What is the solution set to log:(4x + 2) + 4 2 5?
(-0.5, 00)
[1.5, 00)
thy5.c)
[5,00)

Answer:

350 pairs

Step-by-step explanation:

If the ratio of Athletics, Boots, and Dress shoes is 5 to 2 to 3, it means that for every 2 pairs of Boots they have 5 pairs of Athletics shoes and 3 pairs of dress shoes.

So, if they have 70 pairs of boots, we can calculate the number of Athletics as:

[tex]\frac{5*70}{2} =175[/tex]

And if they have 70 pairs of boots, the number of dress shoes are:

[tex]\frac{3*70}{2}=105[/tex]

Finally, they have 70 pairs of boots, 175 pairs of athletics, and 105 pairs of dress shoes. It means that they have 350 pairs in total.

70 + 175 + 105 = 350

Answer:

Step-by-step explanation:

From the summary of the given statistics;

The null and the alternative hypothesis for confirming that the new engine has a mean emission level below the current standard can be computed as follows:

Null hypothesis:

[tex]H_0: \mu = 0.60[/tex]

Alternative hypothesis:

[tex]H_a: \mu < 0.60[/tex]

Type I  error: Here, the null hypothesis which is the new engine has a mean level equal to  .6g/ml is rejected when it is true.

Type II error:  Here, the alternative hypothesis which is the new engine has a mean level less than.6g/ml is rejected when it is true.

Similarly;

From , A sample of 64 engines tested yields a mean emission level of = .5 g/mi. Assume that σ = .4.

Sample size n = 64

sample mean [tex]\overline x[/tex] = .5 g/ml

standard deviation σ = .4

From above, the normal standard test statistics can be determined by using the formula:

[tex]z = \dfrac{\bar x- \mu}{\dfrac{\sigma}{\sqrt{n}}}[/tex]

[tex]z = \dfrac{0.5- 0.6}{\dfrac{0.4}{\sqrt{64}}}[/tex]

[tex]z = \dfrac{-0.1}{\dfrac{0.4}{8}}[/tex]

z = -2.00

The p-value = P(Z ≤ -2.00)

From the normal z distribution table

P -value = 0.0228

Decision Rule: At level of significance ∝ = 0.05, If P value is less than or equal to level of significance ∝ , we reject the null hypothesis.

Conclusion: SInce the p-value is less than the level of significance , we reject the null hypothesis. Therefore, we can conclude that there is enough evidence that a new engine design has reduced the emissions to a level below this standard.

Answer:

the dimensions of the poster with the smallest area is 36cm by 54cm

Step-by-step explanation:

✓Let us represent the WIDTH of the printed material on the poster as "x"

✓Let us represent the HEIGHT of the printed material on the poster as "y"

✓ The given AREA is given as 864 cm2

Then we have

864 cm2= xy ...................eqn(1)

We can make "y" subject of the formula.

y= 864/x .......................eqn(2)

✓The total height the big poster which includes the 9cm margin that is at the bottom as well as the top is

(y+18)

✓The total width of the poster which includes the 6cm margin that is at the bottom as well as the top is

(x+12)

✓Then AREA OF THE TOTAL poster

A= (y+18)(x+12) ...................eqn(3)

Substitute eqn (2) into eqn(3)

A= ( 18+ 864/x)(x+12)

We can now simplify by opening the bracket, as

A=18x +1080 +10368/x

A= 18x +10368/x +1080

Let us find the first derivative of A which is A'

A'= 18-(10368/x²)

If we set A' =0

Then

0= 18- (10368/x²)

18= (10368/x²)

x²= 10368/18

x²= 576

x=√576

x=24

The second derivatives will be A"= 2(10368)/x³ and this will be positive for x> 0, and here A is concave up and x=24 is can be regarded as a minimum

The value of "y" when x=24 can now be be calculated using eqn(2)

y= 864/x

y= 864/24

y=36cm

✓The total width of the poster= (x+12)

= 24+12=36cm

✓The total height big the poster= (y+18)=36+18=54cm

the dimensions of the poster with the smallest area is 36cm by 54cm

Answer:

The total width of the paper [tex]=36 cm.[/tex]

The total height of the paper [tex]=54cm[/tex]

Step-by-step explanation:

Given information:

Top margin of the paper = 9 [tex]cm\\[/tex]

Bottom margin of the paper = 6 [tex]cm\\[/tex]

Area of the printed material = [tex]864[/tex] [tex]cm^2[/tex]

Let, the width of the printed material = [tex]x[/tex]

And the height of the printed material = [tex]y[/tex]

So, Area [tex]x \times y=864[/tex] [tex]cm^2[/tex]

After including margins;

Width of the paper [tex]= (x+12)[/tex]

Height of the paper [tex]= (y+18)[/tex]

Area [tex](A) = (y+18) (x+12)[/tex]

[tex]A=18x+(10368/x)+1080\\[/tex]

Take first derivative:

[tex]A'= 18- (10368/x^2)[/tex]

When [tex]A'=0[/tex]

Then,

[tex]18-(10368/x^2)=0\\x^2=576\\x=24[/tex]

Now ,when we take second derivative and check if it is positive or not ,

We find that it is grater than zero so the obtained value can be consider as minimum and can be proceed for further solution.

Hence ,

[tex]x \times y=864\\y=864/24\\y=36\\[/tex]

Now ,

The total width of the paper

[tex]= 24+12\\=36 cm.[/tex]

And , total height of the paper

[tex]=36+18\\=54 cm.[/tex]

For more information visit:

https://brainly.com/question/14261130

Step-by-step explanation:

a. The point estimate is the mean, 47 days.

b. The margin of error is the critical value times the standard error.

At 31 degrees of freedom and 98% confidence, t = 2.453.

The margin of error is therefore:

MoE = 2.453 × 10.2 / √32

MoE = 4.42

c.  The confidence interval is:

CI = 47 ± 4.42

CI = (42.58, 51.42)

d. We can conclude with 98% confidence that the true mean is between 42.58 days and 51.42 days.

e. We can reduce the margin of error by either increasing the sample size, or using a lower confidence level.

Answer:

1.56 seconds

Step-by-step explanation:

When the ball hits the ground, h = 0.

0 = 20 − 5t − 5t²

Divide both sides by -5.

0 = t² + t − 4

Solve with quadratic formula.

t = [ -1 ± √(1² − 4(1)(-4)) ] / 2(1)

t = (-1 ± √17) / 2

The time must be positive, so:

t = (-1 + √17) / 2

t ≈ 1.56

Answer:

The painting is [tex]t = 10911.1 \ years \ old[/tex]

Step-by-step explanation:

From the question we are told that

    The amount of carbon present after t year is

           [tex]y(t) = y_o * e ^{-0.00012t}[/tex]    {Note ; This is the function }

Here  [tex]y(t)[/tex] is the amount of carbon-14 after time t

        [tex]y_o[/tex] the original amount of carbon-14

Now given that the paint as at now contain 27% of the original carbon-14

  Then it mean that

            [tex]y(t) = 0.27 y_o[/tex]

So the equation is represented as

       [tex]0.27 y_o = y_o * e ^{-0.00012t}[/tex]

=>    [tex]0.27 = * e ^{-0.00012t}[/tex]

=>    [tex]ln(0.27) = -0.00012t[/tex]

=>  [tex]- 1.30933 = -0.00012t[/tex]

=>  [tex]t = \frac{-1.30933}{-0.00012}[/tex]

=>   [tex]t = 10911.1 \ years[/tex]

Answer:

AC = 25.5 or 1.5

Step-by-step explanation:

If they are on a line and they are in the order ABC

AB + BC = AC

12+13.5 = AC

25.5 = AC

If they are on a line and they are in the order CAB

CA + AB = BC

AC + 12 =13.5

AC = 13.5 -12

AC = 1.5

If they are on a line and they are in the order ACB

That would mean that AB is greater than BC and that is not the case

Answer:

$7200

Step-by-step explanation:

The interest rate on $5,000 accumulated by Edgar is 20%.

He does not make any payment for 2 years and the interests are compounded continuously.

The amount of money he owes after 2 years is the original $5000 and the interest that would have accumulated after 2 years.

The formula for compound amount is:

[tex]A = P(1 + R)^T[/tex]

where P = amount borrowed = $5000

R = interest rate = 20%

T = amount of time = 2 years

Therefore, the amount he will owe on his debt is:

[tex]A = 5000 (1 + 20/100)^2\\\\A = 5000(1 + 0.2)^2\\\\A = 5000(1.2)^2\\[/tex]

A = $7200

After 2 years, he will owe $7200

Answer:7,434.57

Explanation: A= 5000(1+0.2/12)^12•2

Answer:

The meadian decreases by 1.5 when the outlier is removed.

Step-by-step explanation:

Well first we need to find the median of the following data set,

(75, 63, 58, 59, 63, 62, 56, 59)

So we order the set from least to greatest,

56, 58, 59, 59, 62, 63, 63, 75

Then we cross all the side numbers,

Which gets us 59 and 62.

59 + 62 = 121.

121 / 2 = 60.5

So 65 is the median before the outlier is removed.

Now when we remove the outlier which is 75.

Then we order it again,

56, 58, 59, 59, 62, 63, 63

Which gets us 59 as the median.

Thus,

the median height decreases by 1.5 units when the outlier is removed.

Hope this helps :)

Answer:

9a^2sqrt(ab)

Step-by-step explanation:

The first noticable thing is that 81 has a perfect square of 9.

So it is now 9sqrt(a^5b)

you can split the a^5, to a^4 × a.

you can now take the sqrt of a^4, which is a^2, and pull it out from the sqrt

You are now left with 9a^2sqrt(ab)

Answer:

9a^2sqrt(ab)

Step-by-step explanation:

Answer:

  growth rate = 0.1733 per day, or 17.33% per day

Step-by-step explanation:

Since the doubling time is 4 days, the growth factor over a period of t days is ...

  2^(t/4)

Then the growth factor for 1 day is

  2^(1/4) ≈ 1.189207

The instantaneous growth rate is the natural log of this:

  ln(1.189207) ≈ 0.1733 . . . per day

Answer:

-7/5

Step-by-step explanation:

We can find the slope using the slope formula

m = ( y2-y1)/(x2-x1)

   = ( -3 -4)/(8-3)

   = -7/5

Answer:

-7/5

Step-by-step explanation:

Hey there!

To find the slope of a line with 2 given points we'll use the following formula,

[tex]\frac{y^2-y^1}{x^2-x^2}[/tex]

-3 - 4 = -7

8 - 3 = 5

-7/5

Hope this helps :)

Answer:

A 98% confidence interval estimate for the difference in mean speed of the films is [-0.042, 0.222].

Step-by-step explanation:

We are given that Eight samples of each film thickness are manufactured in a pilot production process, and the film speed (in microjoules per square inch) is measured.

For the 25-mil film, the sample data result is: Mean Standard deviation 1.15 0.11 and For the 20-mil film the data yield: Mean Standard deviation 1.06 0.09.

Firstly, the pivotal quantity for finding the confidence interval for the difference in population mean is given by;

                     P.Q.  =  [tex]\frac{(\bar X_1 -\bar X_2)-(\mu_1- \mu_2)}{s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }[/tex]  ~  [tex]t__n_1_+_n_2_-_2[/tex]

where, [tex]\bar X_1[/tex] = sample mean speed for the 25-mil film = 1.15

[tex]\bar X_1[/tex] = sample mean speed for the 20-mil film = 1.06

[tex]s_1[/tex] = sample standard deviation for the 25-mil film = 0.11

[tex]s_2[/tex] = sample standard deviation for the 20-mil film = 0.09

[tex]n_1[/tex] = sample of 25-mil film = 8

[tex]n_2[/tex] = sample of 20-mil film = 8

[tex]\mu_1[/tex] = population mean speed for the 25-mil film

[tex]\mu_2[/tex] = population mean speed for the 20-mil film

Also,  [tex]s_p =\sqrt{\frac{(n_1-1)s_1^{2}+ (n_2-1)s_2^{2}}{n_1+n_2-2} }[/tex] = [tex]\sqrt{\frac{(8-1)\times 0.11^{2}+ (8-1)\times 0.09^{2}}{8+8-2} }[/tex] = 0.1005

Here for constructing a 98% confidence interval we have used a Two-sample t-test statistics because we don't know about population standard deviations.

So, 98% confidence interval for the difference in population means, ([tex]\mu_1-\mu_2[/tex]) is;

P(-2.624 < [tex]t_1_4[/tex] < 2.624) = 0.98  {As the critical value of t at 14 degrees of

                                             freedom are -2.624 & 2.624 with P = 1%}  

P(-2.624 < [tex]\frac{(\bar X_1 -\bar X_2)-(\mu_1- \mu_2)}{s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }[/tex] < 2.624) = 0.98

P( [tex]-2.624 \times {s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }[/tex] < [tex]2.624 \times {s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }[/tex] <  ) = 0.98

P( [tex](\bar X_1-\bar X_2)-2.624 \times {s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }[/tex] < ([tex]\mu_1-\mu_2[/tex]) < [tex](\bar X_1-\bar X_2)+2.624 \times {s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }[/tex] ) = 0.98

98% confidence interval for ([tex]\mu_1-\mu_2[/tex]) = [ [tex](\bar X_1-\bar X_2)-2.624 \times {s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }[/tex] , [tex](\bar X_1-\bar X_2)+2.624 \times {s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }[/tex] ]

= [ [tex](1.15-1.06)-2.624 \times {0.1005 \times \sqrt{\frac{1}{8}+\frac{1}{8} } }[/tex] , [tex](1.15-1.06)+2.624 \times {0.1005 \times \sqrt{\frac{1}{8}+\frac{1}{8} } }[/tex] ]

 = [-0.042, 0.222]

Therefore, a 98% confidence interval estimate for the difference in mean speed of the films is [-0.042, 0.222].

Since the above interval contains 0; this means that decreasing the thickness of the film doesn't increase the speed of the film.

Answer:

1. 115 chocolates were added

2. 861 visitors in September

Step-by-step explanation:

1.

Initially:

m=number of mints

60% of 230 sweets were mints  =>

m = 230*0.6 = 138 mints

initial number of chocolates, c1  =  230 - 138 = 92

Now chocolates were added

138 mints represented 40% of the total number of sweets, so

total number of sweets  = 138 / 0.4 = 345

Number of chocolates added = 345 - 230 = 115

2.

In August 36% were locals, 64% were from elsewhere.

In suptember,

locals decreassed by 25% to 36*0.75=27% (of August total)

foreigner increased by 50% to 64*1.5=96% (of August total)

Total Inrease = 96+27-100 = 23% of August total  = 161 visitors

August total = 161/0.23 = 700 visitors

September total = 700 + 161 = 861 visitors

You are absolutely right.

Answer:

  28 • 28 • 28 • 28 • 28 • 28

Step-by-step explanation:

The exponent signifies the number of times the base appears as a factor in the product. Here, the base 28 is a factor 6 times:

  28×28×28×28×28×28

Answer:

  $1025

Step-by-step explanation:

We can use the 2-point form of the equation of a line to write a function that gives Justin's salary as a function of his sales.

We start with (sales, salary) = (400, 500) and (700, 575)

__

The 2-point form of the equation of a line is ...

  y = (y2 -y1)/(x2 -x1)(x -x1) +y1

  salary = (575 -500)/(700 -400)(sales -400) +500

  salary = 75/300(sales -400) +500

For sales of 2500, this will be ...

  salary = (1/4)(2500 -400) +500 = (2100/4) +500 = 1025

Justin's salary after selling $2500 in merchandise is $1025.

Answer:

dI/dt = 0.0001(2000 - I)I

I(0) = 20

[tex]I(t)=\frac{2000}{1+99e^{-0.2t}}[/tex]

Step-by-step explanation:

It is given in the question that the rate of spread of the disease is proportional to the product of the non infected and the infected population.

Also given I(t) is the number of the infected individual at a time t.

[tex]\frac{dI}{dt}\propto \textup{ the product of the infected and the non infected populations}[/tex]

Given total population is 2000. So the non infected population = 2000 - I.

[tex]\frac{dI}{dt}\propto (2000-I)I\\\frac{dI}{dt}=k (2000-I)I, \ \textup{ k is proportionality constant.}\\\textup{Since}\ k = 0.0001\\ \therefore \frac{dI}{dt}=0.0001 (2000-I)I[/tex]

Now, I(0) is the number of infected persons at time t = 0.

So, I(0) = 1% of 2000

            = 20

Now, we have dI/dt = 0.0001(2000 - I)I  and  I(0) = 20

[tex]\frac{dI}{dt}=0.0001(2000-I)I\\\frac{dI}{(2000-I)I}=0.0001 dt\\\left ( \frac{1}{2000I}-\frac{1}{2000(I-2000)} \right )dI=0.0001dt\\\frac{dI}{2000I}-\frac{dI}{2000(I-2000)}=0.0001dt\\\textup{Integrating we get},\\\frac{lnI}{2000}-\frac{ln(I-2000)}{2000}=0.0001t+k \ \ \ (k \text{ is constant})\\ln\left ( \frac{I}{I-222} \right )=0.2t+2000k[/tex]

[tex]\frac{I}{I-2000}=Ae^{0.2t}\\\frac{I-2000}{I}=Be^{-0.2t}\\\frac{2000}{I}=1-Be^{-0.2t}\\I(t)=\frac{2000}{1-Be^{-0.2t}}\textup{Now we have}, I(0)=20\\\frac{2000}{1-B}=20\\\frac{100}{1-B}=1\\B=-99\\ \therefore I(t)=\frac{2000}{1+99e^{-0.2t}}[/tex]

The required expressions are presented below:

[tex]\frac{dI}{dt} = 0.0001\cdot I\cdot (2000-I)[/tex] [tex]\blacksquare[/tex]

[tex]I(0) = \frac{1}{100}[/tex] [tex]\blacksquare[/tex]

[tex]I(t) = \frac{20\cdot e^{\frac{t}{5} }}{1+20\cdot e^{\frac{t}{5} }}[/tex] [tex]\blacksquare[/tex]

After reading the statement, we obtain the following differential equation:

[tex]\frac{dI}{dt} = k\cdot I\cdot (n-I)[/tex] (1)

Where:

Then, we solve the expression by variable separation and partial fraction integration:

[tex]\frac{1}{k} \int {\frac{dI}{I\cdot (n-I)} } = \int {dt}[/tex]

[tex]\frac{1}{k\cdot n} \int {\frac{dl}{l} } + \frac{1}{kn}\int {\frac{dI}{n-I} } = \int {dt}[/tex]

[tex]\frac{1}{k\cdot n} \cdot \ln |I| -\frac{1}{k\cdot n}\cdot \ln|n-I| = t + C[/tex]

[tex]\frac{1}{k\cdot n}\cdot \ln \left|\frac{I}{n-I} \right| = C\cdot e^{k\cdot n \cdot t}[/tex]

[tex]I(t) = \frac{n\cdot C\cdot e^{k\cdot n\cdot t}}{1+C\cdot e^{k\cdot n \cdot t}}[/tex], where [tex]C = \frac{I_{o}}{n}[/tex] (2, 3)

Note - Please notice that [tex]I_{o}[/tex] is the initial infected population.

If we know that [tex]n = 2000[/tex], [tex]k = 0.0001[/tex] and [tex]I_{o} = 20[/tex], then we have the following set of expressions:

[tex]\frac{dI}{dt} = 0.0001\cdot I\cdot (2000-I)[/tex] [tex]\blacksquare[/tex]

[tex]I(0) = \frac{1}{100}[/tex] [tex]\blacksquare[/tex]

[tex]I(t) = \frac{20\cdot e^{\frac{t}{5} }}{1+20\cdot e^{\frac{t}{5} }}[/tex] [tex]\blacksquare[/tex]

To learn more on differential equations, we kindly invite to check this verified question: https://brainly.com/question/1164377

Work Shown:

Replace every x with 4. Use the order of operations PEMDAS to simplify

f(x) = x + 21

f(4) = 4 + 21

f(4) = 25

The input 4 leads to the output 25.

Answer:

x < -( 8-2b) /a  a > 0

Step-by-step explanation:

−ax + 2b > 8

Subtract 2b from each side

−ax + 2b-2b > 8-2b

-ax > 8 -2b

Divide each side by -a, remembering to flip the inequality ( assuming a>0)

-ax/-a < ( 8-2b) /-a

x < -( 8-2b) /a  a > 0

Answer: [tex]x<\frac{-8+2b}{a}[/tex]

                  [tex]a>0[/tex]

Step-by-step explanation:

[tex]-ax+2b>8[/tex]

[tex]\mathrm{Subtract\:}2b\mathrm{\:from\:both\:sides}[/tex]

[tex]-ax>8-2b[/tex]

[tex]\mathrm{Multiply\:both\:sides\:by\:-1\:\left(reverse\:the\:inequality\right)}[/tex]

[tex]\left(-ax\right)\left(-1\right)<8\left(-1\right)-2b\left(-1\right)[/tex]

[tex]ax<-8+2b[/tex]

[tex]\mathrm{Divide\:both\:sides\:by\:}a[/tex]

[tex]\frac{ax}{a}<-\frac{8}{a}+\frac{2b}{a};\quad \:a>0[/tex]

[tex]x<\frac{-8+2b}{a};\quad \:a>0[/tex]

Answer:

Step-by-step explanation:

x+y>0, x>0, when y=0

x+y<-5 x<-5 when y=0

since the sign is only< then it is dotted line, and since one is greater and is less than they actually do not intersect

Answer:

No solution with slanted lines

Step-by-step explanation:

Answer:

Design thinking skills

Step-by-step explanation:

The design thinking skills is observable in individuals who can effectively use Intuition to create prototypes of abstract objects.

Jessie thus shows that she possess design thinking skills by been able to imagine abstract concepts at the same and she applies advanced mathematical formulas which in turn provides solutions to problems.

Answer:

Step-by-step explanation:

40.50/2 = 20.25

You have to divide by 2. This way both of the people will get the same amount of money.

Answer:

each friend will get

Step-by-step explanation:

20 .25

as 40 .50 ÷ 2 = 20 .25

hope this helps

pls can u heart and like and give my answer brainliest pls i beg u thx !!! : )

Answer:

a) The five ordered pairs are:-

(1,60) , (2,120) , (3,180) , (4,240) , (5,300)

b)When You divide the y value by x value for each ordered pair u find the slope.

c)The graph shows a proportional relationship.Because as x-value increases so does y-value.

d)Y=mx+b-->  Y=60x (No y-intercept because it starts from 0)

e)If a person hiked for 9 hours then the distance would be 540. Because If u plug in the number of hours in the x value of the equatione then u will get 540. Here's the work:-

Y=60(9)

Y=540

Step-by-step explanation:

Hope it helps u. And if u get it right pls give me brainliest.

Answer:

An architect is designing a house for the Frazier family. In the design he must consider the desires of the family and the local building codes. The rectangular lot on which the house will be built has 91 feet of frontage on a lake and is 158 feet deep.

The building codes states that one can build no closer than 10 ft. to the lot line. Write an inequality and solve to see how long the front of the house facing the lake may be.

------

length = 91 - 2*10 = 71 ft.

-------------------------------

The Fraziers requested that the house contain no less 2800 ft square and no more than 3200 ft square of floor sample. Write an inequality to represent the range of permissible widths for the house.

---------

2800 <= area <= 3200

2800 <= (length)(width) <= 3200

2800 <= 71w <= 3200

39.44 <= width <= 45.07

hope it helpsss

Step-by-step explanation:

Answer: An architect is designing a house for the Frazier family. In the design he must consider the desires of the family and the local building codes. The rectangular lot on which the house will be built has 91 feet of frontage on a lake and is 158 feet deep.

The building codes states that one can build no closer than 10 ft. to the lot line. Write an inequality and solve to see how long the front of the house facing the lake may be.

------

length = 91 - 2*10 = 71 ft.

-------------------------------

The Fraziers requested that the house contain no less 2800 ft square and no more than 3200 ft square of floor sample. Write an inequality to represent the range of permissible widths for the house.

---------

2800 <= area <= 3200

2800 <= (length)(width) <= 3200

2800 <= 71w <= 3200

39.44 <= width <= 45.07

Answer:

[tex]\large \boxed{\sf 15 \ \ and \ \ 24 \ \ }[/tex]

Step-by-step explanation:

Hello,

We can write the following, x being the second number.

[tex](2x-6)^2+x^2=801\\\\6^2-2\cdot 6 \cdot 2x + (2x)^2+x^2=801\\\\36-24x+4x^2+x^2=801\\\\5x^2-24x+36-801=0\\\\5x^2-24x-765=0\\\\[/tex]

Let's use the discriminant.

[tex]\Delta=b^4-4ac=24^2+4*5*765=15876=126^2[/tex]

There are two solutions and the positive one is

[tex]\dfrac{-b+\sqrt{b^2-4ac}}{2a}=\dfrac{24+126}{10}=\dfrac{150}{10}=15[/tex]

So the solutions are 15 and 15*2-6 = 30-6 = 24

Hope this helps.

Do not hesitate if you need further explanation.

Thank you

Answer:

[tex](-\infty, 0) \cup (0,2) \cup (2,\infty)[/tex]

Step-by-step explanation:

Remember that the domain of the product of functions is the intersection of domains, therefore when you intercept them you get the following interval.

[tex](-\infty, 0) \cup (0,2) \cup (2,\infty)[/tex]