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Schach [20]
3 years ago
13

9.4 rounded to the nearest tenths

Mathematics
2 answers:
Julli [10]3 years ago
8 0
The answer is going to be 9
myrzilka [38]3 years ago
7 0

it would be the same bc u don't have the hundredths in the equation of the number


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Find the value of the following expression
Misha Larkins [42]

Looks like the expression is

(2⁸ • 5⁻⁵ • 19⁰) • (5⁻² / 2³)⁴ • 2²⁸

First, any positive number raised to 0 is 1:

(2⁸ • 5⁻⁵ • 1) • (5⁻² / 2³)⁴ • 2²⁸

(2⁸ • 5⁻⁵) • (5⁻² / 2³)⁴ • 2²⁸

Exponents in a denominator can be rewritten as negative exponents in the numerator:

(2⁸ • 5⁻⁵) • (5⁻² • 2⁻³)⁴ • 2²⁸

Distribute the 4th power across the product 5⁻² • 2⁻³ :

(5⁻² • 2⁻³)⁴ = 5⁻⁸ • 2⁻¹²

so we have

(2⁸ • 5⁻⁵) • (5⁻⁸ • 2⁻¹²) • 2²⁸

Multiplication is associative, so we can ignore the parentheses. Combine factors with equal bases and simplify the exponents:

2⁸ • 5⁻⁵ • 5⁻⁸ • 2⁻¹² • 2²⁸

2⁸⁻¹²⁺²⁸ • 5⁻⁵⁻⁸

2²⁴ • 5⁻¹³

Rewrite the negative exponent as a positive one:

2²⁴ / 5¹³

6 0
3 years ago
What’s the missing side length in this right triangle?
Kruka [31]

Answer:

your correct answer is 15

i just did wrong so the person below me is correct

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
What is 15 divided by 3/4
Vikentia [17]

Answer:

Factor the numerator and denominator and cancel the common factors.

20

Step-by-step explanation:

3 0
3 years ago
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In this problem, y = c1ex + c2e−x is a two-parameter family of solutions of the second-order de y'' − y = 0. find a solution of
Nonamiya [84]
<span>y(1) = 0 y'(1) = e y" = c1ex + c2e-x y' = c1ex - c2e-x for solving c1, 0 = c1e1 + c2e-1 this implies that c1 = - (c2/e2) and to solve c2 e1 = (-c2e-2)e1 - c2e-1 e1 = (-2c2e-1) c2= - (e1/2e-1) = - (e2/2) c1 = - (c2/e2) = (e2/2e2) Therefore y =(e2/2e2)ex - (e2/2)e-x</span>
3 0
3 years ago
Choose yes or no to indicate which equation I’m an be used to describe the pattern. Pls help
mafiozo [28]
A=5 B=0 that the answer if you want the answer to be 5...
hope it helps! Sorry if it doesn’t help.
8 0
3 years ago
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