All categories

2 years ago

Type SSS, SAS, ASA, SAA, or HL to describe these triangles.

Mathematics

1 answer:

Bingel [31]2 years ago

5 0

Answer:

SAS

Step-by-step explanation:

There is one common side (S)

both the triangles have 90° common (A)

Opposite sides are equal which is given (S)

They both are right angled triangles

You might be interested in

Please help me with this

cupoosta [38]

Can you please write what the question is? I need to know what the category is to figure out which one does not belong.

5 0

3 years ago

I need help creating a real world problem

marshall27 [118]

Jasmine goes to McDonalds to order food. A big mac costs $5 and large fries cost $3. If Jasmine was to buy one big mac with one batch of large fries, how much would she pay in total if a taxes of $10 was added on.

Answer: 5+3=8
8+10=18

The answer is $18 dollars.

7 0

2 years ago

PLEASE HELP!! Factor 6x^2-42x-54 Please show work!

skad [1K]

first, rewrite 54 as 6•9

next, rewrite 42 as 6•7

then, that gets you 6x^2 -6•7x- 6•9

lastly, factor out the common term (6)

and then you get 6(x^2-7x-9)

please mark as brainliet <3

6 0

2 years ago

Expand using the properties and rules for logarithms

malfutka [58]

Consider expression $\log_{\frac{1}{2}}\left(\dfrac{3x^2}{2}\right).$

1. Use property

$\log_a\dfrac{b}{c}=\log_ab-\log_ac.$

Then

$\log_{\frac{1}{2}}\left(\dfrac{3x^2}{2}\right)=\log_{\frac{1}{2}}3x^2-\log_{\frac{1}{2}}2.$

2. Use property

$\log_abc=\log_ab+\log_ac.$

Then

$\log_{\frac{1}{2}}\left(\dfrac{3x^2}{2}\right)=\log_{\frac{1}{2}}3x^2-\log_{\frac{1}{2}}2=\log_{\frac{1}{2}}3+\log_{\frac{1}{2}}x^2-\log_{\frac{1}{2}}2.$

3. Use property

$\log_ab^k=k\log_ab.$

Then

$\log_{\frac{1}{2}}\left(\dfrac{3x^2}{2}\right)=\log_{\frac{1}{2}}3+\log_{\frac{1}{2}}x^2-\log_{\frac{1}{2}}2=\log_{\frac{1}{2}}3+2\log_{\frac{1}{2}}x-\log_{\frac{1}{2}}2.$

4. Use property

$\log_{a^k}b=\dfrac{1}{k}\log_ab.$

Then

$\log_{\frac{1}{2}}\left(\dfrac{3x^2}{2}\right)=\log_{\frac{1}{2}}3+2\log_{\frac{1}{2}}x-\log_{\frac{1}{2}}2=\log_{\frac{1}{2}}3+2\log_{\frac{1}{2}}x-\log_{2^{-1}}2=\\ \\=\log_{\frac{1}{2}}3+2\log_{\frac{1}{2}}x+\log_22=\log_{\frac{1}{2}}3+2\log_{\frac{1}{2}}x+1.$

Answer: correct option is B.

7 0

3 years ago

PLZ HELP Which equation could have been used to create this table?

Snowcat [4.5K]

Y=6x because once you plug in the x value you’ll get the same y values as in the table

5 0

2 years ago

Read 2 more answers