First step, write down the expression on one line, using necessary parentheses.
(2^8*5^(-5)*19^0)^(-2) * (5^(-2)/2^3)^4 *2^28
The apply the PEMDAS rule to simplify according to priority:
1. Parentheses
2. Exponentiation
3. Multiplication and Division (left to right)
4. Addition and Subtraction (left to right)
For exponentiation, it is convenient to convert all negative exponents to positive by taking the reciprocal, for example,
a^(-2)= 1/(a^2)
So
(2^8*5^(-5)*19^0)^(-2) * (5^(-2)/2^3)^4 *2^28
=1/[(2^8*5^(-5)*19^0)^(2)] * (1/[5^2*2^3])^4 *2^28
=5^5/[(2^8*19^0)^(2)] * (1/[5^2*2^3])^4 *2^28
Remember positive numbers raised to power zero equals 1, e.g.
a^0=a for any number a.
so 19^0=1
=5^10/[(2^8)^(2)] * (1/[5^2*2^3])^4 *2^28
Now, powers raised to a power equals the product of the powers, e.g.
(2^8)^2=2^16, (5^2)^4=5^8, (2^3)^4=2^12...
continuing,
=5^10/[2^16] * 1/[5^8*2^12] *2^28
When we have exponents to the same base in the denominator and numerator, we subtract exponents, e.g.
5^5/5^8=5^(5-8)=5^(-3), 2^28/[2^16*2^12]=2^(28-16-12)=2^0=1
so we get rid of many terms this way
continuing
=5^(10-8)*2^(28-16-12)
=5^2*2^0
=25*1
=25
Answer:
Multiply 1/4 six times.
Start from zero and move 1/4 six times.
Product: 1 2/4 <u>Other equivalent fractions:</u> 2/3, 1 4/8, 18/12
Step-by-step explanation:
1. 1/4
2. 1/2 (2/4)
3. 3/4
4. one whole (4/4)
5. 5/4 ( 1 1/4)
6. 6/4 ( 1 2/4) (1 1/2)
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
1. 1/4 + 1/4 = 2/4
2. 2/4 + 1/4 = 3/4
3. 3/4 + 1/4 = 4/4
4. 4/4 + 1/4 = 5/4
5. 5/4 + 1/4 = 6/4
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
1/4 times 6. In other ways..
6/4 1/4 times 6/1 = 6/4
<h3>
Answer: sometimes true</h3>
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Explanation:
The plane P can be thought of a perfectly flat ground. Now imagine a flag pole which represents line GH. If AB is drawn in chalk on the pavement, and this line AB intersects the base of the flagpole, then we've made AB and GH intersect. However, this example shows that GH is <u>not</u> on the plane P.
Is it possible to have GH be in the the plane? Yes. We could easily draw another chalk line on the ground to have it intersect AB somewhere. But as the previous paragraph says, it's also possible that GH is not in the plane.
Therefore, the statement is sometimes true
C is the answer, looking at point N, after it is reflected across the y 1, you will see the point ends up at (4,1)
