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3 years ago

Eight thirds x plus one third x equals 8 plus five thirds x

Mathematics

1 answer:

Svetach [21]3 years ago

8 0

Isolate the variable by dividing each side by factors that don't contain the variable.

Exact Form:

x=29/9

Decimal Form:

x=3.2

Mixed Number Form:

x=3 2/9

You might be interested in

What are the solutions of the quadratic equation 49x2 = 9?

pishuonlain [190]

49x^2 - 9 = 0
As there is no x term, we can pretty much guess we have a situation where we factlrise by something known aa difference of two squares, so to factorise it:
49 = 7^2
9 = 3^2
x^2 = (x)^2
so...
(7x - 3)(7x + 3) = 0
7x - 3 = 0 7x + 3 = 0
x = 3/7 x = -3/7

7 0

3 years ago

Read 2 more answers

There are 12 boys and 16 girls in Marco's class (including Marco).What is the probability that a student chosen at random from M

Len [333]

Answer: 3:4 ratio or 3/4

Step-by-step explanation:

5 0

3 years ago

What is 12 percent of 120?

love history [14]

12% of 120 is 14.4

Change the percentage into a decimal by dividing over 100:

12 / 100 = 0.12

Multiply:

0.12 × 120 = 14.4

5 0

3 years ago

Type the correct answer in each box. Use numerals instead of words.

lubasha [3.4K]

Answer:

System A has 4 real solutions.

System B has 0 real solutions.

System C has 2 real solutions

Step-by-step explanation:

System A:

x^2 + y^2 = 17 eq(1)

y = -1/2x eq(2)

Putting value of y in eq(1)

x^2 +(-1/2x)^2 = 17

x^2 + 1/4x^2 = 17

5x^2/4 -17 =0

Using quadratic formula:

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

a = 5/4, b =0 and c = -17

$x=\frac{-(0)\pm\sqrt{(0)^2-4(5/4)(-17)}}{2(5/4)}\\x=\frac{0\pm\sqrt{85}}{5/2}\\x=\frac{\pm\sqrt{85}}{5/2}\\x=\frac{\pm2\sqrt{85}}{5}$

Finding value of y:

y = -1/2x

$y=-1/2(\frac{\pm2\sqrt{85}}{5})$

$y=\frac{\pm\sqrt{85}}{5}$

System A has 4 real solutions.

System B

y = x^2 -7x + 10 eq(1)

y = -6x + 5 eq(2)

Putting value of y of eq(2) in eq(1)

-6x + 5 = x^2 -7x + 10

=> x^2 -7x +6x +10 -5 = 0

x^2 -x +5 = 0

Using quadratic formula:

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

a= 1, b =-1 and c =5

$x=\frac{-(-1)\pm\sqrt{(-1)^2-4(1)(5)}}{2(1)}\\x=\frac{1\pm\sqrt{1-20}}{2}\\x=\frac{1\pm\sqrt{-19}}{2}\\x=\frac{1\pm\sqrt{19}i}{2}$

Finding value of y:

y = -6x + 5

y = -6(\frac{1\pm\sqrt{19}i}{2})+5

Since terms containing i are complex numbers, so System B has no real solutions.

System B has 0 real solutions.

System C

y = -2x^2 + 9 eq(1)

8x - y = -17 eq(2)

Putting value of y in eq(2)

8x - (-2x^2+9) = -17

8x +2x^2-9 +17 = 0

2x^2 + 8x + 8 = 0

2x^2 +4x + 4x + 8 = 0

2x (x+2) +4 (x+2) = 0

(x+2)(2x+4) =0

x+2 = 0 and 2x + 4 =0

x = -2 and 2x = -4

x =-2 and x = -2

So, x = -2

Now, finding value of y:

8x - y = -17

8(-2) - y = -17

-16 -y = -17

-y = -17 + 16

-y = -1

y = 1

So, x= -2 and y = 1

System C has 2 real solutions

4 0

4 years ago

Read 2 more answers

A random sample of 10 parking meters in a beach community showed the following incomes for a day. Assume the incomes are normall

Vlad1618 [11]

Answer: (2.54,6.86)

Step-by-step explanation:

Given : A random sample of 10 parking meters in a beach community showed the following incomes for a day.

We assume the incomes are normally distributed.

Mean income : $\mu=\dfrac{\sum^{10}_{i=1}x_i}{n}=\dfrac{47}{10}=4.7$

Standard deviation : $\sigma=\sqrt{\dfrac{\sum^{10}_{i=1}{(x_i-\mu)^2}}{n}}$

$=\sqrt{\dfrac{(1.1)^2+(0.2)^2+(1.9)^2+(1.6)^2+(2.1)^2+(0.5)^2+(2.05)^2+(0.45)^2+(3.3)^2+(1.7)^2}{10}}$

$=\dfrac{30.265}{10}=3.0265$

The confidence interval for the population mean (for sample size <30) is given by :-

$\mu\ \pm t_{n-1, \alpha/2}\times\dfrac{\sigma}{\sqrt{n}}$

Given significance level : $\alpha=1-0.95=0.05$

Critical value : $t_{n-1,\alpha/2}=t_{9,0.025}=2.262$

We assume that the population is normally distributed.

Now, the 95% confidence interval for the true mean will be :-

$4.7\ \pm\ 2.262\times\dfrac{3.0265}{\sqrt{10}} \\\\\approx4.7\pm2.16=(4.7-2.16\ ,\ 4.7+2.16)=(2.54,\ 6.86)$

Hence, 95% confidence interval for the true mean= (2.54,6.86)

7 0

4 years ago