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Blababa [14]
1 year ago
6

Dahlia is simplifying (3 t) Superscript 4 using these steps: (3 t) Superscript 4 = 3 t times 3 t times 3 t times 3 t = 3 times 3

times 3 times 3 times t times t times t times t Although Dahlia is correct so far, which step could she have used instead to simplify the expression (3 t) Superscript 4? (3 t) Superscript 4 = 3 t Superscript 4 3 Superscript 4 Baseline t = 81 t 4 (3 t) = 12 t 3 Superscript 4 Baseline t Superscript 4 Baseline = 81 t Superscript 4
Mathematics
2 answers:
sergey [27]1 year ago
7 0

Answer:

The answer is 3 to the power of 4 t = 81 t

Step-by-step explanation:

so it is B

mojhsa [17]1 year ago
6 0

Answer:

it is b

Step-by-step explanation:

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2 years ago
Triangle G F E is cut by line segment H J. Line segment H J goes from side G E to F E. The length of G F is 4 x minus 4, the len
alexira [117]

Answer:

HJ = 8  JE = 4

Step-by-step explanation:

it is given that H is the midpoint of GE and J is the midpoint of FE. According to the midpoint theorem the line segment connecting the midpoint of two sides is parallel to the three side and its length is half of the third side. since JH is connecting the midpoints.

HJ= 1/2 (GF)

x + 3 = 1/2 (4x - 4)

x + 3 = 2x - 2

x = 5

^ Thus meaning the value of x is 5.

Now you just fill into your equations:

HJ = x + 3 = (5)  + 3 = 8

JE = x - 1 = (5) - 1 = 4

Therefore, HJ = 8; JE = 4.

5 0
1 year ago
Sometimes a change of variable can be used to convert a differential equation y′=f(t,y) into a separable equation. One common ch
enot [183]

Answer:

y=\frac{-7t^2+22t-7}{7t-22}

Step-by-step explanation:

We are given that

Initial value problem

y'=(t+y)^2-1, y(3)=4

Substitute the value z=t+y

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z=3+4=7

y'=z^2-1

Differentiate z w.r.t t

Then, we get

\frac{dz}{dt}=1+y'

z'=1+z^2-1=z^2

z^{-2}dz=dt

Integrate on both sides

-\frac{1}{z}dz=t+C

z=-\frac{1}{t+C}

Substitute t=3 and z=7

Then, we get

7=-\frac{1}{3+C}

21+7C=-1

7C=-1-21=-22

C=-\frac{22}{7}

Substitute the value of C then we get

z=-\frac{1}{t-\frac{22}{7}}

z=\frac{-7}{7t-22}

y=z-t

y=\frac{-7}{7t-22}-t

y=\frac{-7-7t^2+22t}{7t-22}

y=\frac{-7t^2+22t-7}{7t-22}

8 0
2 years ago
Which statements about the local maximums and minimums for the given function are true? Choose three options.
Nostrana [21]

Answer:

Over the interval [2, 4], the local minimum is –8.

Over the interval [3, 5], the local minimum is –8.

Over the interval [1, 4], the local maximum is 0.

Step-by-step explanation:

The true statements are:

Over the interval [2, 4], the local minimum is –8.

Over the interval [3, 5], the local minimum is –8.

Over the interval [1, 4], the local maximum is 0.  

Lets discuss each option one by one:

Over the interval [1, 3], the local minimum is 0

This is a false statement. Look at the graph. The minimum point given is (3.4,-8). Therefore the local minimum is -8 not 0

Over the interval [2, 4], the local minimum is –8.

This statement is true because the given minimum point is(3.4, -8). Thus the  local minimum is -8 which is true

Over the interval [3, 5], the local minimum is –8.

According to the given minimum point, the local minimum  is -8 which is true

Over the interval [1, 4], the local maximum is 0.

Look at the graph. The maximum point given is (2,0). Thus this statement is true because local maximum is 0.

Over the interval [3, 5], the local maximum is 0.

This is a false statement because there is no maximum point

8 0
2 years ago
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Answer:

a

Step-by-step explanation:

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