Can anybody help with these two problems for me please this for algebra finding slope

Answer:

104

Step-by-step explanation:

89x4=356 90x5=450

450=356=104

Answer:

94

Step-by-step explanation:

((89x4)+94) divided by 5 = 90

Answer:

Step-by-step explanation:

The question is incomplete. Here is the complete question.

The online program at a certain university had an enrollment of 580 students at its inception and an enrollment of 1725 students 3 years later. Assume that the enrollment increases by the same percentage per year. a) Find the exponential function E that gives the enrollment t years after the online program's inception. b) Find E(11), and interpret the result. c) When will the program's enrollment reach 5500 students?

Let E0 be the number of students that enrolled at inception

E(t) be the number of students that enrolled at any time t

At inception i.e at t = 0, E0 = 580

The exponential function E that gives the enrollment at inception will be expressed as;

E(t) = E0e^kt (note that the exponential power is positive since number of students keep increasing yearly).

At inception:

E(0) = 580e^k(0)

E(0) = 580(1)

E(0) = 580

If after 3years, enrollment increases to 1725 students i.e at t = 3, E(t) = 1725. The equation becomes;

E(3) = P0e^kt

E(3) = 580e^3k

1725 = 580e^3k

Make k the subject of the formula:

e^3k = 1725/580

e^3k = 2.9741

Apply ln to both sides of the equation

ln(e^3k) = ln(2.9741)

3k = 1.08995

k = 1.08995/3

k = 0.3633

The exponential function E that gives the enrollment t years after the online program's inception will be expressed as;

E(t) = E0e^kt

Given E0 = 580 and k = 0.3633

E(t) = 580e^0.3633t

b) To get E(11), we will substitute t = 11 into the resulting expression in (a) above to have;

E(t) = 580e^0.3633t

E(11) = 580e^0.3633(11)

E(11) = 580e^3.9965

E(11) = 580×54.4074

E(11) = 31,556.287

E(11) ≈ 31,556 students

This means that about 31,556students enrolled 11years later.

c) To know when the program's enrollment reach 5500 students, we will set E(t) to 5500, E0 to 580 and k to 0.3633 and calculate the value of t

From the equation E(t) = 580e^0.3633t

5500 = 580e^0.3633t

e^0.3633t = 5500/580

e^0.3633t = 9.4828

Apply ln to both sides

ln(e^0.3633t) = ln(9.4828)

0.3633t = 2.2495

t = 2.2495/0.3633

t = 6.19

t ≈ 6years

This means that the program's enrollment will reach 5500 students 6 years later.

Answer:

30 divided by 6=5

Step-by-step explanation:

a=5

5 times 6 is the product of 30

Answer: D

Step-by-step explanation:

Answer:

The answer is

Step-by-step explanation:

Equation of a line is y = mx + c

where

m is the slope

c is the y intercept

To find the equation of the parallel line we must first find the slope of the original line.

The equation is

y = 1/4x + 6

Comparing with the general equation above

Slope = 1/4

Since the lines are parallel their slope are also the same

That's

Slope of parallel line = 1/4

So the equation of the parallel line using point (-4 , 3) and slope 1/4 is

We have the final answer as

Hope this helps you

Answer:

1/5

Step-by-step explanation:

I put it in a graphing calculator and converted it into a fraction

Answer:

Step-by-step explanation:

Part A ;

[tex]\sqrt{7} +\sqrt{3} +\sqrt{98} -\sqrt{18} \\\\\sqrt{98}=7\sqrt{2}\\\sqrt{18}=3\sqrt{2}\\\\=\sqrt{7}+\sqrt{3}+7\sqrt{2}-3\sqrt{2}\\\\\mathrm{Add\:similar\:elements:}\:7\sqrt{2}-3\sqrt{2}=4\sqrt{2}\\\\=\sqrt{7}+\sqrt{3}+4\sqrt{2}[/tex]

Part B ;

[tex]3\sqrt{5} - 3\sqrt{11} + 2\sqrt{121} -3\sqrt{90} \\\\2\sqrt{121}=22\\3\sqrt{90}=9\sqrt{10}\\\\=3\sqrt{5}-3\sqrt{11}+22-9\sqrt{10}[/tex]

Answer:

[tex]g(x) = x - 11[/tex]

Step-by-step explanation:

Given

[tex]f(x) = x - 5[/tex]

Required

Translate to the right by 6 units

When a function is translated to the right, the resulting function becomes

[tex]g(x) = f(x - h)[/tex]

Where h is the number of units moved;

Since; h = 6

[tex]g(x) = f(x - h)[/tex]

becomes

[tex]g(x) = f(x - 6)[/tex]

Solving for f(x - 6);

[tex]f(x) = x - 5[/tex]

Substitute x - 6 for x

[tex]f(x - 6) = x - 6 - 5[/tex]

[tex]f(x - 6) = x - 11[/tex]

This implies that [tex]g(x) = x - 11[/tex] is the result of the translation

Answer:

4 2/4 (NOT SIMPLIFIED)

Step-by-step explanation:

Comprising means made of (aka including). You just add 1 1/4+1 1/4+1 1/4.

=3 3/4

The 1/8 is for the front and back cover. The bookworm ate through 3 volumes so there is 6 covers. So times 1/8 by 6. I turned the 6 into a fraction by putting it over 1. So now you have 1/8×6/1=6/8. Now you are going to add the 3 3/4 to the 6/8. You are going to turn the whole fraction to an improper. You are going to multiply the whole number(3) to the denominator(4), add that to the numerator(3) , and put it over the original denominator (4)= 15/4. Now you are going to set up the problem were you add the 6/8 to the 15/4. To add fractions the denominator have to be the same. With this problem you have two options: you can either do an equation to change the 6/8 denominator equal to 4 or an equation to change the 15/4 denominator equal to 8. Either one would work fine. I'm going to show you how to set the 6/8 denominator to 4.

1. Set it up as an equation                       6/8=x/4

2. Find what gets the first denominator to the second in this case dividing by 2                                                          6/8=x/4

3. What you do to the bottom is what you have to do with the top. So to find x, you need to divide 6 by 2= 3

So now your new equation is now 3/4+15/4. Adding this should give you 18/4, but most likely you have to write it as a proper fraction. To do this you are going to divide the 18 by 4. Your remainder will become your new numerator and the 4 will still be you denominator. The number times 4 fully goes into 18 will be your new whole number.

18/4=4.5

Answer:

D. 15cm2

Step-by-step explanation:

To find the area of a triangle, you multiply the base (10cm) and height of the triangle (3cm) and divide that by 2. So 10×3=30. Then, you need to divide this by 2. Since 30 is an even number, you won't get a decimal. So 30/2=15. And dont forget your units!

Answer:

c) none of the above

Step-by-step explanation:

this is because we should put the equation like:

1/3 - (-4/3) because the distance is 5 units and

1/3 + 4/3 = 5/3

this is because (-) × (-) = (+)

Answer:

y = 3

Step-by-step explanation:

Step 1: Write equation

-2(7 - y) + 4 = -4

Step 2: Subtract 4 on both sides

-2(7 - y) = -8

Step 3: Distribute

-14 + 2y = -8

Step 4: Add 14 to both sides

2y = 6

Step 5: Divide both sides by 2

y = 3

Complete Question

The  complete question is shown on the first uploaded image

Answer:

A

is dimensionally consistent

B

is not dimensionally consistent

C

is dimensionally consistent

D

is not dimensionally consistent

E

is not dimensionally consistent

F

is dimensionally consistent

G

is dimensionally consistent

H

is not dimensionally consistent

Step-by-step explanation:

From the question we are told that

   The equation are

                        [tex]A) \   \  a^3  =  \frac{x^2 v}{t^5}[/tex]

                       [tex]B) \   \  x  =  t [/tex]

                       [tex]C \ \ \ v  =  \frac{x^2}{at^3}[/tex]

                      [tex]D \ \ \ xa^2 = \frac{x^2v}{t^4}[/tex]

                      [tex]E \ \ \ x  = vt+ \frac{vt^2}{2}[/tex]

                     [tex]F \ \ \  x = 3vt[/tex]

                    [tex]G \ \ \  v =  5at[/tex]

                    [tex]H \ \ \  a  =  \frac{v}{t} + \frac{xv^2}{2}[/tex]

Generally in dimension

     x - length is represented as  L

     t -  time is represented as T

     m = mass is represented as M

Considering A

           [tex]a^3  =  (\frac{L}{T^2} )^3 =  L^3\cdot T^{-6}[/tex]

and    [tex]\frac{x^2v}{t^5 } =  \frac{L^2 L T^{-1}}{T^5}  =  L^3 \cdot T^{-6}[/tex]

Hence

           [tex]a^3  =  \frac{x^2 v}{t^5}[/tex] is dimensionally consistent

Considering B

            [tex]x =  L[/tex]

and      

            [tex]t = T[/tex]

Hence

      [tex]x  =  t[/tex]  is not dimensionally consistent

Considering C

     [tex]v  =  LT^{-1}[/tex]

and  

    [tex]\frac{x^2 }{at^3} =  \frac{L^2}{LT^{-2} T^{3}}  =  LT^{-1}[/tex]

Hence

   [tex]v  =  \frac{x^2}{at^3}[/tex]  is dimensionally consistent

Considering D

    [tex]xa^2  = L(LT^{-2})^2 =  L^3T^{-4}[/tex]

and

     [tex]\frac{x^2v}{t^4}  = \frac{L^2(LT^{-1})}{ T^5} =  L^3 T^{-5}[/tex]

Hence

    [tex] xa^2 = \frac{x^2v}{t^4}[/tex]  is not dimensionally consistent

Considering E

   [tex]x =  L[/tex]

;

   [tex]vt  =  LT^{-1} T =  L[/tex]

and  

    [tex]\frac{vt^2}{2}  =  LT^{-1}T^{2} =  LT[/tex]

Hence

   [tex]E \ \ \ x  = vt+ \frac{vt^2}{2}[/tex]   is not dimensionally consistent

Considering F

     [tex]x =  L[/tex]

and

    [tex]3vt = LT^{-1}T =  L[/tex]      Note in dimensional analysis numbers are

                                                       not considered

  Hence

       [tex]F \ \ \  x = 3vt[/tex]  is dimensionally consistent

Considering G

    [tex]v  =  LT^{-1}[/tex]

and

    [tex]at =  LT^{-2}T =  LT^{-1}[/tex]

Hence

      [tex]G \ \ \  v =  5at[/tex]   is dimensionally consistent

Considering H

     [tex]a =  LT^{-2}[/tex]

,

       [tex]\frac{v}{t}  =  \frac{LT^{-1}}{T}  =  LT^{-2}[/tex]

and

    [tex]\frac{xv^2}{2} =  L(LT^{-1})^2 =  L^3T^{-2}[/tex]

Hence

    [tex]H \ \ \  a  =  \frac{v}{t} + \frac{xv^2}{2}[/tex]  is not dimensionally consistent

We want to see which ones of the given expressions are dimensionally consistent. We will see that the correct options are:

This means that we have the same units in the left and in the right side of the equation.

The units are:

Now we can analyze the expressions to see the units in each one, I will show you how to do it:

a) a = v/t + x*v^2

Replacing the units we have:

[m/s^2] = [m/s]/[s] + [m]*[m^2/s^2]

[m/s^2] = [m/s^2] + [m^3/s^2]

You can see that we have an m^3 in the right side, so these are not equivalent.

b) x = 3*v*t

Replacing the units we have:

[m] = 3*[m/s]*[s] = 3*[m]

So yes, the units are the same in both sides, so this is dimensionally consistent.

With the same procedure we can see that:

So the correct options are b and h.

If you want to learn more about dimensions, you can read:

https://brainly.com/question/20384972

Answer:

Step-by-step explanation:

Let b

Bill's rate of travel = x

Let Julie's rate of travel = 3x

The time each travels is 2.5 hours

Together they travel a distance of 190 miles

d = r*t

Bill's distance covered is r * t = 2.5x

Julie's distance travel is 2.5 * 3x = 7.5x

The total distance = Bill + Julie

2.5x + 7.5x = 190

10x = 190

x = 190/10

x = 19

Bill's distance = 2.5 * 19 = 47.5

Julie's distance = 19 * 7.5 = 142.5

Rate = distance / time

Bill's rate = 47.5 / 2.5 = 19 miles per hour.

Julie's rate = 142.5/2.5 = 57 miles per hour

Is there an easier way?

You should notice that Bill's rate is just x which is 19 miles / hour

You can also notice that Julie is going 3 times as fast or 19*3 = 57

Answer:

The displacement of the object on this intervals is 1.33 m.

Step-by-step explanation:

Given that,

The function of velocity is

[tex]v=\dfrac{1}{2t+4}\ m/s[/tex]

For 0 ≤ t ≤8 , n = 2

We need to calculate the intervals

Using formula for intervals

For, n = 1

[tex]\Delta x=\dfrac{t_{f}-t_{i}}{n}[/tex]

[tex]\Delta x=\dfrac{8-0}{2}[/tex]

[tex]\Delta x=4[/tex]

So, The intervals are (0,4), (4,8)

We need to calculate the velocity

Using given function

[tex]v=\dfrac{1}{2t+4}[/tex]

For first interval (0,4),

Put the value into the formula

[tex]v_{0}=\dfrac{1}{2\times0+4}[/tex]

[tex]v_{0}=\dfrac{1}{4}[/tex]

For first interval (4,8),

Put the value into the formula

[tex]v_{4}=\dfrac{1}{2\times4+4}[/tex]

[tex]v_{4}=\dfrac{1}{12}[/tex]

We need to calculate the total displacement

Using formula of displacement

[tex]D=(v_{0}+v_{4})\times(\Delta x)[/tex]

Put the value into the formula

[tex]D=(\dfrac{1}{4}+\dfrac{1}{12})\times4[/tex]

[tex]D=1.33\ m[/tex]

Hence, The displacement of the object on this intervals is 1.33 m.

The displacement of the object whose velocity function is given is 1.33 m

The given parameters are:

[tex]\mathbf{v = \frac{1}{2t + 4},\ 0 \le t \le 8; n =2}[/tex]

The end point of intervals is calculated as:

[tex]\mathbf{\triangle t = \frac{b - a}{n}}[/tex]

So, we have:

[tex]\mathbf{\triangle t= \frac{8 - 0}{2}}[/tex]

[tex]\mathbf{\triangle t = \frac{8}{2}}[/tex]

[tex]\mathbf{\triangle t= 4}[/tex]

So, the intervals are (0,4) and (4,8)

Calculate the velocity at the beginning of each interval

[tex]\mathbf{v_0 = \frac{1}{2(0) + 4} = \frac 14}[/tex]

[tex]\mathbf{v_4 = \frac{1}{2(4) + 4} = \frac 1{12}}[/tex]

Calculate the displacement (S) using:

[tex]\mathbf{S = (v_0 + v_4) \times \triangle t}[/tex]

So, we have:

[tex]\mathbf{S = (1/4 + 1/12) \times 4}[/tex]

Expand

[tex]\mathbf{S = 1 + 1/3}[/tex]

Add

[tex]\mathbf{S = 1 \frac 13}[/tex]

Express as decimals to 2 decimal places

[tex]\mathbf{S = 1.33}[/tex]

Hence, the displacement is 1.33 m

Read more about displacement at:

https://brainly.com/question/17131235

Answer:

Step-by-step explanation:

Hello,

[tex](goh)(x)=g(h(x))=2\left( \sqrt{x^2+1}\right)^2=2(x^2+1)\\\\x=(goh)((goh)^{-1}(x))=2((goh)^{-1}(x)^2+1)\\ \\\left((goh)^{-1}(x)\right)^2=\dfrac{x}{2}-1=\dfrac{x-2}{2}\\ \\(goh)^{-1}(x)=\sqrt{\dfrac{x-2}{2}}[/tex]

And this is a function defined for x-2 [tex]\geq[/tex] 0, meaning x [tex]\geq[/tex] 2

Thanks

Steps to solve:

(5 - 3)^3 + (-3)^2(6 - 3)

~Simplify

2^3 + (-3)^2(3)

2^3 + (-3)^6

~Solve exponents

8 + (-729)

~Subtract

-721

Best of Luck!

Answer:

see I need help for chemistry pls anyone here helps me I must submit before 10:30

Answer:

when x goes opposite side of equals to sign of x changes and same goes with 4 then x=5+4 then the answer will be 9

Answer:

- 5

Step-by-step explanation:

Step 1:

4 ( 2y + 1 ) = 2 ( y - 13 )

Step 2:

8y + 4 = 2y - 26

Step 3:

6y + 4 = - 26

Step 4:

6y = - 30

Answer:

y = - 5

Hope This Helps :)

Answer:

4x-3y+20

Step-by-step explanation:

x=6.875

y=2.5

2x+y=30

x=12.5

y=-5

Answer:

Kindly check explanation

Step-by-step explanation:

From the relative frequency below:

For NORTH - SOUTH:

Monday - Thursday = 115 ; has a relative frequency of 75.16%,

Hence, we can obtain the row total since it amounts to 100% thus;

75.16% of row total = 115

0.7516× row total = 115

Row total = 115 / 0.7516 = 153.00

Hence, Friday - sunday:

Row total - (Monday-Thursday)

153 - 115 = 38

FOR EAST - WEST:

Monday - Thursday = 21 ; has a relative frequency of 25.30%,

Hence, we can obtain the row total since it amounts to 100% thus;

25.30% of row total = 21

0. 253 × row total = 21

Row total = 21 / 0.253 = 83.00

Hence, Friday - sunday:

Row total - (Monday-Thursday)

83 - 21 = 62

x = (38 / 100) × 100%

x = 0.38 × 100%

x = 38.00 %

Answer:

The basic explanation:

North - South/Friday - Sunday- 38 // Row total- 153

East - West/Friday - Sunday- 62 // Row total- 83

x- 38

Answer:

[tex]\huge\boxed{-4}[/tex]

Step-by-step explanation:

=> [tex]\sf -1 + (-3)[/tex]

According to the rule [tex]\sf + * - = -[/tex]

=> [tex]\sf -1-3[/tex]

Answer:

-4

Step-by-step explanation:

-1 + (-3)

To add two negative numbers, add their absolute values, and make the answer negative.

The absolute values of -1 and -3 are 1 and 3. We add 1 + 3 and get 4.

Then we make the answer negative.

-1 + (-3) = -4

Answer:

Step-by-step explanation:

a)the pre image us twice the same size as the image

Answer:

Difference between upper and lower limits is : 1,816

Step-by-step explanation:

A CI (confidence interval ) for  t student distribution is:

( μ₀  - t(α/2)* s/√n  ;    μ₀  + t(α/2)* s/√n )

Where:

μ₀  is the mean and s the standard deviation of the dstribution

n size of the sample

CI = 90 %     means   α = 10 %    α = 0,1      α/2  = 0,05

and degree of freedom   df = n - 1   df = 40

From t student table we get:

tα/2  =  1,6839

Then:

t(α/2)* s/√n  =  1,6839* 3,41/√40

t(α/2)* s/√n  =  0,908

8,73 - 0,908   =  7,822

8,73 + 0,908  = 9,638

CI (90%)  =  ( 7,822  ;  9,638 )

Difference between upper and lower cut-offs points is:

Δ = 1,816

Answer:

D:{x∈R}

R:{y∈R}

Step-by-step explanation:

This is just a linear function. I know this because the degree of the x-variable is 1.

Domain and range are sets of possible values the function can have - though not necessarily at the same time.

Thus, there are no restrictions to the domain and range unless context is given.

Therefore, the domain and range is:

D:{x∈R}

R:{y∈R}