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miskamm [114]
3 years ago
7

What's p-(8+m) over -1+p if m is 1 1/9 and p is 1 7/8 please help

Mathematics
1 answer:
djyliett [7]3 years ago
8 0
(p-8-m)/(-1+p)=1 1/9;p=1 7/8
(p+1)/(p+8+m)=1 1/9;p=1 7/8
1/(8+m)=10/9;p=15/8
that's as far as it can be solved
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ella [17]

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3 years ago
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Give two examples of addition of two mixed numbers with different denominators <br> SHOW ALL STEPS
Lady bird [3.3K]

Answer:

First Example: 3 1/2 + 4 3/4, Second Example: 6 3/8 + 7 9/15

Extra Example: 8 4/20 + 3 5/10

Step-by-step explanation:

First Example:

1/2 + 3/4

1/2 is equal to 2/4 so it is now compatible to be added to 3/4.

2/4 + 3/4

= 5/4

Now for the mixed numbers since its 3 and 4, 3 + 4 = 7.

Final answer is 7 5/4.

Second Example:

3/8 + 9/15

9/15 can be reduced to 3/5

Now the equation is 3/8 + 3/5

= 15/40 + 24/40 is an equivalent equation

15/40 + 24/40  

= 39/40

Now for the mixed numbers since its 6 and 7, 6 + 7 = 13

Final answer is 13 39/40.

I am going to include one last example just in case you need one:

Third Example:

4/20 + 5/10

We can reduce these to

1/5 + 1/2

= 2/10 + 5/10 is the equivalent equation

2/10 + 5/10

= 7/10

Now for the mixed numbers since its 8 and 3, 8 + 3 = 11.

Final answer is 11 7/10.

I Hope this helps!

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3 years ago
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Write an equation of the line that passes through (18, 2) and is parallel to the line 3y−x=−12
miskamm [114]

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\bf 3y-x=-12\implies 3y=x-12\implies y=\cfrac{x-12}{3}\implies y = \cfrac{x}{3}-\cfrac{12}{3} \\\\\\ y = \stackrel{\stackrel{m}{\downarrow }}{\cfrac{1}{3}}x-4\qquad \impliedby \begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array}

so we're really looking for the equation of a line whose slope is 1/3 and runs through (18,2)

\bf (\stackrel{x_1}{18}~,~\stackrel{y_1}{2})~\hspace{10em} \stackrel{slope}{m}\implies \cfrac{1}{3} \\\\\\ \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-\stackrel{y_1}{2}=\stackrel{m}{\cfrac{1}{3}}(x-\stackrel{x_1}{18}) \\\\\\ y-2=\cfrac{1}{3}x-6\implies y=\cfrac{1}{3}x-4

3 0
3 years ago
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