All categories

3 years ago

What combination of transformations is shown below?

Mathematics

1 answer:

melamori03 [73]3 years ago

4 0

Answer:

Rotated then translated

Step-by-step explanation:

You might be interested in

A bicycle wheel has an inside radius of 12 inches. Which expression could be used to find the inside circumference of this wheel

il63 [147K]

Step-by-step explanation:

We have given that,

The inner radius of a bicycle wheel is 12 inches.

It is required to find the expression to find the inside circumference of this wheel.

The outer surface of the wheel is equal to its circumference. It is given by :

$C=2\pi r$

r is radius of wheel

$C=2\times 3.14\times 12\\\\C=75.36\ \text{inches}$

This is the required explanation.

5 0

4 years ago

Use substitution to solve the linear system of equations. 2y = x-6 x= 9

xxTIMURxx [149]

2y = x-6, x=9
2y = 9-6; plug 9 in for x
2y = 3; evaluate 9-6
y = 3/2; divide each side by 2

x = 9
y = 3/2

Hope this helps!

6 0

4 years ago

Prove that 3^17−3^15+3^13 is divisible by 73.

Debora [2.8K]

Answer:

3^13*73

Step-by-step explanation:

3^17-3^15+3^13 So you factor out 3^13

=3^13(3^4-3^2+1)

3^4-3^2+1=73

3^13*73

Is this RSM?

5 0

3 years ago

What must be the length of XY if triangle ABC is

lbvjy [14]

Answer:

x=12.8

Step-by-step explanation:

xy / 8 = 16 /10

xy = (8 * 16) / 10

xy = 12.8

7 0

3 years ago

Plz help with full process

BabaBlast [244]

Answer:

$\displaystyle \frac{(a - b)^{2} - {c}^{2} }{{a}^{2} - {(b+ c)}^{2} } + \frac{(b - c)^{2} - {a}^{2} }{{b}^{2} - {(c+ a)}^{2} } + \frac{(c - a)^{2} - {b}^{2} }{{c}^{2} - {(a+ b)}^{2} }=1$

Step-by-step explanation:

<u>Algebra Simplifying</u>

We are given the expression:

$\displaystyle T=\frac{(a - b)^{2} - {c}^{2} }{{a}^{2} - {(b+ c)}^{2} } + \frac{(b - c)^{2} - {a}^{2} }{{b}^{2} - {(c+ a)}^{2} } + \frac{(c - a)^{2} - {b}^{2} }{{c}^{2} - {(a+ b)}^{2} }$

We need to repeatedly use the following identity:

$(a^2-b^2)=(a-b)(a+b)$

For example, the first numerator has a difference of squares thus we factor as:

$(a - b)^{2} - {c}^{2} =(a-b-c)(a-b+c)$

The first denominator can be factored also:

${a}^{2} - {(b+ c)}^{2} = (a-b-c)(a+b+c)$

Applying the same procedure to all the expressions:

$\displaystyle T=\frac{(a - b- c)(a-b+c) }{a-b-c)(a+b+c) } + \frac{(b - c - a)(b-c+a) }{(b-c-a)(b+c+a) } + \frac{(c - a-b)(c-a+b) }{(c-a-b)(c+a+b)}$

Simplifying all the fractions:

$\displaystyle T=\frac{a - b- c}{a+b+c} + \frac{b-c+a}{b+c+a } + \frac{c-a+b }{c+a+b}$

Since all the denominators are equal:

$\displaystyle T=\frac{a - b+ c+b-c+a+c-a+b}{a+b+c}$

Simplifying:

$\displaystyle T=\frac{a +c+b}{a+b+c}$

Simplifying again:

T = 1

Thus:

$\boxed{\displaystyle \frac{(a - b)^{2} - {c}^{2} }{{a}^{2} - {(b+ c)}^{2} } + \frac{(b - c)^{2} - {a}^{2} }{{b}^{2} - {(c+ a)}^{2} } + \frac{(c - a)^{2} - {b}^{2} }{{c}^{2} - {(a+ b)}^{2} }=1}$

8 0

3 years ago

Read 2 more answers