MATHEMATICS COLLEGE

38 = y/2.
What equals y?

Answers

Answer 1
Answer: 38 = y/2.
y = 38 * 2
y  = 76

answer
y = 76
Answer 2
Answer: Multiply 2 both side
38*2 = (y/2)*2
76 = Y

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WILL MARK BRAINLIEST!!!
PLEASE HELP AND NO LINKS

Answers

Answer:

1000 = 450 + 50w

Step-by-step explanation:

She starts with $450 so that is why you add it. The amount she started with will not change.

She earns $50 per week so you add 50 multiplied by the number of weeks she works.

You do this until it equals 1000

Brian has $88 and he saves $3.50 a week how many more weeks will it take him to get to 200​

Answers

Answer:

32 weeks

Step-by-step explanation:

First you should write an equation for this word problem. Since you know Brian wants to save to $200, the equation will equal 200. Then you figure out how much he has already, which is 88. Right now, the equation we have is 88+(weekly savings)= 200. He saves 3.50 a week, so the equation will finally look like 88+3.50x=200. X is how many weeks it will take for him to reach 200. We then move everything so that it all equals X. You would move the 88 first which will then give 3.50x=112. Dividing by 3.50 on both sides would give X as 32. SO the number of weeks it will take for Brian to save to $200 is 32 weeks.

Need help with calculus question

Answers

The value of the derivative at x=5 is the slope of the tangent line at the point (x,g(x))=(5,-4).

So the tangent line has equation

y-g(5)=g'(5)(x-5)\implies y+4=6(x-5)\implies\boxed{y=6x-34}

Find the number of positive integers not exceeding 108 that are not divisible by 5 or by 7.

Answers

Answer:

75

Step-by-step explanation:

The set of positive integer not exceeding 108 divisible by 5 is

D_5=\{5 \quad 10 \quad 15 \quad 20 \quad 25 \quad 30\quad 35 \quad 40 \quad 45 \quad 50 \quad 55 \quad 60 \quad 65\n \quad 70 \quad 75 \quad 80 \quad 85 \quad 90 \quad 95 \quad100 \quad 105\}

and the set of positive integer not exceeding 108 divisible by 7 is

D_7=\{7 \quad 14 \quad 21 \quad 28 \quad 35 \quad 42 \quad 49 \quad 56 \quad 63 \quad 70 \quad 77 \quad 84 \quad 91 \quad 98 \quad 105\}

Moreover, there are exactly three positive numbers not exceedng 108 that are divisible by both 5 and 7, i.e,

D_5 \cap D_7=\{37 \quad 70 \quad 105\}.

Also note that the size of D_5 is \#D_5=21 , the size of D_7 is \#D_5=15 and \# D_7 \cap D_5 = 3.

On the other hand, If a positive integer not exceding 108 is not divisible by 5 or 7, then it doesn't belong to any of this sets. Therefore, the number of positive interges not exceding 108 that are not divisible by 5 or 7 is equal to

108 -(\#D_7 + \# D_5 - \# D_7 \cap D_5)=108 -(21+15-3)=75

A bookshelf is 3 feet long. Each book on the shelf is 3/4 inches wide. How many books will fit on the shelf? (Remember, there are 12 inches in a foot.)

Answers

Answer:

48

Step-by-step explanation:

3 feet is 36 inches. 3/4 of an inch is .75 inches. So you have a very simple equation of 36/.75=48

A class has 32 students. in how many different ways can five students form a group for an​ activity? (assume the order of the students is not​ important.)

Answers

The combination is a way of selecting items from a collection where the order of selection does not matter.

The number of ways five students form a group for an​ activity is 6944.

What is a combination?

The combination is a way of selecting items from a collection where the order of selection does not matter.

The formula for combination is given:

^nC_r = (n!)/(r!(n-r)!)

We have,

Number of students = 32

Number of students in a group = 5

The number of ways five students form a group:

= ^nC_r

n = 32 and r = 5

= ^(32)C_5

= 32! / 5! (32 - 5)!

= 32! / 5! 27!

= 32 x 31 x 30 x 29 x 28 / 5 x 4 x 3 x 2

= 6944

Thus,

The number of ways five students form a group for an​ activity is 6944.

Learn more about combination here:

brainly.com/question/2970011

#SPJ2
This is a combination problem (35C5)
In general, it is the number of arrangements of 5 out of 35 (35P5) divided by the number of arrangements of 5 out of 5 (5P5), which reduces to 
35P5/5P5 = 35!/((30-5)!)  /  (5!/(5-5)!) = 35!/(5!(30-5)!) = 35C5.

Substituting values
35C5 = 35!/(5!30!)=35*34*33*32*31/(1*2*3*4*5) = 324632 ways
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