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2 years ago

If a translation of T2, -7(x, y) is applied to AABC, what

Mathematics

2 answers:

Mrac [35]2 years ago

7 0

If a translation of T(2, -7) is applied to ΔABC, then the coordinates of B' is (3, -12).

<h3>What are coordinates?</h3>

A coordinate system in geometry is a system that employs one or more integers, or coordinates, to define the position of points or other geometric components on a manifold such as Euclidean space.

Given the coordinate of B are (1, -5), also given the rule for the translation T that is (2, -7). Therefore, the translation of the point B will be,

B' = (1+2, -5-7) = (3, -12)

Hence, If a translation of T(2, -7) is applied to ΔABC, then the coordinates of B' is (3, -12).

Learn more about Coordinates:

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Step2247 [10]2 years ago

5 0

Answer:

D 3,-12

Step-by-step explanation:

Did it on edge, person above is correct.

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PLZ HELP ME WITH THIS PROBLEM I DON'T HAVE THAT MUCH TIME!!

pantera1 [17]

A. -20,-12,-8,-1,(1),5,10,x,x........ur other 2 numbers have to be greater then 1

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8 0

3 years ago

Read 2 more answers

Add 5+[-8] what is the awenser or sum.

Arte-miy333 [17]

Answer:

Your answer is -3

Step-by-step explanation:

We know your answer is going to be the answer because 8 is a bigger number than 5, so you take 5 away from (-8)

Which leaves you with -3

Hope this helps!

-Payshence

7 0

3 years ago

Write a polynomial function of least degree with the given zero. -1+ √ 2, √ 3

kirill115 [55]

Answer:

f(x) = $x^{4}$ + 2x³ - 4x² - 6x + 3

Step-by-step explanation:

Note that radical zeros occur in conjugate pairs, thus

- 1 + $\sqrt{2}$ is a zero then - 1 - $\sqrt{2}$ is also a zero

$\sqrt{3}$ is a zero then - $\sqrt{3}$ is also a zero

Thus the corresponding factors are

(x - (- 1 + $\sqrt{2}$ ) ), (x - (- 1 - $\sqrt{2}$ ) ), (x - $\sqrt{3}$ ), (x - (- $\sqrt{3}$ )), that is

(x + 1 - $\sqrt{2}$ ), (x + 1 + $\sqrt{2}$ ), (x - $\sqrt{3}$ ), (x + $\sqrt{3}$ )

The polynomial is then the product of the roots

f(x) = (x + 1 - $\sqrt{2}$ )(x + 1 + $\sqrt{2}$ )(x - $\sqrt{3}$ )(x + $\sqrt{3}$ )

= ((x + 1)² - ( $\sqrt{2}$ )²)((x² - ( $\sqrt{3}$ )²)

= (x² + 2x + 1 - 2)(x² - 3)

= (x² + 2x - 1)(x² - 3) ← distribute

= $x^{4}$ - 3x² + 2x³ - 6x - x² + 3

= $x^{4}$ + 2x³ - 4x² - 6x + 3

6 0

3 years ago

Use Euler's method with step size 0.2 to estimate y(1), where y(x) is the solution of the initial-value problem y' = x2y − 1 2 y

irina [24]

Answer:

Therefore the value of y(1)= 0.9152.

Step-by-step explanation:

According to the Euler's method

y(x+h)≈ y(x) + hy'(x) ....(1)

Given that y(0) =3 and step size (h) = 0.2.

$y'(x)= x^2y(x)-\frac12y^2(x)$

Putting the value of y'(x) in equation (1)

$y(x+h)\approx y(x) +h(x^2y(x)-\frac12y^2(x))$

Substituting x =0 and h= 0.2

$y(0+0.2)\approx y(0)+0.2[0\times y(0)-\frac12 (y(0))^2]$

$\Rightarrow y(0.2)\approx 3+0.2[-\frac12 \times3]$ [∵ y(0) =3 ]

$\Rightarrow y(0.2)\approx 2.7$

Substituting x =0.2 and h= 0.2

$y(0.2+0.2)\approx y(0.2)+0.2[(0.2)^2\times y(0.2)-\frac12 (y(0.2))^2]$

$\Rightarrow y(0.4)\approx 2.7+0.2[(0.2)^2\times 2.7- \frac12(2.7)^2]$

$\Rightarrow y(0.4)\approx 1.9926$

Substituting x =0.4 and h= 0.2

$y(0.4+0.2)\approx y(0.4)+0.2[(0.4)^2\times y(0.4)-\frac12 (y(0.4))^2]$

$\Rightarrow y(0.6)\approx 1.9926+0.2[(0.4)^2\times 1.9926- \frac12(1.9926)^2]$

$\Rightarrow y(0.6)\approx 1.6593$

Substituting x =0.6 and h= 0.2

$y(0.6+0.2)\approx y(0.6)+0.2[(0.6)^2\times y(0.6)-\frac12 (y(0.6))^2]$

$\Rightarrow y(0.8)\approx 1.6593+0.2[(0.6)^2\times 1.6593- \frac12(1.6593)^2]$

$\Rightarrow y(0.6)\approx 0.8800$

Substituting x =0.8 and h= 0.2

$y(0.8+0.2)\approx y(0.8)+0.2[(0.8)^2\times y(0.8)-\frac12 (y(0.8))^2]$

$\Rightarrow y(1.0)\approx 0.8800+0.2[(0.8)^2\times 0.8800- \frac12(0.8800)^2]$

$\Rightarrow y(1.0)\approx 0.9152$

Therefore the value of y(1)= 0.9152.

4 0

3 years ago

Can someone help me with 1-3

mina [271]

1 would be D.
2. Is A since 3^2=9 and -3^2=9 And 9-9=0. Im pretty sure A. For number 3. Just because the two x-values are 3 and -3. I’m sorry if I’m wrong on #3.

8 0

3 years ago