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kolbaska11 [484]
4 years ago
11

How manly 1/3 cups are in 1 whole cup?

Mathematics
2 answers:
zepelin [54]4 years ago
7 0
There are three 1/3 cups in one whole cup
Gnesinka [82]4 years ago
4 0
3 1/3 cups are in 1 whole cup
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Write 2 difference between 810,329 and 340,329
disa [49]
Well for one 810,329 is greater than 340,329. for two in the ten thousands place 40,329 is greater than 10,329
3 0
4 years ago
Solve the inequality and graph the solution on a number line. -x/2+3<5/2
Contact [7]

Check the picture below.

7 0
4 years ago
the greater of two numbers is 1 less than 4 times the lesser. three times the lesser number is 4 less than the greater. find the
Komok [63]
<h2>Steps:</h2>

So for this, we will be doing a system of equations. Using the info from our question, our two equations are:

(let x = greater number and y = lesser number)

x=4y-1\\3y=x-4

Now, for this I will be using the substitution method. Since we know that x = 4y - 1, substitute that into the second equation:

3y=4y-1-4

From here we can solve for y. Firstly, combine like terms:

3y=4y-5

Next, subtract 4y on both sides of the equation:

-y=-5

Next, multiply both sides by -1:

y=5

Now that we know the value of y, substitute it into either equation to solve for x:

x=4(5)-1\\x=20-1\\x=19\\\\3(5)=x-4\\15=x-4\\19=x

<h2>Answer:</h2>

In short, <u>5 is the lesser number and 19 is the greater number.</u>

4 0
4 years ago
What is the equation of the line that passes through the points (-1, 7) and (2, 10) in Standard Form?
Usimov [2.4K]

bearing in mind that standard form for a linear equation means

• all coefficients must be integers, no fractions

• only the constant on the right-hand-side

• all variables on the left-hand-side, sorted

• "x" must not have a negative coefficient

\bf (\stackrel{x_1}{-1}~,~\stackrel{y_1}{7})\qquad (\stackrel{x_2}{2}~,~\stackrel{y_2}{10}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{10-7}{2-(-1)}\implies \cfrac{3}{2+1}\implies \cfrac{3}{3}\implies 1

\bf \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-7=1[x-(-1)]\implies y-7=x+1 \\\\\\ y=x+8\implies \boxed{-x+y=8}\implies \stackrel{\textit{standard form}}{x-y=-8}

just to point something out, is none of the options, however -x + y = 8, is one, though improper.

3 0
3 years ago
Consider functions f and g
alexdok [17]

Answer:

A

Step-by-step explanation:

f(x) = \dfrac{x - 16}{x^2 + 6x - 40}

g(x) = \dfrac{1}{x + 10}

f(x) + g(x) =  \dfrac{x - 16}{x^2 + 6x - 40} + \dfrac{1}{x + 10}

f(x) + g(x) =  \dfrac{x - 16}{x^2 + 6x - 40} + \dfrac{(x - 4)}{(x + 10)(x - 4)}

f(x) + g(x) =  \dfrac{x - 16 + x - 4}{x^2 + 6x - 40}

f(x) + g(x) =  \dfrac{2x - 20}{x^2 + 6x - 40}

Answer: A

6 0
2 years ago
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