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2 years ago

5

List numbers which have letter A in their spelling from 1-100?

2 answers:

Agata [3.3K]2 years ago

6 0

Trick question, none of those numbers have an A in them.

so the answer is None.

Have an awesome day!

erastovalidia [21]2 years ago

6 0

Question

List numbers which have letter A in their spelling from 1-100?

Answer:

There is no numbers with "a" from 1-100.

you find the first "a" in 1000.

Step-by-step explanation:

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Jessica rides the bus 8% miles each day.how many miles does she ride in 10 days?

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Answer:

80 miles

Step-by-step explanation:

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2 years ago

HELP PLEASE MATH!! A company is testing tires for wear and tear. A given tire is said to either pass the test (P) or fail the te

Xelga [282]

Answer:

PPF, PFF

Step-by-step explanation:

There are several ways you can list all the possible combinations. A couple of my favorite are a) use a binary counting sequence; b) use a gray code counting sequence.

Using the first method, the binary numbers 000 to 111 can be listed in numerical order as 000, <em>001</em>, 010, <em>011</em>, 100, 101, 110, 111. Letting 0=P and 1=F, the ones missing from your list are the ones in italics in my list.

Using the second method, we change the right-most character, then the middle one, and finally the left-most character so there is one change at a time: 000, <em>001</em>, <em>011</em>, 010, 110, 111, 101, 100.

After you have a list of all possible combinations, it is a simple matter to compare the given list to the list of possibilities to see which are missing.

5 0

3 years ago

Read 2 more answers

I’m Really lost if I could get a answer it would be greatly appreciated

Nataly [62]

<h3>Answers:</h3>

$f(g(x)) = \sqrt{x^2+5}+5\\\\g(f(x)) = x+30+10\sqrt{x-1}$

================================================

Work Shown:

Part 1

$f(x) = \sqrt{x-1}+5\\\\f(g(x)) = \sqrt{g(x)-1}+5\\\\f(g(x)) = \sqrt{x^2+6-1}+5\\\\f(g(x)) = \sqrt{x^2+5}+5\\\\$

Notice how I replaced every x with g(x) in step 2. Then I plugged in g(x) = x^2+6 and simplified.

------------------

Part 2

$g(x) = x^2+6\\\\g(f(x)) = \left(f(x)\right)^2+6\\\\g(f(x)) = \left(\sqrt{x-1}+5\right)^2+6\\\\g(f(x)) = \left(\sqrt{x-1}\right)^2+2*5*\sqrt{x-1}+\left(5\right)^2+6\\\\g(f(x)) = x-1+10\sqrt{x-1}+25+6\\\\g(f(x)) = x+30+10\sqrt{x-1}\\\\$

In step 4, I used the rule (a+b)^2 = a^2+2ab+b^2

In this case, a = sqrt(x-1) and b = 5.

You could also use the box method as a visual way to expand out $\left(\sqrt{x-1}+5\right)^2$

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