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Bogdan [553]
3 years ago
9

I need help on these 3 questions please! ​

Mathematics
1 answer:
Flauer [41]3 years ago
8 0

Answer:

3/4*4=3

3/5*6=3.6

2/6*5=1.7

You might be interested in
RS = 12, ST = 2x, RT = 34
Harrizon [31]

Answer:

x=11

Step-by-step explanation:

34(RT) minus 12(RS)=22. 22 divided by 2 equals 11.

6 0
3 years ago
I need help on number 4?
olga nikolaevna [1]
They would have 2 because is it 10:1 just double it to be 20:2
8 0
3 years ago
A community bird-watching society makes and sells simple bird feeders to raise money for its conservation activities. The materi
valentina_108 [34]

Answer:

N(x) = 40 - 2x

P(x) = -2x² + 52 x - 240

maximum profit = 13

Step-by-step explanation:

given data

feeder cost = $6

average sell = 20 per week

price = $10 each

solution

we consider here price per feeder = x

and profit per feeder  id here formula   = x - 6

so that here

total profit will be

P (x)  = ( x - 6 ) Nx

here N(x) is number of feeders sold at price =  x

so formula for N (x)  is here

N(x) = 20 - 2 ( x - 10 )    

N(x) = 40 - 2x

so that

P(x) = (x-6) ( 40 - 2x)

P(x) = -2x² + 52 x - 240

since here

a = -2

b = 52

c = -240

a < 0

so quadratic function have maximum value of c - \frac{b^2}{4a}

so it will be

maximum value = -240 - \frac{52^2}{4(-2)}

maximum value = 98

so here maximum profit attained at

x = \frac{-b}{2a}

x = \frac{-52}{2(-2)}

x = 13

maximum profit = 13

3 0
3 years ago
4) You randomly select one card from a​ 52-card deck. Find the probability of selecting the 6 of hearts or the ace of diamonds.
MakcuM [25]

Answer:

1/26

Step-by-step explanation:

There is only one 6 of hearts and only one ace of diamonds.

The probability of selecting the 6 of hearts is thus 1/52, and that of selecting the ace of diamonds is also 1/52.

The probability of selecting the 6 of hearts or the ace of diamonds is the SUM of these two results:  1/52 + 1/52 = 2/52 = 1/26.

8 0
3 years ago
Read 2 more answers
If perpendiculars from any point within an angle on its arms are equal, prove that it lies on the bisector of that angle
Oxana [17]
Your question can be quite confusing, but I think the gist of the question when paraphrased is: P<span>rove that the perpendiculars drawn from any point within the angle are equal if it lies on the angle bisector?

Please refer to the picture attached as a guide you through the steps of the proofs. First. construct any angle like </span>∠ABC. Next, construct an angle bisector. This is the line segment that starts from the vertex of an angle, and extends outwards such that it divides the angle into two equal parts. That would be line segment AD. Now, construct perpendicular line from the end of the angle bisector to the two other arms of the angle. This lines should form a right angle as denoted by the squares which means 90° angles. As you can see, you formed two triangles: ΔABD and ΔADC. They have congruent angles α and β as formed by the angle bisector. Then, the two right angles are also congruent. The common side AD is also congruent with respect to each of the triangles. Therefore, by Angle-Angle-Side or AAS postulate, the two triangles are congruent. That means that perpendiculars drawn from any point within the angle are equal when it lies on the angle bisector

4 0
3 years ago
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