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4 years ago

An arithmetic sequence is defined by the recursive formula t1 = 9, tn = tn - 1 - 4, where n ∈N and n > 1. The sequence is

Mathematics

1 answer:

VikaD [51]4 years ago

4 0

T2 = t1 - 4 = 9 - 4 = 5
t3 = 5 - 4 = 1
The common difference = -4 so:-

The sequence is 9, 5, 1, -3, -7 ....

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Question: Consider the similarity transformation of ΔABC to produce ΔEDF.

german

Answer:

see the explanation

Step-by-step explanation:

we have triangle ΔABC

step 1

Rotate 90 degrees clockwise ΔABC about point C to obtain ΔA'B'C'

Remember that

A rotation is a rigid transformation

An object and its rotation are the same shape and size, but the figures may be turned in different directions

ΔABC and ΔA'B'C' are congruent

ΔABC≅ ΔA'B'C

step 2

Dilate the triangle ΔA'B'C' to obtain triangle ΔEDF

Remember that

A dilation is a non rigid transformation

A dilation produces similar figures

If two figures are similar, then the ratio of its corresponding angles is proportional and its corresponding angles are congruent

Find the scale factor of the dilation

The scale factor is equal to the ratio of corresponding sides

In this problem

Let

z ----> the scale factor

$z=\frac{ED}{AB}=\frac{DF}{BC}=\frac{EF}{AC}$

Multiply the length sides of triangle ΔA'B'C' by the scale factor z to obtain the length sides of triangle ΔEDF

Note: in this problem the scale factor z is less than 1

That means ----> the dilation is a reduction

3 0

4 years ago

I really need help here ASAP

AlexFokin [52]

Answer:

D. $x=\frac{-4-\sqrt{31}}{3}$ or $x=\frac{-4+\sqrt{31}}{3}$

Step-by-step explanation:

The given equation is:

$3x^2+8x=5$

Divide through by 3;

$x^2+\frac{8}{3}x=\frac{5}{3}$

Add the square of half the coefficient of x to both sides.

$x^2+\frac{8}{3}x+(\frac{4}{3})^2=\frac{5}{3}++(\frac{4}{3})^2$

$x^2+\frac{8}{3}x+\frac{16}{9}=\frac{5}{3}+\frac{16}{9}$

The left hand side is now a perfect square:

$(x+\frac{4}{3})^2=\frac{31}{9}$

Take square root

$x+\frac{4}{3}=\pm \sqrt{ \frac{31}{9}}$

$x=-\frac{4}{3}\pm \sqrt{ \frac{31}{9}}$

$x=-\frac{4}{3}\pm \frac{\sqrt{31}}{3}$

D. $x=\frac{-4-\sqrt{31}}{3}$ or $x=\frac{-4+\sqrt{31}}{3}$

5 0

3 years ago

Solve for x round to the nearest tenth

lidiya [134]

Answer:

47.0

Step-by-step explanation:

In this right angle triangle, we are faced with a challenge of two sides. The opposite side and the adjacent side, hence the tangent is used.

Where it is the opposite side and hypothenus side, the sine is used and when it is the hypothenus side and adjacent side, the cosine is used.

Hence, we have tan62°=x/25

We cross multiply, to have

25(tan 62°)= x

x = 47.01816

In rounding up numbers, number 1 to 4 will be rounded up to zero, while numbers 5 to 9 will be rounded up to 1.

Rounding up 47.01816 to the nearest tenth. The tenth value is the figure is 0, before it we have 1, which is to hundredth. 1 will be rounded up to zero.

So we have 47.0

3 0

3 years ago

Does the equation below have infinitely many or no solutions? <br><br> x ≤ −1

alina1380 [7]

It has infinitely solutions because x could be any number below or equal to -1

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3 years ago