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GarryVolchara [31]

2 years ago

15

Set R contains all integers from 10 to 125, inclusive, and Set T contains all integers from 82 to 174, inclusive. How many integ

ers are included in R, but not in T?

1 answer:

SCORPION-xisa [38]2 years ago

3 0

Answer: PLEASE GIVE BRAINLIEST

72 integers

Step-by-step explanation:

Starting From 82-174 every integer is included in T so every integer that is not included in T but included in R would be 82-10 so every integer from 10 to 82.

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2 years ago

Ritika bought 5 meters of cloth from the market. She cut off 2 3/4 meters and gave it to Aditi. What length of cloth is left wit

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I think its 3/2

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2 years ago

In college basketball, some teams get most of their points from just one player, and other teams are more balanced in scoring. C

lutik1710 [3]

Answer:

(B) At Maryland, the mean number of points per player per game is greater than the median number of points per player per game.

(D)The mean number of points per player per game is the same at all 3 schools.

Step-by-step explanation:

<u>Baylor University</u>

6 players who each score 12 points per game

6 players who each score 0 points per game.

The Scores are: 0,0,0,0,0,0,12,12,12,12,12,12

$Mean=\frac{0+0+0+0+0+0+12+12+12+12+12+12}{12} \\Mean=6\\Median=\frac{12+0}{2}=6$

<u>University of Maryland</u>

1 player scores 58 points per game,

1 player scores 14 points per game,

The rest(10) of the players score 0 points per game.

The Scores are: 0,0,0,0,0,0,0,0,0,0,14,58

$Mean=\frac{0+0+0+0+0+0+0+0+0+0+14+58}{12} \\Mean=6\\Median=\frac{0+0}{2}=0$

<u> Dartmouth College</u>

4 players score 5 points per game,

4 players score 6 points per game,

4 players score 7 points per game.

The Scores are: 5,5,5,5,6,6,6,6,7,7,7,7

$Mean=\frac{5+5+5+5+6+6+6+6+7+7+7+7}{12} \\Mean=6\\Median=\frac{6+6}{2}=6$

The following applies:

(B) At Maryland, the mean number of points per player per game is greater than the median number of points per player per game.

(D)The mean number of points per player per game is the same at all 3 schools.

7 0

2 years ago

Steve likes to entertain friends at parties with "wire tricks." Suppose he takes a piece of wire 60 inches long and cuts it into

Alex_Xolod [135]

Answer:

a) the length of the wire for the circle = $(\frac{60\pi }{\pi+4}) in$

b)the length of the wire for the square = $(\frac{240}{\pi+4}) in$

c) the smallest possible area = 126.02 in² into two decimal places

Step-by-step explanation:

If one piece of wire for the square is y; and another piece of wire for circle is (60-y).

Then; we can say; let the side of the square be b

so 4(b)=y

b= $\frac{y}{4}$

Area of the square which is L² can now be said to be;

$A_S=(\frac{y}{4})^2 = \frac{y^2}{16}$

On the otherhand; let the radius (r) of the circle be;

2πr = 60-y

$r = \frac{60-y}{2\pi }$

Area of the circle which is πr² can now be;

$A_C= \pi (\frac{60-y}{2\pi } )^2$

$=( \frac{60-y}{4\pi } )^2$

Total Area (A);

A = $A_S+A_C$

= $\frac{y^2}{16} +(\frac{60-y}{4\pi } )^2$

For the smallest possible area; $\frac{dA}{dy}=0$

∴ $\frac{2y}{16}+\frac{2(60-y)(-1)}{4\pi}=0$

If we divide through with (2) and each entity move to the opposite side; we have:

$\frac{y}{18}=\frac{(60-y)}{2\pi}$

By cross multiplying; we have:

2πy = 480 - 8y

collect like terms

(2π + 8) y = 480

which can be reduced to (π + 4)y = 240 by dividing through with 2

$y= \frac{240}{\pi+4}$

∴ since $y= \frac{240}{\pi+4}$ , we can determine for the length of the circle ;

60-y can now be;

= $60-\frac{240}{\pi+4}$

= $\frac{(\pi+4)*60-240}{\pi+40}$

= $\frac{60\pi+240-240}{\pi+4}$

= $(\frac{60\pi}{\pi+4})in$

also, the length of wire for the square (y) ; $y= (\frac{240}{\pi+4})in$

The smallest possible area (A) = $\frac{1}{16} (\frac{240}{\pi+4})^2+(\frac{60\pi}{\pi+y})^2(\frac{1}{4\pi})$

= 126.0223095 in²

≅ 126.02 in² ( to two decimal places)

4 0

2 years ago

For ΔABC, ∠A = 4x - 10, ∠B = 5x + 10, and ∠C = 7x + 20. If ΔABC undergoes a dilation by a scale factor of 1 3 to create ΔA'B'C'

VLD [36.1K]

We know that the angles of a triangle sum to 180°. For ΔABC, this means we have:
(4x-10)+(5x+10)+(7x+20)=180

Combining like terms,
16x+20=180

Subtracting 20 from both sides:
16x=160

Dividing both sides by 16:
x=10
This means ∠A=4*10-10=40-10=30°; ∠B=5*10+10=50+10=60°; and ∠C=7*10+20=70+20=90.

For ΔA'B'C', we have
(2x+10)+(8x-20)+(10x-10)=180

Combining like terms,
20x-20=180

Adding 20 to both sides:
20x=200

Dividing both sides by 20:
x=10

This gives us ∠A'=2*10+10=20+10=30°; ∠B'=8*10-20=80-20=60°; and ∠C'=10*10-10=100-10=90°.

Since the angle are all congruent, ΔABC~ΔA'B'C' by AAA.

5 0

2 years ago

Read 2 more answers