Solve for (4x+6)+(-x+2) please i need help

Answer:

There are 35 adult dogs and 20 puppies

Step-by-step explanation:

Represent the puppies with P and the Adult dogs with A

Given

[tex]A : P = 7 : 4[/tex]

Dogs = 55

Required

Determine A and P

First, we need to sum the ratios;

[tex]Total = A + P[/tex]

[tex]Total = 7 + 4[/tex]

[tex]Total = 11[/tex]

Adult Dogs is then calculated as follows;

[tex]A = \frac{A\ Ratio}{Total} * Dogs[/tex]

[tex]A = \frac{7}{11} * 55[/tex]

[tex]A = \frac{385}{11}[/tex]

[tex]A = 35[/tex]

Puppies is calculated by subtracting the number of adult dogs from the total

[tex]P = Dogs - Adult\ Dogs[/tex]

[tex]P = 55 - 35[/tex]

[tex]P = 20[/tex]

Hence;

There are 35 adult dogs and 20 puppies

Answer:

x=3/8

Step-by-step explanation:

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:

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

Steps to solve:

x^2 + 2x + 3 when x = 4

~Substitute

4^2 + 2(4) + 3

~Simplify

16 + 8 + 3

~Add

24 + 3

~Add

27

Best of Luck!

Answer:45

Step-by-step explanation:

[9(7-3)+13]-[11-(6+9)]

[9(4)+13]-[11-(15)]

[36+13]-[4]

49-4

45

Answer: The answer is 53

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:

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: 6/11

Work Shown:

22 people total, 10 men, so 22-10 = 12 women are in the class.

12/22 = (2*6)/(2*11) = 6/11 is the fraction of women in the class

Answer:

I'm pretty sure the answer they want is 12/22 or 6/11 students are women although some people are not men or women which technically makes the question impossible to solve.

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}

Answer:

1/5

Step-by-step explanation:

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

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:

3.5

Step-by-step explanation:

total pretzel=10 1/2 or10.5

no. of brother = 3

now,

10.5 pretzel for each borther= 10.5/3

=3.5 ans

Answer:

complementary

Step-by-step explanation:

yes

The word that describes their measures of ∠1 and ∠2 is: complementary.

Recall:

From the information given, we know that:

m∠1 and m∠2 forms a right angle.

Since a right angle = 90 degrees, therefore:

m∠1 + m∠2 = 90° (complementary)

Therefore, the word that describes their measures of ∠1 and ∠2 is: complementary.

Learn more about complementary angles on:

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Answer:

y=400h

Where y is distance and h is time.

Answer:

c

Step-by-step explanation:

They are not commutative because the domains of f(x) and g(x) are different.

The functions  [tex]f(x) = x^{3}[/tex] and [tex]g(x) = \frac{1}{x^{3} }[/tex] are not commutative because domains of f(x) and g (x) are different.

Here,

The functions are [tex]f(x) = x^{3}[/tex] and [tex]g(x) = \frac{1}{x^{3} }[/tex] and graph of the function are shown in figure.

We have to check the function are composite or not.

What is Domain of function?

The domain of a function is the set of all possible inputs for the function.

Now,

Domain of function  [tex]f(x) = x^{3}[/tex] is ( -∞, ∞).

And, Domain of function [tex]g(x) = \frac{1}{x^{3} }[/tex] is ( -∞,0 ) U ( 0, ∞).

Hence, the domain of the functions are different so they are not commutative.

So, The functions  [tex]f(x) = x^{3}[/tex] and [tex]g(x) = \frac{1}{x^{3} }[/tex] are not commutative because domains of f(x) and g (x) are different.

Learn more about the Domain of function visit:

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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:

Step-by-step explanation:

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

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:

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Answer:

$220 is the amount of simple interest you will earn

Step-by-step explanation:

I = p * r * t

Principle = $2000

Rate = 5.5% (you need to change this to a decimal by dividing by 100) or

         0.055

Time = 2 years

I = 2000 * 0.055 * 2

I = $220

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!

Step-by-step explanation:

then put x valule into KL equation

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:

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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

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Answer:

x=1

Step-by-step explanation:

Anytime there is a vertical line, the y value becomes infinite as the line goes on forever. The only restriction on the line is on the x value of your coordinate. The same goes for a horizontal line, only the x value would become infinite and the y value would be constant.

Vertical Line: (x,∞)

Horizontal Line: (∞,y)

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