Use prime factorization to find the LCM of 6, 25, and 30

If the perimeter of a four sided figure is 124m, determine the maximum area that can be obtained

MAXImum [283]

Just do a square and do 124/4= 31 and then to get the area you do length multiplied by length which is 31 X 31 which equals 961m squared.
Hope this helps

8 0

4 years ago

Laws Of Indicies Can someone please help me with this?

nasty-shy [4]

Answer:

See below

Step-by-step explanation:

$1) : {a}^{5} {b}^{4} {c}^{3} \div {a}^{2} {b}^{3} c \\ \\ = {a}^{5 - 2} {b}^{4 - 3} {c}^{3 - 1} \\ \\ = {a}^{3} {b}^{1} {c}^{2} \\ \\ = {a}^{3} {b} {c}^{2} \\ \\ \\2)\: (x^2 y^3)^3\\\\= x^{2\times 3}y^{3\times 3}\\\|= x^{6}y^{9}\\\\\\3)\:5 {a}^{4} \div ( {a}^{2} \times 3a) \\ \\ = 5 {a}^{4} \div (3 {a}^{2 + 1} ) \\ \\ = 5 {a}^{4} \div (3 {a}^{3} ) \\ \\ = \frac{5}{3} {a}^{4 - 3} \\ \\ = \frac{5}{3} {a} \\ \\ \\ 4)\: \frac{14 {a}^{5} }{2 {a}^{3} \times 7 {a}^{4} } \\ \\ = \frac{14 {a}^{5} }{2 \times 7 {a}^{3 + 4} } \\ \\ = \frac{ \cancel{14} {a}^{5} }{ \cancel{14}{a}^{3 + 4} } \\ \\ = \frac{ {a}^{5} }{ {a}^{7} } \\ \\ = {a}^{5 - 7} \\ \\ = {a}^{ - 2} \\ \\ = \frac{1}{{a}^{2} } \\ \\$

8 0

3 years ago

Read 2 more answers

Write an expression that has four terms. your expression should have three different variables and a constant

Katyanochek1 [597]

A term is a number or variable in a math sentence (such as an expression or equation).

Example:

a + 2b + c + 5

5 is a constant; a,b, c are all variables.

4 0

3 years ago

A password has to use the following format: LDDDDLLL, where L can be any of the upper case letters in the alphabet, and D can be

maw [93]

Answer:

Only vowels and odd numbers:

$P = \frac{5}{26} *\frac{4}{25}* \frac{3}{24}* \frac{2}{23} *\frac{5}{10}* \frac{4}{9} *\frac{3}{8} *\frac{2}{7}$

Spells math:

$P = \frac{1}{26} *\frac{1}{25}* \frac{1}{24}* \frac{1}{23}$

Step-by-step explanation:

We have four letters, so the probability that one letter is a vowel is 5/26 (we have 5 vowels in a total of 26 letters), then the second letter has a probability of 4/25 of being a vowel (1 vowel used), and so on (third letter being vowel = 3/24 and fourth letter being vowel = 2/23)

Then, for the digits, we do the same, one digits has 5/10 probability of being odd, then the second digit has 4/9, the third has 3/8 and the fourth has 2/7.

So the final probability would be:

$P = \frac{5}{26} *\frac{4}{25}* \frac{3}{24}* \frac{2}{23} *\frac{5}{10}* \frac{4}{9} *\frac{3}{8} *\frac{2}{7}$

To find the probability that the password spells the word “MATH", each letter has to be a specific letter, so the first letter has 1/26 probability, the second has 1/25, and so on:

$P = \frac{1}{26} *\frac{1}{25}* \frac{1}{24}* \frac{1}{23}$

6 0

3 years ago

What is the solution to the equation-3(h+5)+2 = 4(h+6)- 9? h= 4 h= -2 h = 2 h = 4

ruslelena [56]

Expanding
-3h-15+2=4h+24-9
-3h-4h=24-9-2+15
-7h=28
h=4

8 0

3 years ago