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Kazeer [188]
2 years ago
9

Abigail mixes 2/3 pounds of walnuts with 3/5 pound of dried fruit. To create more of the same mixture, how many pounds of walnut

s does she need to mix with one pound of dried fruit
Mathematics
1 answer:
nignag [31]2 years ago
7 0
1 1/9

To figure out how much walnuts to a pound of dried fruit you’d need to do proportions. So we cross multiply:

2/3 = X
—————- —————-
3/5 = 1

You multiply diagnols 2/3 x 1 = 2/3
3/5 times x is 3/5x.

2/3 = 3/5x. Isolate the variable. Divide 3/5 on both sides cancel out 3/5. 2/3 divide by 3/5 is 2/3 x 5/3 which is 10/9 which equals 1 1/9.

So for 1 lb of dried fruit you need 1 1/9 lb of walnuts to maintain the same mixture
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A nutrition company is marketing a low calorie snack brownie. A serving size of the snack is 3 brownies and has a total of 50 ca
lozanna [386]

Answer:

350 calories

Step-by-step explanation:

divide 50 by 3 then multiply that number by 21.


3 0
3 years ago
Find the equation of the line that passes through the points (-5,7) and (2,3)
Olegator [25]

Answer:

y=-\frac{4}{7}x+4\frac{1}{7}

Step-by-step explanation:

So, in order to solve this problem, I started off by drawing it out. On my graph that I have attached below, I first started out by locating the points (-5,7) and (2,3). Now, this is an optional step, but I highly encourage practicing your graphing skills by solving this problem on graph paper as well. Next, I connected the two points that I just graphed. This is the line that passes through (-5,7) and (2,3).

Now, here is where the actual solving starts. If you haven't already been taught this yet, I will introduce it to you now. I am going to find the equation of this line by filling in what I know in the equation y=mx+b, where m= the slope of the line, and b= y intercept.

Slope of the line: m= \frac{y_{1} - y_{2}  }{x_{1} - x_{2}  } = \frac{7-3}{-5-2} = \frac{4}{-7}= -\frac{4}{7}

If you haven't been taught how to find the slope of a line I recommend you find out.

Substitute the slope into the equation.

y=-\frac{4}{7} x+b\\

Now, we will solve for the 'b,' or y intercept.

We already have x and y values to use: (-5,7) or (2,3). I'll use x=2 and y=3 to solve for the y intercept.

y=-\frac{4}{7} x+b\\\\3=-\frac{4}{7} *2+b\\\\3=-\frac{8}{7} +b\\b=3+\frac{8}{7} \\b=\frac{21}{7}+\frac{8}{7}=\frac{29}{7} =4\frac{1}{7} \\b=4\frac{1}{7}

Last step: substitute the slope and y intercept into y=mx+b.

y=mx+b\\y=-\frac{4}{7}x+4\frac{1}{7}

That is the answer to this problem.

I hope this helps.

3 0
3 years ago
I'm doing Pre-algebra and I don't know how would I figure out the 13th term of the pattern; 100, 91, 82, 73.... without writing
Over [174]
The way we would answer the question would be to express it as

100-9(n-1) with n being the nth term of the pattern.

Why is this? We know we subtract 9 each time and START from 100.
Remember though that 100 is the first term in the sequence. So, we would have (n-1)*9 to subtract 0 from our 1st term and 9 from our second term.

Also, we have 9*(n-1) because we are just subtracting multiples of 9. It is always going to subtract by 9.


If you needed the answer here it is...

The answer to the problem would be 
100-9(13-1)
= 100-9(12)
= 100-108
= -8

5 0
3 years ago
Helppppppppppppppppppppppppppppppp plz
ss7ja [257]

The answer is the second image from left to right (B). Examples of direct and inverse variations are showed in the image below. :)

6 0
2 years ago
"Quadrilateral ABCD underwent a sequence of transformations to give quadrilateral A′B′C′D′. Which transformations could have tak
suter [353]
<span>The quadrilateral ABCD have vertices at points A(-6,4), B(-6,6), C(-2,6) and D(-4,4).
</span>
<span>Translating  10 units down you get points A''(-6,-6), B''(-6,-4), C''(-2,-4) and D''(-4,-6).
</span>
Translaitng <span>8 units to the right you get points A'(2,-6), B'(2,-4), C'(6,-4) and D'(4,-6) that are exactly vertices of quadrilateral A'B'C'D'.
</span><span>
</span><span>Answer: correct choice is B.
</span>

6 0
3 years ago
Read 2 more answers
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