x)}) = cos^4(x)*( \frac{sin^4(x)}{cos^4(x)} + \frac{sin^2(x)}{cos^2(x)})\\\\1 - cos^2(x) = sin^4(x) + cos^2(x)*sin^2(x)\\\\1 - cos^2(x) = sin^2(x)*sin^2(x) + cos^2(x)*sin^2(x)$