All categories

3 years ago

Is Triangle UVW cong Triangle XYZ? If so, name the postulate that applies

Mathematics

1 answer:

rosijanka [135]3 years ago

3 0

Answer:

Might not be congruent

Step-by-step explanation:

Because look carefully and use the measurment and you can compare my answer to yours

You might be interested in

Lara baked a pan of brownies for a potluck. The number of squares she cuts the brownies into will depend on the number of people

Feliz [49]

Answer: i would say s

Step-by-step explanation:

i hope this helps

7 0

3 years ago

A phone company charges for service according to the formula y=18+0.03x, where x is the number of minutes talked, and y is the m

irina [24]

The answer is dollars per minute is the slope

7 0

3 years ago

Read 2 more answers

Can someone help my sister Kathy with this

Svetllana [295]

The three tenth values that would cause the inequality to be true would be 4, 5, and 6 because each make the other decimal greater than 24.34.

I hope this helped! :-)

3 0

4 years ago

Read 2 more answers

perry thinks that 1/2 and 2/4 are equivalent fractions. but when he draws the number lines to the right he sees that 1/2 and 2/4

kodGreya [7K]

He is correct but he might be doughting him self

3 0

4 years ago

Explain me PLEASEEEE!!!

kirza4 [7]

Answer:

$\frac{(a^{2})^{3}.a^{-3}}{a^{10}}=\frac{a^{6}.a^{-3}}{a^{10}}=\frac{a^{3}}{a^{10}}=a^{-7}=\frac{1}{a^{7}}$

Step-by-step explanation:

Let us revise the properties of exponents

$a^{m}.a^{n}=a^{m+n}$

$\frac{a^{m}}{a^{n}}=a^{m-n}$

$(a^{m})^{n}=a^{m.n}$

$a^{-m}=\frac{1}{a^{m} }$

Let us use these properties to solve the question

→ By using the 3rd property above

∵ $(a^{2})^{3}=a^{2.3}=a^{6}$

∴ $\frac{(a^{2})^{3}.a^{-3}}{a^{10}}=\frac{a^{6}.a^{-3}}{a^{10} }$

→ By using the 1st property above

∵ $a^{6}.a^{-3}=a^{6+-3}=a^{6-3}=a^{3}$

∴ $\frac{(a^{2})^{3}.a^{-3}}{a^{10}}=\frac{a^{6}.a^{-3}}{a^{10}}=\frac{a^{3}}{a^{10}}$

→ By using the 2nd property above

∵ $\frac{a^{3}}{a^{10}}=a^{3-10}=a^{-7}$

∴ $\frac{(a^{2})^{3}.a^{-3}}{a^{10}}=\frac{a^{6}.a^{-3}}{a^{10}}=\frac{a^{3}}{a^{10}}=a^{-7}$

→ By using the 4th property above

∵ $a^{-7}=\frac{1}{a^{7}}$

∴ $\frac{(a^{2})^{3}.a^{-3}}{a^{10}}=\frac{a^{6}.a^{-3}}{a^{10}}=\frac{a^{3}}{a^{10}}=a^{-7}=\frac{1}{a^{7}}$

8 0

3 years ago