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2 years ago

Which transformations map the strip onto itself?

Mathematics

1 answer:

Pie2 years ago

4 0

Answer:

<h2>A horizontal translation and a 180° rotation only.</h2>

Step-by-step explanation:

Notice those shapes are positioned as a reflection.

So, one way to map them onto is self is by rotating 180° and a horizontal translation. Or it could work if we translate it horizontally, the we reflect it vertically. Those transformations would map the strip onto itself.

Therefore, the second choice is the best answer.

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Marcia bought a 50 foot roll she used two 8 and 5/6 how much if left

inna [77]

Answer:

vvvvv

Step-by-step explanation:

I suspect two 8 means 8x2, but if it means 2+8, I'll add another answer.

1. 8x2= 16 + 5/6(.83) = 16.83 = 50-16.83= 33.17

ORRRRR

2. 2+8=10+.83= 10.83 = 39.17

8 0

3 years ago

A curve is given by y=(x-a)√(x-b) for x≥b, where a and b are constants, cuts the x axis at A where x=b+1. Show that the gradient

ankoles [38]

Answer:

A curve is given by y=(x-a)√(x-b) for x≥b. The gradient of the curve at A is 1.

Solution:

We need to show that the gradient of the curve at A is 1

Here given that ,

$y=(x-a) \sqrt{(x-b)}$ --- equation 1

Also, according to question at point A (b+1,0)

So curve at point A will, put the value of x and y

$0=(b+1-a) \sqrt{(b+1-b)}$

0=b+1-c --- equation 2

According to multiple rule of Differentiation,

$y^{\prime}=u^{\prime} y+y^{\prime} u$

so, we get

${u}^{\prime}=1$

$v^{\prime}=\frac{1}{2} \sqrt{(x-b)}$

$y^{\prime}=1 \times \sqrt{(x-b)}+(x-a) \times \frac{1}{2} \sqrt{(x-b)}$

By putting value of point A and putting value of eq 2 we get

$y^{\prime}=\sqrt{(b+1-b)}+(b+1-a) \times \frac{1}{2} \sqrt{(b+1-b)}$

$y^{\prime}=\frac{d y}{d x}=1$

Hence proved that the gradient of the curve at A is 1.

7 0

2 years ago

In this problem, how would I find the value of Z ?

BabaBlast [244]

Answer:

z = 30

Step-by-step explanation:

WX and VU are congruent

2z + 10 = z + 45

subtract z from both sides (z + 10 = 45)

subtract 10 from both sides (z = 30)

the answer is z = 30

3 0

3 years ago

∑from 1to infinity of n^2* e^-2n

Naily [24]

∑ from 1 to infinity of n^2* e^(-2n) = (e^2 + e^4)/(e^2 - 1)^3 = 0.23768

3 0

2 years ago

<img src="https://tex.z-dn.net/?f=z%5E%7B7%7D%3D128i" id="TexFormula1" title="z^{7}=128i" alt="z^{7}=128i" align="absmiddle" cla

abruzzese [7]

If z ⁷ = 128i, then there are 7 complex numbers z that satisfy this equation.

$z^7 = 128i = 2^7i = 2^7e^{i\frac\pi2}$

$\implies z=\sqrt[7]{2^7} e^{i\frac17\left(\frac\pi2+2n\pi\right)}$

(where n = 0, 1, 2, …, 6)

$\implies z = 2 e^{i\left(\frac\pi{14}+\frac{2n\pi}7\right)}$

$\displaystyle\implies z = 2 \left(\cos\left(\frac\pi{14}+\frac{2n\pi}7\right)+i\sin\left(\frac\pi{14}+\frac{2n\pi}7\right)\right)$

3 0

2 years ago