What is the area of the parallelogram represented by the vertices A(-6, -1), B(-2, -1), C(-1,-4) and D(-5, -4)?

Answer:

a = 60°

Step-by-step explanation:

From the picture attached,

In the given triangle,

NE and ML are the perpendicular bisectors of two sides of the triangle.

Therefore, m∠MEK = m∠MLK = 90°

Since, sum of interior angles of polygon KLME = 360°

m∠MEK + m∠MLK + m∠EML + m∠EKL = 360°

90° + 90° + 120° + a = 360°

a = 360° - 300°

a = 60°

The ratio 2:3 means for every 2 inches on the original, the photocopy would be 3 inches.

3/2 = 1.5

The photocopied image is 1.5 times larger than the original.

Side BG on the original is side FG on the copy:

14 x 1.5 = 21 meters

FG = 21 meters

Answer:

FG = 21

Step-by-step explanation:

The ratio is 2:3

2               BC

----- = ----------------

3                FG

2              14

----- = ----------------

3                FG

Using cross products

2FG = 3*14

2FG = 42

Divide by 2

FG = 21

Answer:

Step-by-step explanation:

Relation 1 is a function

Relation 2 is not a function

Relation 3 is a function

Realation 4 is not a function

Answer:

See explanation

Step-by-step explanation:

Given

[tex]Flat = 20[/tex]

[tex]Visit = 5[/tex]

Required

The function to represent x visits

This is calculated as:

[tex]f(x) = Flat + Visit * x[/tex]

So, we have:

[tex]f(x) = 20 + 5 * x[/tex]

[tex]f(x) = 20 + 5x[/tex]

The second question is incomplete; however, I will explain how to calculate the horizontal asymptote of a rational function.

For polynomials with the same degree (i.e. m = n), the horizontal asymptote is calculated by dividing the coefficients of the highest degrees.

e.g.

[tex]f(x) = \frac{6x^2 + 1}{3x^2 + 4}[/tex] ---the degrees of both is 2

So, the horizontal asymptote is:

[tex]y = 6/3[/tex]

[tex]y =2[/tex]

If the numerator has a higher degree, then there is no horizontal asymptote.

If the denominator has a higher degree, then the horizontal asymptote is:

[tex]y = 0[/tex]

Answer:

First one is B, Second is 5

Step-by-step explanation:

got it right on edge

Answer:

26/35

Step-by-step explanation:

So we gotta add it together, so 1/7+0.6 or 1/7+3/5. Common denominator is 35, so the answer is 26/35

Answer:

A) 4.7 < x < -3

B) 1.14 < x < 1.09

Step-by-step explanation:

a)

x + 14 < 4x + 2 < 3x + 11

x + 14 < 4x

14 < 4x - x

14 < 3x

4.7 < x

2 < 3x + 11

2 - 11 < 3x

-9 < 3x

-3 < x

4.7 < x - 3 < x

4.7 < x < -3

b)

x + 8 < 8x - 6 < 5x + 12

x + 8 < 8x

8 < 8x - x

8 < 7x

1.14 < x

-6 < 5x + 12

12 < 5x + 6

12 < 11x

1.09 < x

1.14 < x 1.09 < x

1.14 < x < 1.09

Answer:

Step-by-step explanation:

15. = 2.39

and jus use mathaway lma

Answer:

70 cents or 0.70 dollars

Step-by-step explanation:

one dime is 10 cents so if you have 7 than you have 70 cents

Answer: [tex]864\ m^2,\ 24\ m[/tex]

Step-by-step explanation:

Given

Perimeter of the rhombus is [tex]120\ m[/tex]

Length of one of the diagonal is [tex]d_1=36\ m[/tex]

All the sides of the rhombus are equal

[tex]\Rightarrow 4a=120\\\Rightarrow a=30\ m[/tex]

Area of the rhombus with side and one diagonal is

[tex]\Rightarrow \text{Area=}\dfrac{1}{2}d\sqrt{4a^2-d^2}[/tex]

Insert the values

[tex]\Rightarrow \text{Area=}\dfrac{1}{2}\times 36\times \sqrt{4\cdot 30^2-36^2}\\\\\Rightarrow \text{Area= }18\sqrt{3600-1296}\\\Rightarrow \text{Area= }18\times 48\\\Rightarrow \text{Area= }864\ m^2[/tex]

Area with two diagonals length can be given by

[tex]\Rightarrow \text{Area =}0.5\times d_1\times d_2 \\\text{Insert the values}\\\Rightarrow 864=36\times d_2\\\Rightarrow d_2=24\ m[/tex]

Thus, the area of the rhombus is [tex]864\ m^2[/tex] and the length of the other diagonal is [tex]24\ m[/tex]

Answer:

x = 60.6 units

Step-by-step explanation:

Hi there!

In this right triangle, we're given the measure of an angle, the side opposite that angle and another side adjacent to that angle (that is not the hypotenuse). In this circumstance, we can use the tangent ratio to help us solve for the missing side:

[tex]tan\theta=\frac{opp}{adj}[/tex]

Plug in the given information:

opp = 27, adj = x, θ = 24

[tex]tan(24)=\frac{27}{x}\\x=\frac{27}{tan(24)} \\x=60.6[/tex]

Therefore, the length of the missing side is 60.6 units when rounded to the nearest tenth.

I hope this helps!

Answer:

4,1,-6

3,-3,-4

x-axis,5,0

2,-7,3

1,4,2

y-axis,0,-4

Step-by-step explanation:

Answer:

siis  corretoed

Step-by-step explanation:

Answer: Is this a question? Or statement Yes they can be angles

Step-by-step explanation:

Can you put picture, it is not clear

Answer:

1. 22:24 and 33:36

2. Yes, because for every time x move two times y will move three times in a graph, just like if a ratio between boys and girls was 2:3 it would mean that for every 2 boys there are 3 girls.

Step-by-step explanation:

11:12 = 22:24 and 33:36

Answer:

see graph

Step-by-step explanation:

The function that is shown below is the graph of the given function [tex]y = log_{4}(x+3)[/tex] .

"A function from a set X to a set Y assigns to each element of X exactly one element of Y. The set X is called the domain of the function and the set Y is called the codomain of the function."

The given function is:

[tex]y = log_{4}(x+3)[/tex]

For [tex]x = -2[/tex], [tex]y = log_{4}(-2+3) = log_{4}1 = 0[/tex]

For [tex]x = -1[/tex], [tex]y = log_{4}(-1+3) = log_{4}2 = 0.5[/tex]

For [tex]x = 0[/tex], [tex]y = log_{4}(0+3) = log_{4}3 = 0.793[/tex]

For [tex]x = 1[/tex], [tex]y = log_{4}(1+3) = log_{4}4 = 1[/tex]

For [tex]x = 2[/tex], [tex]y = log_{4}(2+3) = log_{4}5 = 1.161[/tex]

By putting the values of (x, y) in the graph, we get the graph of [tex]y = log_{4}(x+3)[/tex].

Learn more about a function here:

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Answer:

Hello,

Answer 8a²c(a+3c) =8a³c+24a²c

Step-by-step explanation:

2a²+6ac=2a(a+3c)

4a²c+12ac²=4ac(a+3c)

H.C.F=2a(a+3c)*4ac=8a²c(a+3c)

Answer:

1st expression:2a×a+2×3×a×c

2nd expression:2×2×a×a+2×2×3×a×c×c

here,

2×a×a+2×3×a×c

2a^a+6ac

Answer:

Step-by-step explanation:

In order to write the equation of this line, we need to pick 2 points on the graph where the line goes right through the intersection of the grids. Actually, you only need 1 of these, because the line goes through at (0, 15). Another point then can be (6, 6). Locate this point so you know what I means when I say that the line goes right through where the grids intersect at x = 6 and y = 6 (as opposed to somewhere in the middle of one of these grids). Find the slope between these 2 points:

[tex]m=\frac{6-15}{6-0}=-\frac{9}{6}=-\frac{3}{2}[/tex]

Since there's only one choice with that slope, that is the choice you want.

Answer:

9(n + 22) ≤ -15

Step-by-step explanation:

9 times the sum of a number and 22 is at most −15

Let n = number

Times = multiplication

Sum = Addition

9 * n + 22

We must add parenthesis on the end on 22 and before n

We must do this because it says 9 times the sum of a number and 22 which means that 9 is being multiplied by both n and 22

We would have 9(n + 22)

9(n + 22) is at most -15

At most meaning that -15 is the highest number that 9(n + 22) can equal

Because it says at most 9(n + 22) can equal -15 or anything less than -15

So at most can be replaced with ≤

9(n + 22) ≤ -15

Answer:

1/27

Step-by-step explanation:

- substitute the variable with the given datas

[3^-5 * 3^4]^3 * [3^-4 * (-9)^3]^0

- use  the properties of power

(3^-1)^3 * 1

3^-3

(1/3)^3

1/27

The answer is number 3 i.e. 1/27

Answer:

-3/2

Step-by-step explanation:

Using the slope formula, we can find the slope of the line shown

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

    = ( -1-3)/(-3-3)

   = -4/-6

  = 2/3

A line that is perpendicular is the negative reciprocal

-1/ (2/3) = -3/2

Answer:

the slope is 0

Step-by-step explanation:

y2-y1/x2-x1=-1-3/-3-3

y=0

Answer:

A(-2,1)》A1(-8, 4)

B(-3,3)》B1(-12,12)

C(-1,3)》C1(-4,12)

HOPE IT HELPS....

Answer:

[tex]TD \cong GS[/tex]

Step-by-step explanation:

See comment for complete question

Given:

[tex]TM \cong GL[/tex]

[tex]MD \cong LS[/tex]

Required

The information that shows [tex]\triangle MTD \cong \triangle LGS[/tex] by SSS

By SSS implies that, the three sides of both triangles are congruent

Already, we have:

[tex]TM \cong GL[/tex]

[tex]MD \cong LS[/tex]

The third side of [tex]\triangle MTD[/tex] is [tex]TD[/tex]

The third side of [tex]\triangle LGS[/tex] is [tex]GS[/tex]

So, for both to be congruent by SSS, the third sides must be congruent

i.e.

[tex]TD \cong GS[/tex]

Answer:

UV = 10

HG = 25

Step-by-step explanation:

UV = (11+9)/2 = 20/2 = 10

HG = (28+22)/2 = 50/2 = 25

Answer:

Below in bold.

Step-by-step explanation:

First problem:

The length of the mid segment is the mean of the 2 outer segments.

So it is (11+9)/2

= 20/2

= 10.

Second problem:

As above, it is (22+28) / 2

= 50/2

= 25.

Answer:

The correct answer is B. $13,875.

Step-by-step explanation:

Since X and Y are in partnership with capital contributions of $ 50000 and $ 30000 respectively, and the partnership agreement provides that profits are to be shared in proportion to capital contributions and each partner is entitled to 10% interest on capital, and profit for the year was $ 37000, to determine what was the total amount credited to Y’s current account at the end of the year the following calculation must be performed:

50,000 + 30,000 = 80,000

80,000 = 100

30,000 = X

30,000 x 100 / 80,000 = X

37.5 = X

37,000 x 0.375 = X

13.875 = X

Answer:

the correct answer is option 1.

The correct explanation is: All equilateral triangles are congruent and therefore similar, with side lengths in a 1:1 ratio.

Those triangles look the same but are different in size.

And in similar triangles,

the corresponding sides are in proportion to each other and the corresponding angles are equal to each other.

In an equilateral triangle, all three sides are congruent, and all three angles are congruent.

Therefore, any equilateral triangle can be transformed into any other equilateral triangle through a combination of translations, rotations, and reflections, without changing the size or shape of the triangle.

Thus, all equilateral triangles are similar, with side lengths in a 1:1 ratio, since they have the same shape but may differ in size.

To learn more about similar triangles;

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Answer:

x>4

Step-by-step explanation:

3/(x-4) >0

Divide each side by 3

3/(x-4) * 1/3 >0*1/3

1/(x-4) >0

We know if 1/(x-4) >0 then x-4 > 0

x-4>0

Add 4 to each side

x-4+4 >0+4

x>4

[tex]\boxed{\large{\bold{\textbf{\textsf{{\color{blue}{Answer}}}}}}:)}[/tex]

[tex]:\implies{\dfrac{3}{x-4}>0}\\\\:\hookrightarrow{\dfrac{3}{x-4}×\dfrac{1}{3}>0×\dfrac{1}{3}}\\\\:\longrightarrow{x-4>0}\\\\:\implies{x-4+4>0+4}\\\\ :\dashrightarrow{\sf{x>4}}[/tex]

Answer:

x < -51/10

General Formulas and Concepts:

Pre-Algebra

Equality Properties

Step-by-step explanation:

Step 1: Define

Identify

-2(5x + 1) > 49

Step 2: Solve for x

Answer:

x < -51/10

Step-by-step explanation:

-10x -2 > 49

-10x > 51

x < -51/10

Answer:

A

Step-by-step explanation:

Additive inverse is (5/3) and multiplicative inverse is (-3/5). Their product is (-1)

Inverse means the opposite of something.

In this case the opposite of -5/3 is 5/3.

So the answer is: 5/3

Answer:

Step-by-step explanation:

This is quite a doozy, my friend. We will set up a d = rt table, fill it in...and pray.

The table will look like this before we even fill anything in:

            d        =        r        *        t

SUV

sedan

Ok now we start to pick apart the problem. Motion problems are the hardest of all story problems ever. This is because there are about 100 ways a motion problem can be presented. So far what we KNOW for an indisputable fact is that the distance from Georgetown to Greenville is 120 km. So we fill that in, making the table:

             d      =      r      *      t

SUV     120

sedan  120

The next part is derived from the sentence "After an hour, the SUV was 24 km ahead of the sedan." This tells us the rate of the SUV in terms of the sedan. If the SUV is 24 km ahead of the sedan in 1 hour, that tells us that the rate of the sedan is r and the rate of the SUV is r + 24 km/hr. BUT we have other times in this problem, one of them being 25 minutes. We have a problem here because the times either have to be in hours or minutes, but not both. So we will change that rate to km/min. Doing that:

24 [tex]\frac{km}{hr}[/tex] × [tex]\frac{1hr}{60min}=.4\frac{km}{min}[/tex] So now we can fill in the rates in the table:

            d      =      r      *      t

SUV    120    =   r + .4

sedan 120    =     r

They left at the same time, so now the table looks like this:

             d      =      r      *      t

SUV    120     =   r + .4  *      t

sedan  120    =      r      *      t

We will put in the time difference of 25 minutes in just a sec.

If d = rt, then the equation for each row is as follows:

SUV:   120 = (r + .4)t

sedan:   120 = rt

Since the times are the same (because they left at the same time, we will set the equations each equal to t. The distances are the same, too, I know that, but if we set the distances equal to each other and then solve the equations for a variable, the distances cancel each other out, leaving us with nowhere to go. Trust me, I tried that first! Didn't work.

Solving the first equation for time:

sedan:  [tex]\frac{120}{r}=t[/tex]  That's the easy one. Now the SUV. This is where that time difference of 25 minutes comes in from the last sentence. Let's think about what that sentence means in terms of the times of each of these vehicles. If the sedan arrived 25 minutes after the SUV, then the sedan was driving 25 minutes longer; conversely, if the sedan arrived 25 minutes after the SUV, then the SUV was driving 25 minutes less than the sedan. The latter explanation is the one I used in the equation. Again, if the SUV was driving 25 minutes less than the sedan, and the equations are solved for time, then the equation for the SUV in terms of time is

[tex]\frac{120}{r+.4}=t-25[/tex] and we solve that for t:

[tex]\frac{120}{r+.4}+25=t[/tex]

Again, going off the fact that times they both leave are the same, we set the equations equal to one another and solve for r:

[tex]\frac{120}{r+.4}+25=\frac{120}{r}[/tex]

I began by first multiplying everything through by (r + .4) to get rid of it in the denominator. Doing that:

[tex][r+.4](\frac{120}{r+.4}) +[r+.4](25)=[r+.4](\frac{120}{r})[/tex] which simplifies very nicely to

[tex]120+25(r+.4)=\frac{120}{r}(r+.4)[/tex]  So maybe it's not so nice. Let's keep going:

[tex]120+25r+10=\frac{120r}{r}+\frac{48}{r}[/tex] and keep going some more:

[tex]130+25r=120+\frac{48}{r}[/tex] and now we multiply everything through by r to get rid of THAT denominator:

[tex]r(130)+r(25r)=r(120)+r(\frac{48}{r})[/tex] giving us:

[tex]130r+25r^2=120r+48[/tex] Now we have a second degree polynomial we have to solve by factoring. Get everything on one side and factor using the quadratic formula.

[tex]25r^2+10r-48=0[/tex]

That factors to

r = 1.2 and r = -1.6 and both of those rates are in km/minute. First of all, we cannot have a negative rate (this is not physics where we are dealing with velocity which CAN be negative) so we throw out the -1.6 and convert the rate of 1.2 km/minute back to km/hr:

[tex]1.2\frac{km}{min}[/tex] × [tex]\frac{60min}{1hr}[/tex] and we get

r = 72 km/h, choice B.

Wow...what a pain THAT was, right?!