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irinina [24]
3 years ago
5

On a coordinate plane, 2 trapezoids are shown. The first trapezoid has points A (negative 1, negative 2), B (1, negative 2), C (

2, negative 5), and D (negative 2, negative 5). The second trapezoid has points A prime (negative 4, negative 1), B prime (negative 2, negative 1), C prime (negative 1, negative 4), and D prime (negative 5, negative 4). Which statements are true about trapezoid ABCD and its translated image, A'B'C'D'? Select two options. The rule for the translation can be written as T–3, 1(x, y). The rule for the translation can be written as T–1, 3(x, y). The rule for the translation can be written as (x, y) → (x + 1, y – 3). The rule for the translation can be written as (x, y) → (x – 3, y + 1). Trapezoid ABCD has been translated 3 units to the right and 1 unit up.
Mathematics
2 answers:
geniusboy [140]3 years ago
6 0

Answer:

The rule for the translations can be written as T -3,1(x,y).

The rule for translations can be written as (x,y)➡️(x-3,y+1).

Step-by-step explanation:

I just took the test

Zepler [3.9K]3 years ago
3 0

Answer:

answers 1 and 4 :)

Step-by-step explanation:

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6 0
3 years ago
What is a quick picture for 0.51
Zarrin [17]
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7 0
3 years ago
Electric charge is distributed over the disk x2 + y2 ≤ 16 so that the charge density at (x, y) is rho(x, y) = 2x + 2y + 2x2 + 2y
professor190 [17]

Answer:

Required total charge is 256\pi coulombs per square meter.

Step-by-step explanation:

Given electric charge is dristributed over the disk,

x^2=y^2\leq 16 so that the charge density at (x,y) is,

\rho (x,y)=2x+2y+2x^2+2y^2

To find total charge on the disk let Q be the total charge and x=r\cos\theta,y=r\sin\theta so that,

Q={\int\int}_Q\rho(x,y) dA                where A is the surface of disk.

=\int_{0}^{2\pi}\int_{0}^{4}(2x+2y+2x^2+2y^2)dA

=\int_{0}^{2\pi}\int_{0}^{4}(2r\cos\theta+2r\sin\theta+2r^2 \cos^{2}\theta+2r^2\sin^2\theta)rdrd\theta

=2\int_{0}^{2\pi}\int_{0}^{4}r^2(\cos\theta+\sin\theta)drd\theta+2\int_{0}^{2\pi}\int_{0}^{4}r^3drd\theta

=\frac{2}{3}\int_{0}^{2\pi}(\sin\theta+\cos\theta)\Big[r^3\Big]_{0}^{4}d\theta+2\int_{0}^{2\pi}\Big[\frac{r^4}{4}\Big]d\theta

=\frac{128}{3}\int_{0}^{2\pi}(\sin\theta+\cos\theta)d\theta+128\int_{0}^{2\pi}d\theta

=\frac{128}{3}\Big[\sin\theta-\cos\theta\Big]_{0}^{2\pi}+128\times 2\pi

=\frac{128}{3}\Big[\sin 2\pi-\cos 2\pi-\sin 0+\cos 0\Big]+256\pi

=256\pi

Hence total charge is 256\pi coulombs per square meter.

3 0
3 years ago
Logs are stacked in a pile. The bottom row has 50 logs and next to bottom row has 49 logs. Each row has one less log than the ro
Mars2501 [29]

Answer:

46 logs on the 5th row.

Step-by-step explanation:

Number of logs on the nth row is

n =  50 - (n-1)

 n = 51 - n    (so on the first row we have  51 - 1 = 50 logs).

So on the 5th row we have 51 - 5 = 46 logs.

8 0
3 years ago
Simplify the expression. Write answer without spaces between terms and signs.
dsp73

Answer:

4h-7

Step-by-step explanation:

Step by Step Solution

STEP1:Equation at the end of step 1 (1 - 2h) + 2 • (3h - 4) STEP2:

Final result :

4h - 7

4 0
3 years ago
Read 2 more answers
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