Can you add a whole number to a number with a variable

1 answer:

7 0

No, u cant. Because they are not like terms

Examples :
1 + 2v....cannot add or subtract because they are not like terms
2v + 4v = 6v...u can if they have exactly the same variable
3x^2 + 2x...cant
x^2 + 4x^2...can...= 5x^2
3 + 3 ....can
2 + 5x...cant

now u can multiply and divide one...u just cant add or subtract them

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Write and solve a real-world additional problem including the sum of a rational number and it's additive inverse?

Harrizon [31]

If Kayla has a red ribbon with length 25 9 m and piece of 5 7 m is cut from it and gave to Sania . Find how much long ribbon left with her?

Kayla has a ribbon of length = 25 9 m.
Part of ribbon give to Sania = 5 7 m.
So, the length of the ribbon left =25 9 m - 5 7 m = 175 63 = 130 63 m
So , the ribbon left with Kayla is 2.06 m .

7 0

3 years ago

grade 5 math. How would I write a ratio for a project when asked to compare prices from two stores, IE 32 slices of texas toast

REY [17]

To compare, think of the ratios as a fraction. so it is 32 slices per $7.99.
You can also say $7.99 per 32 slices.
so it can be ̲7̲.̲9̲9̲ ̲ or the other way around.
32
To find the unit price, or the price of 1 item, you have to divide the price by the amount.
So 7.99 divided by 32 is 0.249687, but we don't have thousandths of a cent, so we round to the hundredth.
$0.25 is the unit price, or the price of 1 toast, rounded to the nearest cent.

So let's try another one, this time we have 2 different rates, and we will compare them.

So let's say we are buying cheese. Store A sells 12 blocks of cheese for $14.99, and Store B sells 20 blocks for $23.99.
So first we have to find the unit rate for each.

Store A Store B
14.99 per 12 23.99 per 20

 ̲1̲4̲.̲9̲9̲ ̲ ̲2̲3̲.̲9̲9̲ ̲
12 20

14.99 / 12 is 1.24916 23.99 / 20 is 1.1995
$1.25, if rounded to the cent. $1.20, if rounded to the cent
1 block is $1.25 in Store A. 1 block is $1.20 in Store B

Store B sells for cheaper.

Hope that helps.

7 0

3 years ago

Solve 10e^2x - 5 =23^x for x.

yawa3891 [41]

$10e^{2x}-5=23e^x\qquad\text{subtract}\ 23e^x\ \text{from both sides}\\\\10e^{2x}-23e^x-5=0\\\\10(e^x)^2-23(e^x)-5=0\\\\\text{substitution:}\ e^x=t > 0\\\\10t^2-23t-5=0\\\\10t^2-25t+2t-5=0\\\\5t(2t-5)+1(2t-5)=0\\\\(2t-5)(5t+1)=0\iff2t-5=0\ \vee\ 5t+1=0\\\\2t=5\ \vee\ 5t=-1\\\\t=\dfrac{5}{2} > 0\ \vee\ t=-\dfrac{1}{5} < 0\\\\\text{therefore}\ e^x=\dfrac{5}{2}\to\ln e^x=\ln\left(\dfrac{5}{2}\right)\\\\\boxed{x=\ln\left(\dfrac{5}{2}\right)}$