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Iteru [2.4K]
3 years ago
6

HELP PLS MY TEACHER WANTS TO KNOW THE ANSWER!!!

Mathematics
1 answer:
valentina_108 [34]3 years ago
6 0

Answer:

e ≥ 114  where e is the number of envelopes

Step-by-step explanation:

Let the number of envelopes for invitations to address be --------e

Hence the inequality of the number of envelopes is : e ≥ 140

Number of addressed envelopes is : 26

Remaining number of envelopes to address is : 140 -26 = 114 or more

The  inequality that describes how many more invitations must be addressed is ;

e ≥ 114

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What is (5,1) reflected on the y axis
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Answer:

(-5 , 1)

Step-by-step explanation:

If you are reflecting over the x-axis, you are changing the sign of the y.

If you are reflecting over the y-axis, you are changing the sign of the x.

In this case, you have the point (5 , 1). You are reflecting over the y-axis, which means that you are flipping the sign of the x value.

(5 , 1) reflected over the y-axis is (-5 , 1)

(-5 , 1)

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2 years ago
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Lisa is buying a shirt and a hat at the mall. The shirt costs $30.98, and the hat costs $20.01. If she gave the sales clerk $100
Paha777 [63]

$49.01 would be her change

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Last summer there were 88 players at coach rodriguez basketball camp. this year there are 125% of this number of players. how ma
ludmilkaskok [199]
To determine the number of those who joined the basketball camp this year, we just have to multiply the number of those who joined last year by the decimal equivalent of the percentage given. That is,
                                   x = 88(1.25)
                                   x = 110
Therefore, there are a total of 110 players for this years basketball camp. 
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3 years ago
Mean Value theorem help! BRAINLIEST if you can explain it clearly!!!
Korolek [52]
\bf f(x)=ln(x-5)\qquad [6,8]\qquad \cfrac{df}{dx}=\cfrac{1}{x-5}\\\\
-----------------------------\\\\
\textit{mean value theorem}\qquad f'(c)=\cfrac{f(b)-f(a)}{b-a}\\\\
-----------------------------\\\\
f'(c)=\cfrac{1}{c-5}\qquad thus\quad \cfrac{1}{c-5}=\cfrac{f(8)-f(6)}{8-6}
\\\\\\
\cfrac{1}{c-5}=\cfrac{ln(3)-ln(1)}{2}\implies \cfrac{1}{c-5}=\cfrac{ln(3)-0}{2}
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\cfrac{2}{ln(3)}=c-5\implies \boxed{\cfrac{2}{ln(3)}+5=c}
7 0
3 years ago
Shishir bought 4000 orange at 70 paisa each. But 400 of them were rotten. He sold 2000 oranges at 90 paisa each.If he plans to m
Ipatiy [6.2K]

Answer:

He needs to sell the rest of the oranges at 75 paisa each.

Step-by-step explanation:

Consider the given information that, Shishir bought 4000 orange at 70 paisa each.

Note: 1 rupees = 100 paisa

Thus, 70 paisa = 70/100 rupees = 0.70 rupees

Therefore, the cost price of 4000 oranges is:

4000×0.70 rupees = 2800 rupees

The selling price of 2000 oranges is:

2000×0.90 rupees = 1800 rupees

The number of oranges now Shishir have:

4000 - 2000 - 400 = 1600

He wants to make a profit of RS 200. Thus the selling price of 4000 oranges should be:

2800 rupees + 200 rupees = 3000 rupees

He earned 1800 rupees by selling 2000 oranges at 90 paisa. So, the remaining amount that he needs to make with 1600 oranges is:

3000 rupees - 1800 rupees = 1200 rupees

Therefore, the cost of one orange is:

1600 oranges = 1200 rupees

1 orange = 1200/1600 rupees

1 orange = 0.75 rupees

Hence, he needs to sell the rest of the oranges at 75 paisa each.

4 0
3 years ago
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