Help! Geometry question!! Photo attached!

Mathematics

1 answer:

jonny [76]3 years ago

3 0

The triangles are not the same size so a dilation made the original one smaller and a translation moved it to map ABC to A'B'C'.

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PLEASE HELP! I don't understand this question!

pantera1 [17]

Answer:

Its the first choice: right, scalene.

Step-by-step explanation:

It is a right angled triangle, (indicated by the little square at Q).

The sides have different lengths so it scalene.

8 0

3 years ago

Read 2 more answers

Complete the table of values

Agata [3.3K]

Both problems give you a function in the second column and the x-values. To find out the values of a through f, you need to plug in those x-values into the function and simplify!

You need to know three exponent rules to simplify these expressions:
1) The negative exponent rule says that when a base has a negative exponent, flip the base onto the other side of the fraction to make it into a positive exponent. For example, $3^{-2} =
\frac{1}{3^{2} }$ .
2) Raising a fraction to a power is the same as separately raising the numerator and denominator to that power. For example, $(\frac{3}{4}) ^{3} = \frac{ 3^{3} }{4^{3} }$ .
3) The zero exponent rule says that any number raised to zero is 1. For example, $3^{0} = 1$ .


Back to the Problem:
Problem 1
The x-values are in the left column. The title of the right column tells you that the function is $y = 4^{-x}$ . The x-values are:
1) x = 0
Plug this into $y = 4^{-x}$ to find letter a:
$y = 4^{-x}\\
y = 4^{-0}\\
y = 4^{0}\\
y = 1$

2) x = 2
Plug this into $y = 4^{-x}$ to find letter b:
$y = 4^{-x}\\ 
y = 4^{-2}\\ 
y = \frac{1}{4^{2}} \\ 
y= \frac{1}{16}$

3) x = 4
Plug this into $y = 4^{-x}$ to find letter c:
$y = 4^{-x}\\ 
y = 4^{-4}\\ 
y = \frac{1}{4^{4}} \\ 
y= \frac{1}{256}$


Problem 2
The x-values are in the left column. The title of the right column tells you that the function is $y = (\frac{2}{3})^x$ . The x-values are:
1) x = 0
Plug this into $y = (\frac{2}{3})^x$ to find letter d:
$y = (\frac{2}{3})^x\\
y = (\frac{2}{3})^0\\
y = 1$

2) x = 2
Plug this into $y = (\frac{2}{3})^x$ to find letter e:
$y = (\frac{2}{3})^x\\ y = (\frac{2}{3})^2\\ y = \frac{2^2}{3^2}\\
y = \frac{4}{9}$

3) x = 4
Plug this into $y = (\frac{2}{3})^x$ to find letter f:
$y = (\frac{2}{3})^x\\ y = (\frac{2}{3})^4\\ y = \frac{2^4}{3^4}\\ y = \frac{16}{81}$

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Answers:
a = 1
b = $\frac{1}{16}$ 
c = $\frac{1}{256}$
d = 1
e = $\frac{4}{9}$
f = $\frac{16}{81}$