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3 years ago

5

Compare the modeld for one third and two sixths

1 answer:

DochEvi [55]3 years ago

4 0

1/3 is equal to 2/6
This is because 1/3 converted to 6ths you have to multiply 1/3 times 2 which equals 2/6
1/3 = 2/6

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Solve the exponential function

nikklg [1K]

Answer:

See below.

Step-by-step explanation:

Just like 18, continue with 19-21. Solve for x by rewriting each equation in log form.

$e^x = 4.9\\\\x = ln 4.9\\\\x = 1.59$

When rewriting the form, the base of the exponent becomes the base of the log.

8 0

4 years ago

CAN SOME ONE PLEASE ANSWER THIS RIGHT THANK YOU

Galina-37 [17]

Answer:

2:A number is greater than 17.2

Step-by-step explanation:

the ">" in the inequality represents is greater

so the m> means m is greater than... and then the 17.2 means m is greater than 17.2. The way I learnt it was to imagine the ">" sign is like a crocodile, it wants the bigger number. So the ">" being towards the "m" shows that m is the bigger number.

3 0

2 years ago

Read 2 more answers

Factors and common factors of 24,64,88

babunello [35]

24- 1,2,3,4,6,8,12,24
64-1,2,4,8,16,32,64
88-1,2,4,8,11,22,44,88

Here are all the factors.^^
The common factors are 1,2,3,4,6
Hope this helps!

8 0

3 years ago

dylan started working on homework at 5:45 p.m. it took him 1 hour and 57 minutes to complete it what time did

kkurt [141]

I'm assuming its asking what time he finished? I added it up and it seems to be 7:42 pm.

5 0

3 years ago

It's a question from real and complex numbers which I can't solve. so someone PLZ HeLp

Semmy [17]

Answer:

$\frac{1}{5}$

Step-by-step explanation:

Using the rules of exponents

$a^{m}$ × $a^{n}$ = $a^{(m+n)}$ , $\frac{a^{m} }{a^{n} }$ = $a^{(m-n)}$ , $(a^m)^{n}$ = $a^{mn}$

Simplifying the product of the first 2 terms

$\frac{a^{p^2+pq} }{a^{pq+q^2} }$ × $\frac{a^{q^2+qr} }{a^{qr+r^2} }$

= $a^{p^2-q^2}$ × $a^{q^2-r^2}$

= $a^{p^2-r^2}$

Simplifying the third term

5( $(a^p+r)^{p-r}$

= 5 $a^{(p+r)(p-r)}$ = 5 $a^{(p^2-r^2)}$

Performing the division, that is

$\frac{a^{(p^2-r^2)} }{5a^{(p^2-r^2)} }$ ← cancel $a^{(p^2-r^2)}$ on numerator/ denominator leaves

= $\frac{1}{5}$

4 0

3 years ago