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Aliun [14]
3 years ago
10

Can someome please help me solve this my teacher hasn't done a good with teaching us this.

Mathematics
1 answer:
scoundrel [369]3 years ago
5 0

Answer:

  30500 = 3.05·10^4

Step-by-step explanation:

Your calculator can do this for you. You may need to set the display to scientific notation, if that's the form of the answer you want.

__

This can be computed by converting both numbers to standard form:

  (5·10^2) +(3·10^4)

  = 500 +30000 = 30500 = 3.05·10^4

__

Addition of numbers in scientific notation in general requires that they have the same power of 10. It may be convenient to convert both numbers to the highest power of 10.

  5·10^2 + 3·10^4

  = 0.05·10^4 +3·10^4 . . . . now both have multipliers of 10^4

  = (0.05 +3)·10^4

  = 3.05·10^4

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Does the mixed number 4 3/4 make each equation true? Choose Yes or No for each equation.
zhuklara [117]

Answer:

first one: yes

second one: yes

third one: no

fourth one: yes

Step-by-step explanation:

3 0
3 years ago
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HELP PLEASE<br> Solve the system by substitution<br> -4.5x - 2y = -12.5<br> 3.25x - y = -0.75
krek1111 [17]
From the second equation ,we can know that y=3.25x+0.75
plug it into the first equation
-4.5x-2(3.25x+0.75)=-12.5
-4.5x-6.5x-1.5=-12.5
-11x=-11
x=1
plug it back into the second equation, y=3.25x+0.75=4
the answer would be (1,4)
6 0
3 years ago
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Can someone please help
Rashid [163]

Answer:

It shifted 6 units down

Step-by-step explanation:

Do the graph normally without the k and compare them side by side.

8 0
3 years ago
Observe that equation (3) has constant coefficients. If y1(x) and y2(x) form a fun- damental set of solutions of equation (3), s
WARRIOR [948]

The equations (2) and (3) you referred to are unavailable, but it is clear that you are trying to show that two set of solutions y1 and y2, to a (second-order) differential equation are solutions, and form a fundamental set. This will be explained.

Answer:

SOLUTION OF A DIFFERENTIAL EQUATION.

Two functions y1 and y2 are set to be solutions to a differential equation if they both satisfy the said differential equation.

Suppose we have a differential equation

y'' + py' + qy = r

If y1 satisfies this differential equation, then

y1'' + py1' + qy1 = r

FUNDAMENTAL SET OF DIFFERENTIAL EQUATION.

Two functions y1 and y2 are said to form a fundamental set of solutions to a second-order differential equation if they are linearly independent. The functions are linearly independent if their Wronskian is different from zero.

If W(y1, y2) ≠ 0

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7 0
3 years ago
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x^4 - (x^2-1)(x^2+1)

Use the formula: (a+b)(a-b) = a^2 - b^2

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No matter what the value of 'x' is, the final answer will always be 1.
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3 years ago
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