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mash [69]
3 years ago
13

Put together, Dulcina and Tremaine have 129 total matchbooks. Tremaine's collection has 39 fewer matchbooks in it than Dulcina's

collection. How many matchbooks are in Tremaine's collection? How many matchbooks are in Dulcina's collection?
Mathematics
1 answer:
Anit [1.1K]3 years ago
7 0

Answer:

Let the Dulcina's collection be 'x'

Let the Tremaine collection be 'x-39'

x + x - 39 =129

2x = 129 +39

2x = 168

x = 168/2

x = 84

Dulcina's collection = x = 84

Tremaine's collection = x - 39 = 84 - 39 = 45

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-12=3+3u Solve for “u”
Kipish [7]

Answer:

- 12 = 3 + 3u \\ 3 + 3u =  - 12 \\ 3u =  - 12 - 3 \\ 3u =  - 15 \\ u =  \frac{ - 15}{3}  \\ u =  - 5

u = -5

6 0
3 years ago
You have three online accounts and three different banks valued at: bank “A” $250.67, 12% interest, bank “B” $765.13, 7% interes
larisa [96]

The average interest gained from three accounts in one year is $913.224

<u>Explanation:</u>

Given:

Bank A:

Principal, P₁ = $250.67

Rate, r₁ = 12 %

Bank B:

Principal, P₂ = $765.13

Rate, r₂ = 7 %

Bank C:

Principal, P₃ = $28500.36

Rate, r₃ = 9 %

Interest from Bank A:

Interest = \frac{p X r X t}{100}\\\\I = \frac{250.67 X 12 X 1}{100} \\\\I = 30.0804

Interest from Bank B:

Interest = \frac{p X r X t}{100}\\\\I = \frac{765.13 X 7 X 1}{100} \\\\I = 53.5591

Interest from Bank C:

Interest = \frac{p X r X t}{100}\\\\I = \frac{28500.36 X 9 X 1}{100} \\\\I = 2565.0324

Average interest gained from three accounts in one year :

I = \frac{30.0804+53.5591+2656.0324}{3} \\\\I = 913.224

Therefore, average interest gained from three accounts in one year is $913.224

7 0
3 years ago
Can someone help me I’m a little lost
Ksivusya [100]

Answer:

14 days

Step-by-step explanation:

Total Problems = 175

Finished Problems = 8

Remaining = 175 - 8 = 167

8 problems in 30 minutes, means:

30/8 = 15/4 = 3.75 minutes per problem

If he works for 45 minutes, at that rate, he can solve:

45/3.75 = 12 problems per day

To finish 167 problems, he will need:

167/12 = 13.91

Rounded, that is <u>14 days more to finish</u>

8 0
4 years ago
A triangle cannot have more than one obtuse angle true or false
anastassius [24]
The answer is true;
 because if a triangle had more than one obtuse angle it would become like a rhombus or something else so it is not possible to remain a triangle and have more than one obtuse angle c:
7 0
3 years ago
Read 2 more answers
How do you do this problem? I need to know how you found the answer.
Alexxandr [17]

to get the equation of a line, we simply need two points, say for the Red one ... notice in the graph the lines passes through (0,2) and (-1,6), so let's use those


\bf (\stackrel{x_1}{0}~,~\stackrel{y_1}{2})\qquad (\stackrel{x_2}{-1}~,~\stackrel{y_2}{6}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{6-2}{-1-0}\implies \cfrac{4}{-1}\implies -4 \\\\\\ \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-2=-4(x-0) \\\\\\ y-2=-4x\implies \blacktriangleright y=-4x+2 \blacktriangleleft


now, for the Blue one, say let's use hmmm it passes through (0,2) and (1.6)


\bf (\stackrel{x_1}{0}~,~\stackrel{y_1}{2})\qquad (\stackrel{x_2}{1}~,~\stackrel{y_2}{6}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{6-2}{1-0}\implies \cfrac{4}{1}\implies 4 \\\\\\ \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-2=4(x-0) \\\\\\ y-2=4x\implies \blacktriangleright y=4x+2 \blacktriangleleft

3 0
3 years ago
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