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3 years ago

Does the equation 3(2x−1)+5=6(x+1) have one, none, or an infinite amount of solutions?

Mathematics

2 answers:

EleoNora [17]3 years ago

8 0

Answer: No solutions

Step-by-step explanation: If you solve the problem all the way, you get 0 = 4 which is not valid so there is simply no solution

prisoha [69]3 years ago

7 0

Answer:

no solution

Step-by-step explanation:

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How to write 100203 in standard form

natima [27]

100203 = 1.00203 x 10^5

8 0

2 years ago

Write the name of the period that has the digits 486

kogti [31]

1. <span>The name of period that has 486 in the given number which is 751 486 is hundreds.
period is considered to a group of places of the digits, like the given number 486 is the group of hundred because the highest number it contains is 400.
=> 400 + 80 + 6
=> 4 hundreds + 8 tens + 6 ones
=> 486
while 751 is the period of hundred thousands.
=> 700 000 + 50 000 + 1 000
=> 751 000

</span>

8 0

3 years ago

I don't know what to write! Help me please

Andrews [41]

Sorry, can not see the full picture, try uploading the full image of the question.

3 0

3 years ago

Find the area bounded by the graphs of the indicated equations over the given interval.

NeX [460]

Answer:

Step-by-step explanation:

$y = 0$

is basically the horizontal axis.

First, find the integral of x^2-25.

Remember that

integral of a constant is that constant times x.

Also that

to take the integral of a power function, add 1 to the degree and divide by that same degree.

$y = {x}^{2} - 25$

We then get

$\frac{ {x}^{3} }{3} - 25x$

Evaluate at -3

$\frac{ - 3 {}^{3} }{3} - 25( - 3)$

$- 27 + 75 = 48$

Then we evaluate at 0

$\frac{0 {}^{3} }{3} - 25(0) = 0$

Next, we subtract the the answer then we get

$48 - 0 = 48$

8 0

2 years ago

According to The Wedding Report, Inc., the mean cost for a wedding in the United States is $28732 (as of November 2008). Suppose

Stella [2.4K]

Answer:

Following are the solution to the given points:

Step-by-step explanation:

$X = \text{cost of wedding}\sim \text{Normal}\ (\mu = 28732, \sigma= 1500)\\\\$

For point a:

$Probability\ = 0.00000359\\\\ \text{(Using Excel function:} =NORMDIST(22000,28732,1500,1)).$

For point b:

$Probability \ = 0.014678\\\\\text{(Using Excel function:} =1-NORMDIST(32000,28732,1500,1))\\\\$

For point c:

$Probability\ = 0.794614436 \\\\$

$\text{(Using Excel function:} \\=NORMDIST (30000,28732,1500,1)-NORMDIST(25000,28732,1500,1))\\\\$

For point d:

$Q_1 = 27720.26537 \\\\\text{(Using Excel function:} =NORMINV(0.25,28732,1500)) \\\\Q_3 = 29743.73463 \\\\\text{(Using Excel function:} =NORMINV(0.75,28732,1500)).$

For point e:

$IQR = Q_3 - Q_1 = 29743.73463 - 27720.26537 = 2023.469251.$

For point f:

$Top\ 10\% = 30654.32735 \\\\\text{(Using Excel function:} =NORMINV(0.9,28732,1500)).$

5 0

3 years ago