The coordinates of the vertices of △JKL are J(1, 4) , K(6, 4) , and L(1, 1) .


The coordinates of the vertices of △J′K′L′ are J′(0, −4) , K′(−5, −4) , and L′(0, −1) .



What is the sequence of transformations that maps △JKL to △J′K′L′ ?


reflection across the x-axis

rotation of 180° about the origin

rotation of 90° counterclockwise about the origin

translation 1 unit left


A sequence of transformations that maps △JKL to △J′K′L′ is a _________ followed by a ___________?

Answer:

55 OZ

Step-by-step explanation:

Answer:

D

Step-by-step explanation:

Answer:

D

Step-by-step explanation:

Answer:

-3x -7

Step-by-step explanation:

Add the numbers:

x-12+ 5-4x

x-7 -4x

Combine like terms:

x-7 -4x

-3x -7

Solution:

-3x -7

Answer:

9/26

Step-by-step explanation:

2/159÷2/477=2/53÷2/159=6/52÷2/53=3

x÷6/52=3

x=18/52

x=9/26

Answer:

The quality standard is violated.

Step-by-step explanation:

In 3 cartons of eggs no more than 2% are crack.

Let's find out how many eggs are in the 3 cartons.

(3)(12) = 36

There are 36 eggs in all 3 cartons. To find how many are supposedly cracked we need to multiply 36 by the decimal for of 2%.

To change a percent to decimal form we move the decimal of the percent twice to the left. So 2% turns in to .02.

Now, multiply:

(36)(.02) = 0.72

Not even 1 egg in all 3 batches should be cracked.

Therefore, the quality standard has been violated.

In order for the quality standard to be upheld there would have to be 100 eggs and only 2 of them be cracked.

I hope this helps!!

- Kay :)

Answer: B.

Step-by-step explanation: Mr. Bates made the pretzel 35% SMALLER. That means that the remains of the pretzel is 65% of the original 100% of the pretzel. Therefore, the answer is B.

The segment joining the midpoints of the diagonals of a trapezoid is parallel to the bases

From the figure, we have the following coordinates

B = (a, 0)

D = (d, c)

The midpoint of line segment BD is calculated as

BD = 1/2 * (x1 + x2, y1 + y2)

So, we have

BD = 1/2 * (a + d, 0 + c)

This gives

BD = ((a + d)/2, c/2)

Also, we have:

Lines MN and AB are horizontal lines

This means that they have a slope of 0

i.e.

Slope of AB = Slope of MN = 0

Parallel lines have equal slope

Hence, the segment joining the midpoints of the diagonals of a trapezoid is parallel to the bases

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Answer:

Ratio: [tex]=\mathbf{\frac{6}{11}}[/tex]

The decimal form (showing up to four  decimal places): 0.5454

The decimal is a repeating decimal i.e 0.54545454..

Step-by-step explanation:

We need to find ratio of hockey players to soccer players

Number of players playing Hockey= 6

Number of players playing Soccer = 11

Ratio: [tex]\frac{Hockey\:Players}{Soccer\:Players} =\mathbf{\frac{6}{11}}[/tex]

The decimal form (showing up to four  decimal places): 0.5454

Tell whether the decimal is a terminating or a repeating decimal.

The decimal is a repeating decimal i.e 0.54545454..

Answer:

c. 35 miles per hour

Answer:

Step-by-step explanation:

Part A: No. The line would have to be straight in order for the graph to be linear.

Part B:

Linear: y=2x-4.

Nonlinear: y=x^2-6.

Answer:

The first option

Step-by-step explanation:

Answer:

1st one

Step-by-step explanation:

Opposite of positive 3 is negative 3.

Answer:

restrict the domain to (-pi/2, pi/2)

Step-by-step explanation:

edge 2020

Answer: c. restrict the domain to (-pi/2, pi/2)

Step-by-step explanation:

Edge :)

Answer:

goood luccccccckkkk...

Answer:

-1

Step-by-step explanation:

y = mx + c

m is the slope/gradient

to find slope = y2 - y1 / x2 - x1

= -8 - -10 / 4 - 6

m = -1

Answer:

x = 3/2

Step-by-step explanation:

Let's solve your equation step-by-step.

(3)(1)+2x=6

Step 1: Simplify both sides of the equation.

(3)(1)+2x=6

3+2x=6

2x+3=6

Step 2: Subtract 3 from both sides.

2x+3−3=6−3

2x=3

Step 3: Divide both sides by 2.

2x/2  3/2

x = 3/2

Hope this helps :)

Answer:

following is the answer for your question this can be done in the manner

Answer:

Step-by-step explanation:

4^-1

Answer:

HAHAHAHH

Step-by-step explanation:

Answer:

x=2

Step-by-step explanation:

15x-9x = 6x

10+2 = 12

6x=12

x=2

Answer: x=2

Step-by-step explanation:

9x+10 = 15x-2

-9x         -9x

____________

10 = 6x-2

+2        +2

__________

12=6x

Divide

x=2

Answer:

Let the numbers be a, b, c, d

[tex] { \tt{a + b + c + d = 93}}[/tex]

Let a be the multiple of 19, a = 57

And b = c = d

[tex]{ \tt{57+ 3b= 93}} \\ { \tt{3b = 36}} \\ { \tt{b = 12}}[/tex]

[tex]{ \tt{a = 57}} \\ { \tt{b = c = d = 12}} \\ \\ { \boxed{ \rm{12}}} \: \: { \boxed{ \rm{12}}} \: \: { \boxed{ \rm{12}}} \: \: { \boxed{ \rm{57}}}[/tex]

Answer:

12, 12, 12, 57

Step-by-step explanation:

Given:

If one of the four numbers is a multiple of 19 and the other three numbers are equal, then the sum of the other three numbers will be a multiple of 3.

Multiples of 19:  

19, 38, 57, 76, 95, ...

Subtract each multiple of 19 (that is less than 93) from 93:

⇒ 93 - 19 = 74  →  not divisible by 3

⇒ 93 - 38 = 55  →  not divisible by 3

⇒ 93 - 57 = 36  →  divisible by 3

⇒ 93 - 76 = 17  →  not divisible by 3

The only result that is a multiple of 3 is 36.  

Therefore, the multiple of 19 is 57.

To find the other three numbers, simply divide the result of the total less the found multiple of 19 by 3:

⇒ (93 - 57) ÷ 3 = 36 ÷ 3 = 12

Therefore, the four numbers in ascending order are:

12, 12, 12, 57

Answer:

It is a function

Step-by-step explanation:

A function is of the form:

[tex]\left[\begin{array}{c}x_1&x_2&x_3\\-&-&-\\x_n\end{array}\right][/tex]            [tex]\left[\begin{array}{c}y_1&y_2&y_3\\-&-&-\\y_n\end{array}\right][/tex]

Where each of the x values must be distinct.

The x values are referred to as domain while the y values are called range.

Having said that, the given data can be represented as follows:

[tex]\left[\begin{array}{c}-3&1&6\\7&12\end{array}\right][/tex] ----------- [tex]\left[\begin{array}{c}7&-8&-6\\-8&2\end{array}\right][/tex]

From the representation above, none of the x values are repeated.

Hence, it is a function

Answer:

14/21 or 2/3

Step-by-step explanation:

Answer:

Quinn will need 2 3/4 cups of brown sugar for 6 1/2 cups of flour

Step-by-step explanation:

If you change 3/4 into decimal form you get .75. You then multiply .75 by 3, because 2*3=6, to get 2.25 or 2 1/4. Then add the 1/2, or 2/4, to 2 1/4. You get 2 3/4

Answer:

B

Step-by-step explanation:

(look up limit calculator)

a has a limit

b does not exist

c  has a limit

d has a limit

e has a limit

Option D is correct one. Limit doesn't exist for [tex]\lim_{x\to \ 2} \frac{x^{2}-6x +8 }{x-2}[/tex].

The limit of f(x) at x = a is said to be exist if the function approaches the same value from both the sides i.e. LHL =RHL (left hand limit = right hand limit)

According to the question we have,

A. [tex]\lim_{x \to \ 1} \frac{2-x}{x+1}[/tex]

at x =1

LHL = [tex]\lim_{h \to \ 0 } \frac{2- (1-h)}{(1-h)+1}[/tex] = [tex]\frac{1}{2}[/tex]

RHL = [tex]\lim_{h \to \0} \frac{(2-(1+h) }{(1+h)+1)} =\frac{1}{2}[/tex]

Here, RHL = LHL

Hence, limit exist.

B. [tex]\lim_{x \to \ 2} \frac{x^{2} -8}{x-2}[/tex]

at x =2

LHL = [tex]\lim_{h \to \ 0} \frac{(2-h)^{2}-9 }{(2-h)-2}[/tex] =[tex]\lim_{h \to \ 0} \frac{4+h^{2} -9-4h}{2-h-2}[/tex] = [tex]\lim_{h \to \ 0 }\frac{2h-4}{-1} =\frac{-4}{-1} =4[/tex]

RHL = [tex]\lim_{h \to \ 0} \frac{(2+h)^{2} -9}{(2+h)-2}[/tex] = [tex]\lim_{h \to \ 0} \frac{4h +4 +h^{2}-9 }{h} = \lim_{h \to \ 0} \frac{2h+4}{1}[/tex]=4

Here, RHL = LHL

Hence, the limit exist.

C. [tex]\lim_{x \to \ 1} (x^{2} -2x-5)[/tex]

at x =1

LHL = [tex]\lim_{h \to \ 0} (( 1-h)^{2} -2(1-h)-5)= \lim_{h \to \ 0} (1 -2-5)=-6[/tex]

RHL = [tex]\lim_{h \to \ 0} ((1+h)^{2} -2(1+h)-5) = (1-2-5)=-6[/tex]

Here, LHL = RHL = -6

Hence, the limits exist.

D. [tex]\lim_{x \to \ 2} \frac{x^{2}-6x + 8 }{x - 2}[/tex]  

at x =2

LHL = [tex]\lim_{h \to \0} \frac{(2-h)^{2} -(2-h)+8}{(2-h)-2} = \lim_{h \to \ 0} \frac{4+h^{2}-4h-2+h+8 }{(2-h)-2}[/tex] =[tex]\lim_{h \to \ 0} \frac{2h-4+1}{1} =-3[/tex]

RHL = [tex]\lim_{h \to \ 0} \frac{(2+h)^{2} -(2+h) +8}{(2+h)-2} = \lim_{h \to \ 0} \frac{4+h^{2}+4h -2-h+8}{h} = \lim_{h \to \ 0} \frac{2h+4-1}{1} =3[/tex]

Here, RHL≠LHL

Hence, the limit doesn't exist.

E. [tex]\lim_{x \to \ 3} (x^{2} -3x-1)[/tex]

at x = 3

LHL = [tex]\lim_{h \to \ 0} ((3-h)^{2} -3(3-h)-1)[/tex] = [tex]9-9-1=-1[/tex]

RHL= [tex]\lim_{h \to \ 0} ((3+h)^{2} -3(3+h)-1)= 9-9-1 =-1[/tex]

Here, RHL = LHL

Hence, the limit exist.

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