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vampirchik [111]
3 years ago
13

PLS HELP ON A TIMER !!

Mathematics
1 answer:
maw [93]3 years ago
5 0
I think either b or A
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Use a calculator to solve for xin the equation 2e^3x = 400. Round your answer to<br> three decimals.
Paul [167]

the answer is

D. 1.766
3 0
4 years ago
Drag the tiles to the correct boxes to complete the pairs.
padilas [110]

Answer:

One Solution:

-3x+y=7\\21x+4y=-8

Infinitely Many Solution:

31x-y=4\\61x-2y=8

No Solution:

y=-4r+5\\y=-4r+1

Step-by-step explanation:

System 1

-3x+y=7\\21x+4y=-8

Eq(1)\times7+Eq(2)\\11y=41\\y=\frac{41}{11} \\\\From\ Eq\ 1\\-3x+\frac{41}{11} =7\\-3x=7-\frac{41}{11}\\-3x=\frac{36}{11}\\x=\frac{-12}{11}

Hence only one solution (-\frac{12}{11},\frac{41}{11})

System 2

31x-y=4\\61x-2y=8

Both represents the same line. Hence these have infinitely many solutions.

System 3

y=-4r+5\\y=-4r+1

Compare both the lines with y=mx+b, where m is the slope.

m=-4 for both the lines.

Hence these two lines are parallel that means there does not exists any point of intersection

So this system will not have any solution.

4 0
3 years ago
The box plots below show the ages of college students in different math courses
scZoUnD [109]

Answer:

4. The mean and median age are most likely to be same for the students in Math 1.

Step-by-step explanation:

We have been given two box plots that show the ages of college students in different math courses. We are asked to choose the true statement about these box plots.

1. The median age of the students in math 1 is greater than the median age of the students in math 2.  

We can see from our given box plots that median age for both Math 1 and Math 2 is 19 that is represented by vertical middle line of box, therefore, 1st statement is not true.

2. The median age of the students in math 1 is less than the median age of the students in math 2.

Since median ages for students in Math 1 and Math 2 are same (19), therefore, 2nd statement is not true.

3. The mean and median age are most likely same for both sets of data.

We can see that box plot for ages of Math 1 students is symmetric, therefore, both mean and median will be same for Math 1 students.

The box plot representing ages of Math 2 students is skewed right as it has an outlier as 23, therefore, its mean age will be greater than median age.

Therefore, 3rd statement is not true either.

4. The mean and median age are most likely to be same for the students in Math 1.

We can see that box plot representing ages for Math 1 students is symmetric and its whiskers are equidistant from median. So both mean and median age are same that is 19, therefore, 4th statement is true.

5 0
3 years ago
Read 2 more answers
Given that xy=3/2 and both x and y are nonnegative real numbers, find the minimum value of 10x (3y/5)
Stells [14]

The munimum value is, at x = 3/10, y = 5 and 10x+3y/5 = 6.

<h3>What are nonnegative real numbers?</h3>
  • Non-negative real numbers are the set of positive real numbers that are bigger than 0 (zero).
  • That is, the true values are either positive or negative.
  • The collection will include numbers such as 0, 1, 2, 3, 4, 5, and so on.

To find the minimum value of 10x (3y/5):

Given - y = 3/(2x)

So, we want the bare minimum of,

  • = 10x + 3/5 × 3/(2x)
  • = 10x + 9/(10x)

Take the derivative to obtain:

  • = 10 - 9/(10x^2)

When you set it to zero, you get:

  • = 100x^2 - 9 = 0
  • = (10x-3)(10x+3) = 0
  • = x = ± 3/10

So, at x = 3/10, y = 5 and 10x + 3y/5 = 3 + 3 = 6.

Therefore, the munimum value is, at x = 3/10, y = 5 and 10x+3y/5 = 6.

Know more about nonnegative real numbers here:

brainly.com/question/26606859

#SPJ4

5 0
2 years ago
I don’t under stand this subject<br><br> Plz
egoroff_w [7]

I'll break your gadgets haha whaha

\red{\rule{50pt}{30000000000000000000000000000000000000000000000000000000pt}}

6 0
3 years ago
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