Which rigid transformation(s) can map TriangleMNP onto TriangleTSR? reflection only rotation only translation, then reflection translation, then rotation

Answer:

x=3/8+c/4

Step-by-step explanation:

Answer:

Step-by-step explanation:

Answer:

20.25-923

Step-by-step explanation:

20.25-923

912+11=923

Answer: 4

Step-by-step explanation: make me

Answer:

3/32

Step-by-step explanation:

The slope is found by

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

   = ( 19-22)/(-20-12)

   = -3/-32

   = 3/32

Answer:

meeeeeee

2+2+ fish

Step-by-step explanation:

Answer:

I LOVE among us. and 2 + 2 = 4

Step-by-step explanation:

Can you pleaseeee give me brainliest?

Answer:

B,C,E (last one)

Step-by-step explanation:

Answer:

90% of people marry there 7th grade love. since u have read this, u will be told good news tonight. if u don't pass this on nine comments your worst week starts now this isn't fake. apparently if u copy and paste this on ten comments in the next ten minutes you will have the best day of your life tomorrow. you will either get kissed or asked out in the next 53 minutes someone will say i love

Step-by-step explanation:

Answer:

x = 5

x = 2

x = -6

Step-by-step explanation:

Use synthetic division to determine whether x – 4 is a factor of:

–2x5 + 6x4 + 10x3 – 6x2 – 9x + 4

For x – 4 to be a factor, you must have x = 4 as a zero. Using this information, I'll do the synthetic division with x = 4 as the test zero on the left:

completed division

Since the remainder is zero, then x = 4 is indeed a zero of –2x5 + 6x4 + 10x3 – 6x2 – 9x + 4, so:

Yes, x – 4 is a factor of –2x5 + 6x4 + 10x3 – 6x2 – 9x + 4

Find all the factors of 15x4 + x3 – 52x2 + 20x + 16 by using synthetic division.

Remember that, if x = a is a zero, then x – a is a factor. So use the Rational Roots Test (and maybe a quick graph) to find a good value to test for a zero (x-intercept). I'll try x = 1:

completed division

This division gives a zero remainder, so x = 1 must be a zero, which means that  x – 1 is a factor. Since I divided a linear factor (namely, x – 1) out of the original polynomial, then my result has to be a cubic: 15x3 + 16x2 – 36x – 16. So I need to find another zero before I can apply the Quadratic Formula. I'll try x = –2:

completed division

Since I got a zero remainder, then x = –2 is a zero, so x + 2 is a factor. Plus, I'm now down to a quadratic, 15x2 – 14x – 8, which happens to factor as:

(3x – 4)(5x + 2)

Then the fully-factored form of the original polynomial is:

15x4 + x3 – 52x2 + 20x + 16

= (x – 1)(x + 2)(3x – 4)(5x + 2)

Given that  x = -3 + sqrt(11)   is a zero of x4 + 6x3 – 7x2 – 30x + 10, fully solve the

equation x4 + 6x3 – 7x2 – 30x + 10 = 0.

Since they have given me one of the zeroes, I'll use synthetic division to divide it out:

completed division

(You will probably want to use scratch paper for the computations required when manipulating the radical root.) Copyright © Elizabeth Stapel 2002-2011 All Rights Reserved

Since you only get these square-root answers by using the Quadratic Formula, and since the square-root part of the Formula is preceded by a "plus-minus" sign, then these square-root answers must always come in pairs. Thus, if x = -3 + sqrt(11) is a root, then so also must x = -3 - sqrt(11) be a root. So my next step is to divide by x = -3 - sqrt(11):

completed division

I had started with a fourth-power polynomial. After the first division, I was left with a cubic (with very nasty coefficients!). After the second division, I'm now down to a quadratic (x2 + 0x – 5, or just x2 – 5), which I know how to solve:

x = +/- sqrt(5)

Then the full solution is:

x = -3 +/- sqrt(11), +/- sqrt(5)

If you have studied complex numbers, then you may see a problem of the following type.

Given that 2 – i is a zero of x5 – 6x4 + 11x3 – x2 – 14x + 5, fully solve the

equation  x5 – 6x4 + 11x3 – x2 – 14x + 5 = 0.

They have given us a zero, so I'll use synthetic division and divide out 2 – i:

completed division

(You will probably want to use scratch paper for the computations required when doing complex division.)

Recall that, to arrive at a zero of 2 – i, they must have used the Quadratic Formula, which always spits out complex answers in pairs. That is, you get the imaginary part (the part with the "i") from having a negative inside the "plus or minus square-root of" part of the Formula. This means that, since 2 – i is a zero, then 2 + i must also be a zero.  So I'll divide by 2 + i:

completed division

This leaves me with a cubic, so I'll need to find another zero on my own. (That is, I can't apply the Quadratic Formula yet.) I can use the Rational Roots Test to help find potential zeroes, and a quick graph of x3 – 2x2 – 2x + 1 can help. I will try x = –1:

completed division

Now I'm down to a quadratic (x2 – 3x + 1, which happens not to factor), so I'll apply the Quadratic Formula to get:

x = (3 +/- sqrt(5))/2

Then all the zeroes of x5 – 6x4 + 11x3 – x2 – 14x + 5 are given by:

x = 2 - i, 2 + i, (3 - sqrt(5))/2, (3 + sqrt(5))/2, -1

Answer:

15y+7=12-20y (add 20y to both sides)

35y+7=12 (minus 7 from both sides)

35y = 5 (divide both sides by 35) (5/35=1/7)

y= 1/7

Answer:

64/7 = 64/7 yes y = 1/7

Step-by-step explanation:

To prove this we simply have to plug in!

15(1/7) + 7 = 12 - 20(1/7) (simplify)

15/7 + 7 = 12 - 20/7 (simplify further)

64/7 = 64/7

So yes y = 1/7

answer: n = -14

3x + 1 - 9 = 3(x + 2) + n

<=> 3x - 8 = 3x + 6 + n

<=> 3x - 8 - 3x - 6 = n

<=> n = -14

Answer:

I need a picture but if you count 3 up and and left 5 whichever line hits both of those points is the answer

120 is the answer to the problem.

Answer:

[tex]\boxed{ \$0.1032}[/tex]

Step-by-step explanation:

[tex] 100 g \: will \: be \: \to \: 0.1kg \\ hence : if \:1.25 kg \: costs \: \$ 1.29 \\0.1kg \: will \: cost \to \: \frac{ \$ 1.29 \times 0.1kg}{1.25kg} \\ 0.1kg \: will \: cost \to \: 0.08 ( \$ 1.29)\\ = \boxed{ \$0.1032}\\ [/tex]

Answer:

that's fake right?

Step-by-step explanation:

¯\_(ツ)_/¯

Answer:

16 = x

Step-by-step explanation:

Since angle B and E are congruent to each other (because triangle ABC is congruent to triangle DEC), We can set up an equation to find x.

x + 15 = 6x - 65

Add 65 and subtract x to and from both sides.

80 = 5x

Divide by x.

16 = x.

Answer: The  y=2

Step-by-step explanation:

Answer:

1

Step-by-step explanation:

(-4) + (-8) + 11 - (-2)

~Simplify

-4 - 8 + 11 + 2

~Subtract

-12 + 11 + 2

~Add

-1 + 2

~Add

1

Best of Luck!

Answer:

2/3

Step-by-step explanation:

3-1/5-2=

2/3

Answer:

You would multiply $8200 by .075, giving you 615. Then to find the decrease. Take the starting value (8200) by the end value (615). and you would get 7,585.

Answer:

Hey! It's me again!

Answer:

A

Step-by-step explanation:

Answer:

60

Step-by-step explanation:

30 goggles will be sent to their new home. you will need to restock them after losing the 30 to sell to others

Answer:

Thanks for the free points

Answer:

So, the answer for b is 20.

The way to figure this out is:

angle N is congruent to angle S

angle M is congruent to angle M

side NM is 10

because of the three things above, triangle NHM is congruent to triangle SML because of SAS(side angle side postulate)

Hope that helps! Let me know if it does! ;P

Answer:

she decorated two cookies per minute

Step-by-step explanation:

72 cookies divided by 36 minutes

Explanation: 1 / 2 = 9 / 18 Halfway that and 0 would be 4.5 / 18 Since 3 / 18 is below that, we would round to 0

Answer:

30,000

Step-by-step explanation:

3/2 multiplied by 20k is

3 x 20000 over 2

which is 60000 over 2

which is 30000

Answer:

all of that was wrong

Answer:

[tex] \overline{EG} \cong \overline{HK} [/tex]

[tex] \overline{HJ} \cong \overline{EF} [/tex]

[tex] \angle F \cong \angle J [/tex]

[tex] \angle H \cong \angle E [/tex]

∆FGE ≅ ∆JKH

Step-by-step explanation:

Since ∆EFG ≅ ∆HJK, therefore, the corresponding sides and angles of both ∆s are congruent.

Thus:

[tex] \overline{EG} \cong \overline{HK} [/tex]

[tex] \overline{HJ} \cong \overline{EF} [/tex]

[tex] \angle F \cong \angle J [/tex]

[tex] \angle H \cong \angle E [/tex]

∆FGE ≅ ∆JKH (take note of the arrangement of the letters or naming of each vertex of both triangles. Each vertex of in one ∆ must correspond to the vertex of the other for the ∆s to be congruent)

△EFG ≅ △HJK,

From the figure attached we can conclude  that

  [tex]\rm \bar {EG}[/tex] ≅ [tex]\rm \bar {HK}[/tex]

  [tex]\rm \bar HJ[/tex] ≅ [tex]\rm \bar FE[/tex]

   ∠[tex]\rm F[/tex] ≅∠[tex]\rm J[/tex]

   ∠[tex]\rm H[/tex] ≅∠[tex]\rm E[/tex]

△FGE ≅ △JKH

Given that

△EFG ≅ △HJK,

Since △EFG ≅ △HJK is given hence we can conclude that corresponding sides and corresponding angles of both the triangles will be equal.

Two triangles are said to be congruent when they are of similar shape and size means all the angles of one triangle are  equal to the corresponding angles of other triangle and  all the  sides of one triangle are equal to  corresponding sides of other triangle.

From the figure attached we can conclude  that

  [tex]\rm \bar {EG}[/tex] ≅ [tex]\rm \bar {HK}[/tex]

  [tex]\rm \bar HJ[/tex] ≅ [tex]\rm \bar FE[/tex]

   ∠[tex]\rm F[/tex] ≅∠[tex]\rm J[/tex]

   ∠[tex]\rm H[/tex] ≅∠[tex]\rm E[/tex]

△FGE ≅ △JKH

For more information please refer to the link below

https://brainly.com/question/25063327