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2 years ago

3(12x – 9) + 9 = 2x + 4 Would this be infinite solution, one solution or no solution?

Mathematics

1 answer:

timama [110]2 years ago

8 0

Answer:

One solution

Step-by-step explanation:

Let us first expand the left side of the equation to rid of the parenthesis:

$3*12x-3*9+9=2x+4\\36x-27+9=2x+4\\36x-18=2x+4$

Now let us isolate the variable to sove for x:

$36x-2x=4+18\\34x=22\\x=\frac{22}{34}\\x=\frac{11}{17}$

Therefore, x= $\frac{11}{17}$ . As you can see, there is just one solution.

I hope this helps! Please let me know if you have any further questions :)

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I am so confused. Can anyone help me??

natita [175]

Answer:

bro! im in 7th grade i would try to help but im very stupid sorry for the dissopiontment

Step-by-step explanation:

8 0

3 years ago

AYOOO PLZ HELP ASAP!!!

tamaranim1 [39]

Answer:

Step-by-step explanation:

Well we know that

$224=2^{5} *7$

so we can get the 2 outside of the radical

$x^{11} =(x^{5} )^{2} *x$

and we can get the x^2 outside too.

$y^8=y^5*y^3$

and we also can get y outside.

so we have:

$2x^{2}y\sqrt[5]{7xy^3}$

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2 years ago

On his drive from Montpelier, Vermont, to Columbia, South Carolina, Arthur averaged 51 mph. If he had been able to average 60 mp

Llana [10]

Answer:

Distance between Montpelier and Columbia is 1020

Step-by-step explanation:

51 mi/hr average:

d = 51 t

60 mi/hr average:

d t = 60*(t-3)

Substitute (1) into (2)

51 t = 60*(t-3)

51 t = 60 t + 180

9 t = 180

t = 20

d = 51 t

= 51 (20)

= 1020

let see distance is 1020 lets check

51 t = 60 * (t-3)

51 (20)= 60* (20-3)

1020 = 60*17

1020 = 1020

4 0

3 years ago

Use the standard norml distribution or the t-distribution to construct a 99% confidence interval for the population mean. Justif

dalvyx [7]

Answer:

E) we will use t- distribution because is un-known,n<30

the confidence interval is (0.0338,0.0392)

Step-by-step explanation:

Step:-1

Given sample size is n = 23<30 mortgage institutions

The mean interest rate 'x' = 0.0365

The standard deviation 'S' = 0.0046

the degree of freedom = n-1 = 23-1=22

99% of confidence intervals $t_{0.01} =2.82$ (from tabulated value).

$The mean value = 0.0365$

$x±t_{0.01} \frac{S}{\sqrt{n-1} }$

$0.0365±2.82 \frac{0.0046}{\sqrt{23-1} }$

$0.0365±2.82 \frac{0.0046}{\sqrt{22} }$

$0.0365±2.82 \frac{0.0046}{4.690 }$

using calculator

$0.0365±0.00276$

Confidence interval is

$(0.0365-0.00276,0.0365+0.00276)$

$(0.0338,0.0392)$

the mean value is lies between in this confidence interval

(0.0338,0.0392).

Answer:-

using t- distribution because is unknown,n<30,and the interest rates are not normally distributed.

4 0

3 years ago

Please give me the right answer I need help

Nimfa-mama [501]

Area of fountain is 147.58 and area of path is 28.26

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3 years ago

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