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3 years ago

Find the difference 5x1/3-3x2/3

2 answers:

6 0

Hi There

First, you have to multiply the numbers. Start off with 5*1/3, you should get 1 2/3, now multiply 3*2/3 you should get 2. So now you have to subtract 1 2/3 - 2. You should get -1/3.

SO, -1/3 is you're final answer

Hope This Helps :)

8_murik_8 [283]3 years ago

4 0

With this problem, you have to do order of operations.
you do the multiplying first (the part in parenthesis)
(5 x 1/3) - (3 x 2/3)
5/3 - 6/3 = -1/3

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A rectangular box has dimensions 12 by 16 by 21, as shown here:

professor190 [17]

Answer: 29

Step-by-step explanation:

d = √(l2 + w2 + h2)

d = √(12^2+ 16^2+ 21^2) = 29

3 0

2 years ago

Inequality <br> Choose all answers that apply

maxonik [38]

Hi there!

»»————-　★　————-««

I believe your answer is:

Option A; $m \leq\frac{3}{5}$

»»————-　★　————-««

Here’s why:

We can use inverse operations to solve for 'm'.

⸻⸻⸻⸻

$\boxed{\text{Solving for 'm':}}\\\\5m+1\leq4\\--------\\\rightarrow 5m + 1 - 1 \leq 4 - 1\\\\\rightarrow 5m\leq3\\\\\rightarrow \frac{5m\leq3}{5}\\\\\rightarrow \boxed{m\leq\frac{3}{5}}$

⸻⸻⸻⸻

Any number that is less than or equal to the value of three-fifths would work.

The option that is less than or equal to (3/5) is option A.

Assuming that there are more options than the options shown, choose the numbers that are less than or equal to (3/5).

»»————-　★　————-««

Hope this helps you. I apologize if it’s incorrect.

8 0

3 years ago

solution of a student got the following marks in 9 questions of a question paper.3,5,7,3,8,0,1,4and6.find the median of these ma

Sophie [7]

The median of these marks is 4.

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3 years ago

Find the critical numbers of the function. (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE

EleoNora [17]

Answer:

0, 10

Step-by-step explanation:

The given function is:

$g(y) = \frac{y-5}{y^2-3y+15}$

According to the quotient rule:

$d(\frac{f(y)}{h(y)}) = \frac{f(y)*h'(y)-h(y)*f'(y)}{h^2(y)}$

Applying the quotient rule:

$g(y) = \frac{y-5}{y^2-3y+15}\\g'(y)=\frac{(y-5)*(2y-3)-(y^2-3y+15)*(1)}{(y^2-3y+15)^2}$

The values for which g'(y) are zero are the critical points:

$g'(y)=0=\frac{(y-5)*(2y-3)-(y^2-3y+15)*(1)}{(y^2-3y+15)^2}\\(y-5)*(2y-3)-(y^2-3y+15)=0\\2y^2-3y-10y+15-y^2+3y-15\\y^2-10y=0\\y=\frac{10\pm \sqrt 100}{2}\\y_1=\frac{10-10}{2}= 0\\y_2=\frac{10+10}{2}=10$

The critical values are y = 0 and y = 10.

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