All categories

3 years ago

5x-(x+3)=1/3 (9x+18)-5 Please show me step by step

Mathematics

1 answer:

zzz [600]3 years ago

3 0

Step-by-step explanation:

5x-(x+3)=1/3(9x+18)-5

5x-x+3=3x+6-5

1/3 multiplied by 9/1 is three because if you take away the ones and put your problem like this: 9/3, you'll get 3.

1/3 multiplied by 18/1 is six because if you take away the ones and switch to division, it'll look like this: 18/3. 18/3 is 6.

5x-1x+3=3x+6-5

If a variable is alone, you should - in this case - put a one before it, but if the variable is alone you should know that it's automatically a one.

4x+3=3x-1

6-5=1 Simple math problem, probably self explanatory

4x+3=3x-1

Subtract 3x from 4x and you'll get 1x because 4-3 is 1.

1x+3=-1

Then you add 3 to -1 which is 2.

The answer is 1x=2

Or you could've done:

4x+3=3x-1

-4x -4x

3=-1x-1

+-1 +-1

2=1x

It's the same answer, but all I did was subtract the 4x from the 4x and 3x.

You might be interested in

yuradex [85]

9514 1404 393

Answer:

B) y = (x -7)(4x +3)

Step-by-step explanation:

For zeros p and q, factors of the function are (x -p) and (x -q).

For the given zeros, the factors are (x -7) and (x -(-3/4)). By multiplying the latter by 4, we can eliminate the fraction.

y = (x -7)(4x +3)

4 0

3 years ago

Toxaphene is an insecticide that has been identified as a pollutant in the Great Lakes ecosystem. To investigate the effect of t

Likurg_2 [28]

Answer:

This data suggest that there is more variability in low-dose weight gains than in control weight gains.

Step-by-step explanation:

Let $\sigma_{1}^{2}$ be the variance for the population of weight gains for rats given a low dose, and $\sigma_{2}^{2}$ the variance for the population of weight gains for control rats whose diet did not include the insecticide.

We want to test $H_{0}: \sigma_{1}^{2} = \sigma_{2}^{2}$ vs $H_{1}: \sigma_{1}^{2} > \sigma_{2}^{2}$ . We have that the sample standard deviation for $n_{2} = 22$ female control rats was $s_{2} = 28$ g and for $n_{1} = 18$ female low-dose rats was $s_{1} = 51$ g. So, we have observed the value

$F = \frac{s_{1}^{2}}{s_{2}^{2}} = \frac{(51)^{2}}{(28)^{2}} = 3.3176$ which comes from a F distribution with $n_{1} - 1 = 18 - 1 = 17$ degrees of freedom (numerator) and $n_{2} - 1 = 22 - 1 = 21$ degrees of freedom (denominator).

As we want carry out a test of hypothesis at the significance level of 0.05, we should find the 95th quantile of the F distribution with 17 and 21 degrees of freedom, this value is 2.1389. The rejection region is given by {F > 2.1389}, because the observed value is 3.3176 > 2.1389, we reject the null hypothesis. So, this data suggest that there is more variability in low-dose weight gains than in control weight gains.

8 0

3 years ago

Complete the missing value in the solution of the equation. i’m really confused

Lostsunrise [7]

Answer:

x=4

Step-by-step explanation:

4x+y = 10

Let y = -6

4x-6 = 10

Add 6 to each side

4x-6+6 =10+6

4x = 16

Divide each side by 4

4x/4 = 16/4

x = 4

4 0

4 years ago

When constructing a Confidence Interval for the population mean and both the sample standard deviation and the population standa

Bingel [31]

Answer: idk

Step-by-step explanation:

6 0

3 years ago

Plz answer I’ll mark brainliest . In each of the circles below, four angles are formed by the intersection of two secant lines.

timama [110]

9514 1404 393

Answer:

Step-by-step explanation:

The angle where the chords cross is half the sum of the intercepted arcs. The expression below says that. It is the 2nd choice.

8 0

3 years ago