A certain forest can support a population of 800 deer. There are currently 200 deer in the forest and their population is growing at a rate of 2% per year. Complete the sequence below that models this population growth.

The width of the mat around the photo is 3 inches

The combined area of the of the photo and math = 224 square inches

The combined area is represented as:

(2w +8)(2w + 10) = 224

Expand the equation using the distributive rule

[tex]4w^2+20w+16w+80=224\\\\4w^2+36w+80-224=0\\\\4w^2+36w-144=0[/tex]

Divide through by 4

[tex]w^2+9w-36=0[/tex]

Factorize the resulting quadratic equation to solve for w

[tex]w^2-3w+12w-36=0\\\\w(w-3)+12(w-3)=0\\\\(w-3)(w+12)=0[/tex]

w  -  3  =  0

w   =  3

w  +  12  =  0

w   =  -12

The width of the mat around the photo is 3 inches

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The width of the mat around the photo is 3 inches

Combined area:

2w +8)(2w + 10) = 224

where,

w = width of the mat around the photo

(2w +8)(2w + 10) = 224

open parenthesis

4w² + 20w + 16w + 80 = 224

4w² + 36w + 80 - 224 = 0

4w² + 36w - 144 = 0

divide through by 4

w² + 9w - 36 = 0

solve quadratic equation using factorization method

w² + 12w - 3w - 36 = 0

w(w + 12) - 3(w + 12) = 0

(w + 12) (w - 3) = 0

w + 12 = 0 or w - 3 = 0

w = -12 or w = 3

The width of the mat around the photo cannot be a negative value.

Therefore, the width of the mat around the photo is 3 inches

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Answer:

Enter the following transactions into richard’s checkbook register and state his ending balance: date type description amount 03/01 check #204 blue sky apartments $455. 00 03/05 dep payroll automatic deposit $390. 36 03/08 debit benny’s hamburgers and fries $9. 20 03/15 check #205 car payment $251. 59 03/19 dep payroll automatic deposit $390. 36 a. $715. 79 b. $1,009. 81 c. $780. 72 d. $880. 24.

1 km = 1000 m, so:

23,769 km = 23,769 · 1000 m = 23 769 m

Answer:

23769m

Step-by-step explanation:

Basicly, 3.769x 1000= 23769m

Please mark brainliest

Answer:

The answer is b

Step-by-step explanation:

During the 8-9 , Mr hare travel 40miles

For the time 9 onwards they travel concurrently,

Let x be the distance covered by both since they can only meet if they covered the sams distance,

x/50 = x/40 -1 ,where the 1 is the (8-9) 1 hr

x/40 - x/50 = 1

x = 200

Time take by Mr Hare = 200/ 40 = 5

from 8 add 5 hours will be 1

Note, i took this answer from an answer from another answer.

Answer:

huh

Step-by-step explanation:

Answer:

$140.41

Step-by-step explanation:

250 * 1.3687 = 3412.175 Euros.

- 150 Euros = 192.175

192.175  1.3687 = $140.41

Answer:

wat?

Step-by-step explanation:

merry christmas....yey!!

Answer:

The answer is 4.2,

Hope that helped. x

Step-by-step explanation:

3 x 7 (3*7) = 21

21 divided by 5 = 4.2.

Answer:

4.2

Step-by-step explanation:

Using PEMDAS,

We know that to solve (3*7) divided by 5, we will need to solve what is in the parenthesis.

So, we need to figure out what is 7*3, which is 21.

This is what we have right now:

21 divided by 5

The next step is dividing,

In the end, the answer is 4.2, hope this helped!

Using the BMI equation, it is found that weights between 139.4 and 195.2 pounds will keep his BMI between 20 and 28.

The BMI B, for a person of weight w(in pounds) and height h(in inches) is given by:

[tex]BMI = \frac{703w}{h^2}[/tex]

In this problem:

For a BMI of 20 inches, the weight is:

[tex]BMI = \frac{703w}{h^2}[/tex]

[tex]20 = \frac{703w}{70^2}[/tex]

[tex]w = \frac{20 \times 70^2}{703}[/tex]

[tex]w = 139.4[/tex]

For a BMI of 28 inches, the weight is:

[tex]BMI = \frac{703w}{h^2}[/tex]

[tex]28 = \frac{703w}{70^2}[/tex]

[tex]w = \frac{28 \times 70^2}{703}[/tex]

[tex]w = 195.2[/tex]

Weights between 139.4 and 195.2 pounds will keep his BMI between 20 and 28.

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Answer:

Step-by-step explanation:

This is an isosceles triangle. You can tell because 2 of the sides are equal and the third is not equal to either of the other two.

x + x + 56 = 180                  Combine the left

2x + 56 = 180                      Subtract 56 from both sides.

2x = 180 - 56

2x = 124                               Divide by 2

x = 124/2

x = 62

Answer:

n  =  6 ,  − 6

Step-by-step explanation:

Take the root of both sides and solve.

[tex]-\cfrac{1}{2}n^2+18=0\implies 18=\cfrac{1}{2}n^2\implies 18=\cfrac{n^2}{2}\implies 36=n^2 \\\\\\ \stackrel{\textit{taking }\sqrt{~~}\textit{ to both sides}}{\pm\sqrt{36}=\sqrt{n^2}}\implies \pm 6 = n[/tex]

Answer:

196

Step-by-step explanation:

Just plug in a -7 for everywhere there is a t:

[tex]g(t)=4t^2\\g(-7)=4(-7)^2\\g(-7)=4*49=196[/tex]

one hundred seventy nine

Step-by-step explanation:

3/5 are done, that means 2/5 are still to do.

that means these 30 km are these 2/5.

2/5 x = 30

2x = 150

x = 75

the total distance is 75 km.

Answer:

75 km

Step-by-step explanation:

2/5 = 30 km

→ Find 1/5

1/5 = 15 km

→ Now find 5/5 or 1

1 = 75 km

Answer:

-1.25x-5

Step-by-step explanation:

Multiply using Distribution:

1.25(-x-4)

-1.25x-5

My answer can't be simplified anymore, since there isn't any like terms.

Answer:

-8(10x-3) = -80x+24 = A

Step-by-step explanation:

yr welcome

Answer:

If I have to guess, I’ll say

Consecutive angles are congruent.

One pair of parallel sides.

Consecutive angles are supplementary.

Step-by-step explanation:

if I’m wrong, I’m sorry. If I’m right, mark me brainlest!

Answer:

Consecutive angles are supplementary.

Both pairs of opposite of angles are congruent.

Both pairs of opposite sides are congruent.

Step-by-step explanation:

I took the quiz and this was the correct answer.

Answer:

i think that it would be 0

Answer:

$245,000

Step-by-step explanation:

Let the intial cost = x

[tex]100\% + 36\% = 136\% = 1.36[/tex]

[tex]1.36x = 333200[/tex]

[tex]x = \frac{333200}{1.36} [/tex]

[tex]x = 245000[/tex]

Answer:

16

Have a good day <3

Answer:

28

Step-by-step explanation:

8*2=16

3*4=12

Then you add 16+12

hopefully it helps

Answer:

76

Step-by-step explanation:

plug in 4 in the equation:

8(4)2 + 3(4) = 76

Answer:

11%

Step-by-step explanation:

[tex]2\dfrac 29 = -\dfrac 45 m \\\\\implies 2+ \dfrac 29 =- \dfrac 45 m \\\\\implies \dfrac{20}9 = -\dfrac 45 m \\\\\implies 5(20) = (-4m)9\\\\\implies 100 = -36m\\\\\implies m = -\dfrac{100}{36} \\\\\implies m = -\dfrac{25}9\\\\\implies m=-2\dfrac 79[/tex]

Therefore, the answer is m = -25/9.

Hoped this helped.

[tex]-BrainaicUser1357-[/tex]

The way to balance a firm's need to succeed and the need for not asking the workers for perfection is through the offering of training to workers, which allows them to notably improve their performances autonomously, without this being directly instigated by the company's directory.

That is, through the education of workers in the different areas of their jobs, the level of the same will be improved and better results will be obtained.

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Answer:

2(x + 3) <9

Step-by-step explanation:

Two times the quantity of a number increased by 3 is less than 9

       2  ·                                          (x + 3   )                           < 9    

     = 2(x + 3) <9

Using the negative binomial distribution, it is found that:

a) [tex]E(X) = 15[/tex]

b) [tex]Var(X) = 6.67[/tex]

c) [tex]P(X = 12) = 0.0883[/tex]

Negative binomial distribution:

It is the number of trials until q successes of a binomial variable, with p probability of success.

The expected value is:

[tex]E(X) = \frac{q}{p}[/tex]

The variance is:

[tex]Var(X) = \frac{pq}{(1 - p)^2}[/tex]

The probability mass function is:

[tex]P(X = x) = C_{x+q-1,q-1}(1 - p)^xp^q[/tex]

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

In this problem, the parameters are: [tex]q = 6, p = 0.4[/tex].

Item a:

[tex]E(X) = \frac{6}{0.4} = 15[/tex]

Item b:

[tex]Var(X) = \frac{6(0.4)}{(1 - 0.4)^2} = 6.67[/tex]

Item c:

[tex]P(X = 12) = C_{17,5}(0.6)^12(0.4)^6 = 0.0883[/tex]

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The average distance is 300 yards

The ratio is given as:

Driver : 2-iron = 5 : 4

When the iron driver is 240 yards, we have:

Driver : 240 = 5 : 4

Express as fraction

Driver/240 = 5/4

Multiply both sides by 240

Driver= 300

Hence, the average distance is 300 yards

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Answer:

205+2w=195+3w

Step-by-step explanation:

:)

The equation to represent when the weights are the same will be 3w + 195 = 2w + 205 and the number of weeks is 10.

A connection between a number of variables results in a linear model when a graph is displayed. The variable will have a degree of one.

The linear equation is given as,

y = mx + c

Where m is the slope of the line and c is the y-intercept of the line.

Henry and Jose are putting on weight for football. Henry weighs 205 pounds and is acquiring 2 pounds each week Jose weighs 195 pounds and is acquiring 3 pounds each week.

Let 'w' be the number of weeks and 'y' be the total weight. Then the equations are given as,

y = 2w + 205

y = 3w + 195

If the weight becomes the same. Then the number of weeks is given as,

3w + 195 = 2w + 205

Simplify the equation, then we have

3w + 195 = 2w + 205

3w - 2w = 205 - 195

w = 10

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Answer:

5 weeks

Step-by-step explanation:

John deposits $5 per week and has $10, so his equation is 5x + 10.

Jasmine's equation is 3x + 20.

Make them equal to each other and solve for x:

5x + 10 = 3x + 20

Subtract 10 from each side

5x = 3x + 10

Subtract 3x from each side

2x = 10

Divide each side by 2

x = 5

Answer:

5 weeks

Step-by-step explanation:

John:$10,$15,$20,$25,$30,$35

Jasmine:$20,$23,$26,$29,$32,$35

5 weeks

For the first integral, z = π/4 is a pole of order 3 and lies inside the contour |z| = 1. Compute the residue:

[tex]\displaystyle \mathrm{Res}\left(\frac{e^z\cos(z)}{\left(z-\frac\pi4\right)^3}, z=\frac\pi4\right) = \lim_{z\to\frac\pi4}\frac1{(3-1)!} \frac{d^{3-1}}{dz^{3-1}}\left[e^z\cos(z)\right][/tex]

We have

[tex]\dfrac{d^2}{dz^2}[e^z\cos(z)] = -2e^z \sin(z)[/tex]

and so

[tex]\displaystyle \mathrm{Res}\left(\frac{e^z\cos(z)}{\left(z-\frac\pi4\right)^3}, z=\frac\pi4\right) = - \lim_{z\to\frac\pi4} e^z \sin(z) = -\frac{e^{\pi/4}}{\sqrt2}[/tex]

Then by the residue theorem,

[tex]\displaystyle \int_C \frac{e^z\cos(z)}{\left(z-\frac\pi4\right)^3} \, dz = 2\pi j \left(-\frac{e^{\pi/4}}{\sqrt2}\right) = \boxed{-\sqrt2\,\pi e^{\pi/4} j}[/tex]

For the second integral, z = 2j and z = j/2 are both poles of order 2. The second poles lies inside the rectangle, so just compute the residue there as usual:

[tex]\displaystyle \mathrm{Res}\left(\frac{\cosh(2z)}{(z-2j)^2\left(z-\frac j2\right)^2}, z=\frac j2\right) = \lim_{z\to\frac j2}\frac1{(2-1)!} \frac{d^{2-1}}{dz^{2-1}}\left[\frac{\cosh(2z)}{(z-2j)^2}\right] = \frac{16\cos(1)-24\sin(1)}{27}j[/tex]

The other pole lies on the rectangle itself, and I'm not so sure how to handle it... You may be able to deform the contour and consider a principal value integral around the pole at z = 2j. The details elude me at the moment, however.