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3 years ago

Solve for x. 3(x + 5) - 2(x + 2) = 20

2 answers:

7 0

Well, we'll use the distributive property whenever x by itself is being added/subtracted to another number and a number is outside the parentheses :P.

3(x + 5) ------> 3 times x + 3 times 5 -----> 3x + 15.

2(x + 2) ------> 2 times x + 2 times 2 -----> 2x + 2

Now once you get all the variables on the left side of the equation and the positive numbers on the right side of the equation, x = 9. To check, just type this problem into an algebraic calculator and you'll get the same results! c:

IRINA_888 [86]3 years ago

4 0

[tex] 3(x + 5) - 2(x + 2) = 20 \\
3x + 15 - 2x - 4 = 20\\
5x + 11 = 20\\
5x = 9\\
x = 1\frac{4}{5} [tex]

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A rate simplified so that it has a denominator of 1

Yakvenalex [24]

This is called a unit rate.

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3 years ago

(4x+ 1) - 3i = 2+ (3y + 5)i

Vesna [10]

(4x+1)-3i= 2 +(3y+5)i

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3 years ago

A company manager for a tire manufacturer is in charge of making sure there is the least amount of defective tires. If the tire

Anestetic [448]

Answer:

0.347% of the total tires will be rejected as underweight.

Step-by-step explanation:

For a standard normal distribution, (with mean 0 and standard deviation 1), the lower and upper quartiles are located at -0.67448 and +0.67448 respectively. Thus the interquartile range (IQR) is 1.34896.

And the manager decides to reject a tire as underweight if it falls more than 1.5 interquartile ranges below the lower quartile of the specified shipment of tires.

1.5 of the Interquartile range = 1.5 × 1.34896 = 2.02344

1.5 of the interquartile range below the lower quartile = (lower quartile) - (1.5 of Interquartile range) = -0.67448 - 2.02344 = -2.69792

The proportion of tires that will fall 1.5 of the interquartile range below the lower quartile = P(x < -2.69792) ≈ P(x < -2.70)

Using data from the normal distribution table

P(x < -2.70) = 0.00347 = 0.347% of the total tires will be rejected as underweight

Hope this Helps!!!

6 0

3 years ago

Determine the commutators of the operators a and a+,where a = (x + ip)/2 ^1/2 and a+ = (x - ip)/ 2 ^1/2

Vsevolod [243]

Answer:

Given that:

$a = (\frac{x+ip}{2})^{\frac{1}{2}}$ and $a+= (\frac{x-ip}{2})^{\frac{1}{2}}$

if a , a+ commutator, it obeys $aa^+ = a^+a$

First find:

$aa^+ = (\frac{x+ip}{2})^{\frac{1}{2}} (\frac{x-ip}{2})^{\frac{1}{2}}$

= $(\frac{(x)^2-(ip)^2}{4})^{\frac{1}{2}}=(\frac{(x)^2+(p)^2}{4})^{\frac{1}{2}}$

Now;

$a^+a =(\frac{x-ip}{2})^{\frac{1}{2}} (\frac{x+ip}{2})^{\frac{1}{2}} = (\frac{(x)^2-(ip)^2}{4})^{\frac{1}{2}}$

= $(\frac{(x)^2-(ip)^2}{4})^{\frac{1}{2}}=(\frac{(x)^2+(p)^2}{4})^{\frac{1}{2}}$

therefore, $aa^+ = a^+a$ which implies the operators a and a+ are commutators.

7 0

3 years ago

A student desired to invest $1,540 into an investment at 9% compounded semiannually for 6 years. With all else equal, what is th

irga5000 [103]

Answer:

The future value of this initial investment after the six year period is $2611.6552

Step-by-step explanation:

Consider the provided information.

A student desired to invest $1,540 into an investment at 9% compounded semiannually for 6 years.

Future value of an investment: $FV=P(1+r)^n$

Where Fv is the future value, p is the present value, r is the rate and n is the number of compounding periods.

9% compounded semiannually for 6 years.

Therefore, the value of r is: $r=\frac{0.09}{2}=0.045$

Number of periods are: 2 × 6 = 12

Now substitute the respective values in the above formula.

$FV=1540(1+0.045)^{12}$

$FV=1540(1.045)^{12}$

$FV=1540(1.69588)$

$FV=2611.6552$

Hence, the future value of this initial investment after the six year period is $2611.6552

6 0

3 years ago