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Answer:

#(2+76-:1+9)+4-3*9=64#

Explanation:

In solving #(2+76-:1+9)+4-3*9# according to PEMDAS, parentheses should be solved first, but within parentheses one should divide first and then do addition.

Hence #(2+76-:1+9)+4-3*9#

= #(2+76+9)+4-3*9# and now multiplication first

= #87+4-27#

= #64#

Observe that within parentheses, division / multiplication is performed before addition or subtraction.

Answer:

See a solution process below:

Explanation:

First, using the information provided in the problem, we can convert gallons to liters by multiplying each side of the first equation by #4.8#:

#4.8 xx 1" gallon" = 4.8 xx 3.8"L"#

#4.8" gallons" = 18.24"L"#

We can now take the result in Liters, or #18.24#, and multiply each side of the second equation by this result to obtain the number of mL in 4.8 gallons.

However, the second equation is incorrect. There are 1000mL in 1L. Therefore, we will use this equation: #1"L" = 1000"mL"#

#18.24 xx 1"L" = 18.24 xx 1000"mL"#

#18.24"L" = 18240"mL"#

Therefore, 4.8 gallons is approximately 18,240mL

Answer:

#18," "36," "54," "72," "90," "108," "126," "144)#

Explanation:

Multiplying is a short way of writing repeated addition..

Instead of #7+7+7+7+7#, we can write #5xx7#

#7+7+7+7+7 = 5 xx7 =35#

To find the multiples of #18#, just keep adding on another #18#

#18= 1xx18 =color(blue)(18#

#18+18 = 2 xx 18 =color(blue)(36#

#18+18+18 =3xx18 =color(blue)(54#

#18+18+18+18 = 4 xx 18=color(blue)(72#

and so on ....

The first eight multiples are:

#color(blue)(18," "36," "54," "72," "90," "108," "126," "144)#

You might notice they are all the even multiples of #9#

Answer:

Think you want this as a fraction as it is already a decimal.

See a solution process below:

Explanation:

First, we can write:

#x = 0.0bar54#

Next, we can multiply each side by #100# giving:

#100x = 05.4bar54#

Then we can subtract each side of the first equation from each side of the second equation giving:

#100x - x = 5.4bar54 - 0.0bar54#

We can now solve for #x# as follows:

#100x - 1x = (5.4 + 0.0bar54) - 0.0bar54#

#(100 - 1)x = 5.4 + 0.0bar54 - 0.0bar54#

#99x = 5.4 + (0.bar054 - 0.0bar54)#

#99x = 5.4 + 0#

#99x = 5.4#

#(99x)/color(red)(99) = 5.4/color(red)(99)#

#(color(red)(cancel(color(black)(99)))x)/cancel(color(red)(99)) = (10 xx 5.4)/(10 xx 99)#

#x = 54/990#

#x = (18 xx 3)/(18 xx 55)#

#x = (color(red)(cancel(color(black)(18))) xx 3)/(color(red)(cancel(color(black)(18))) xx 55)#

#x = 3/55#

Answer:

See a solution process below:

Explanation:

You are correct, first you must convert kilograms to pounds.

The conversion factor for kilograms to pounds is:

#1"kg" = 2.20462"lbs"#

To find how many pounds in #1.20"kg"# we must multiply each side of the equation by #color(red)(1.20)# giving:

#color(red)(1.20) xx 1"kg" = color(red)(1.20) xx 2.20462"lbs"#

#1.20"kg" = 2.645544"lbs"#

To find the cost of these carrots we must multiply the number of pounds by the cost per pound giving:

#2.645544"lbs" xx ($0.79)/"lb" =>#

#2.645544color(red)(cancel(color(black)("lbs"))) xx ($0.79)/color(red)(cancel(color(black)("lbs"))) =>#

#2.645544 xx $0.79 =>#

#$2.08997976 =>#

#$2.09# rounded to the nearest cent.

Answer:

#37/24 or 1 13/24#

Explanation:

First, you need a common denominator; in this case, you can just multiply the denominators together:

#3 xx 8 = 24#

In other cases, you would need to find the least common multiple between two the denominators.

Now, you just need to add two fractions with that new common denominator we just found. Keep in mind, you can't just do

#2/24 + 7/24#

because that would change the VALUE of the two fractions (e.g. #2/24# is smaller than #2/3#). So to make #2/3# with a denominator of #24# (but with the same VALUE), you have to multiply the numerator and denominator by #8#, which will give you #16/24#.

For the second fraction, you would multiply the numerator and denominator by #3# to get #21/24#.

So we now have

#16/24 + 21/24#

You can safely add the two fractions because they have a common denominator, resulting in the final answer of #37/24#.

This is an improper fraction (e.g. the numerator is higher than the denominator). You can divide numerator and denominator to get a proper fraction, which in this case will be #1 13/24#.

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