All categories

3 years ago

Which pair of triangles is congruent?

Mathematics

1 answer:

daser333 [38]3 years ago

4 0

You are only showing one pair of triangles

You might be interested in

Which set of ordered pairs represents a function?

m_a_m_a [10]

Answer:

{(0, 0), (1, 2), (2, 4), (3, 4)

Step-by-step explanation:

the x in the ordered pair is not suppose to repeat, clearly this doesn't

7 0

4 years ago

What is 516.3606 rounded to the nearest hundredths

liraira [26]

516.36 because the thousandth digit isn't above 5 so you don't round up

6 0

3 years ago

Read 2 more answers

Find the volume of a rectangular prism with length 20 m,

Cerrena [4.2K]

Answer:

V= 270m³

Step-by-step explanation:

4 0

3 years ago

Which expression is not a polynomial

polet [3.4K]

4-x I know this because my teacher just went over this problem

6 0

3 years ago

Which is the simplified form of the expression ((2 Superscript negative 2 Baseline) (3 Superscript 4 Baseline)) Superscript nega

AlekseyPX

Answer:

The option "StartFraction 1 Over 3 Superscript 8" is correct

That is $\frac{1}{3^8}$ is correct answer

Therefore $[(2^{-2})(3^4)]^{-3}\times [(2^{-3})(3^2)]^2=\frac{1}{3^8}$

Step-by-step explanation:

Given expression is ((2 Superscript negative 2 Baseline) (3 Superscript 4 Baseline)) Superscript negative 3 Baseline times ((2 Superscript negative 3 Baseline) (3 squared)) squared

The given expression can be written as

$[(2^{-2})(3^4)]^{-3}\times [(2^{-3})(3^2)]^2$

To find the simplified form of the given expression :

$[(2^{-2})(3^4)]^{-3}\times [(2^{-3})(3^2)]^2$

$=(2^{-2})^{-3}(3^4)^{-3}\times (2^{-3})^2(3^2)^2$ ( using the property $(ab)^m=a^m.b^m$ )

$=(2^6)(3^{-12})\times (2^{-6})(3^4)$ ( using the property $(a^m)^n=a^{mn}$

$=(2^6)(2^{-6})(3^{-12})(3^4)$ ( combining the like powers )

$=2^{6-6}3^{-12+4}$ ( using the property $a^m.a^n=a^{m+n}$ )

$=2^03^{-8}$

$=\frac{1}{3^8}$ ( using the property $a^{-m}=\frac{1}{a^m}$ )

Therefore $[(2^{-2})(3^4)]^{-3}\times [(2^{-3})(3^2)]^2=\frac{1}{3^8}$

Therefore option "StartFraction 1 Over 3 Superscript 8" is correct

That is $\frac{1}{3^8}$ is correct answer

6 0

3 years ago

Read 2 more answers