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Anastaziya [24]
3 years ago
10

What is the value of the fourth power of 10

Mathematics
2 answers:
Shalnov [3]3 years ago
3 0
10^4 = 10\times 10 \times 10 \times 10 = 100 \times 10 \times 10 = 100 \times 100 = \boxed{10,000}
pentagon [3]3 years ago
3 0
10^4 is 10,000. ten thousand
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PLease answer !!! Find constants $A$ and $B$ such that \[\frac{x + 7}{x^2 - x - 2} = \frac{A}{x - 2} + \frac{B}{x + 1}\] for all
Xelga [282]

Answer:

1. (A,B) = (3,-2)

2. The values of t are: -3, -1

Step-by-step explanation:

Given

\frac{x + 7}{x^2 - x - 2} = \frac{A}{x - 2} + \frac{B}{x + 1}

|t| = 2t + 3

Required

Solve for the unknown

Solving \frac{x + 7}{x^2 - x - 2} = \frac{A}{x - 2} + \frac{B}{x + 1}

Take LCM

\frac{x + 7}{x^2 - x - 2} = \frac{A(x+1) + B(x-2)}{(x - 2)(x-1)}

Expand the denominator

\frac{x + 7}{x^2 - x - 2} = \frac{A(x+1) + B(x-2)}{x^2 - 2x + x -2}

\frac{x + 7}{x^2 - x - 2} = \frac{A(x+1) + B(x-2)}{x^2 - x -2}

Both denominators are equal; So, they can cancel out

x + 7 = A(x+1) + B(x-2)

Expand the expression on the right hand side

x + 7 = Ax + A + Bx - 2B

Collect and Group Like Terms

x + 7 = (Ax + Bx)  + (A - 2B)

x + 7 = (A + B)x + (A - 2B)

By Direct comparison of the left hand side with the right hand side

(A + B)x = x

A - 2B = 7

Divide both sides by x in (A + B)x = x

A + B = 1

Make A the subject of formula

A = 1 - B

Substitute 1 - B for A in A - 2B = 7

1 - B - 2B = 7

1 - 3B = 7

Subtract 1 from both sides

1 - 1 - 3B = 7 - 1

-3B = 6

Divide both sides by -3

B = -2

Substitute -2 for B in A = 1 - B

A = 1 - (-2)

A = 1 + 2

A = 3

Hence;

(A,B) = (3,-2)

Solving |t| = 2t + 3

Because we're dealing with an absolute function; the possible expressions that can be derived from the above expression are;

t = 2t + 3    and   -t = 2t + 3

Solving t = 2t + 3

Make t the subject of formula

t - 2t = 3

-t = 3

Multiply both sides by -1

t = -3

Solving -t = 2t + 3

Make t the subject of formula

-t - 2t = 3

-3t = 3

Divide both sides by -3

t = -1

<em>Hence, the values of t are: -3, -1</em>

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3 years ago
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PtichkaEL [24]
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Zigmanuir [339]

Answer:

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Parker drove a total distance of 4 and 1 over 2 miles during the months of June and July. If Parker only drove 1 over 8 of a mil
never [62]

Answer: 4 and 1 over 2 ÷ 1 over 8


Step-by-step explanation:

Given that : Total distance drove by Parker =4 and 1 over 2 miles=4\frac{1}{2} miles

Distance drove by Parker each day = 1 over 8 mile=\frac{1}{8} mile

To find the total number of days , we need to divide the total distance by the distance drove by him in each day.

Thus, the number of days he went driving=4 and 1 over 2 ÷ 1 over 8

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7/7 - 4/7 = 3/7
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