30 The answer is 30. Hope this helps you!
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50)1650
-165
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00
-0
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00
I think the correct answer is B. It is the triangle case SSA that may have one, two, or zero solutions. This case can have either number of solutions but it depends on the sides of the triangle given. Having one solution can be all of the cases except SSS, having 2 solutions can only be applied to SSA.
Alright, so we'd use the combinations with repetition formula, so we choose from 4 schools to distribute to and distribute 8 blackboards. It's then
( 8+4-1)!/8!(4-1)!=11!/(3!*8!)=165
For at least one blackboard, we first distribute 1 to each school and then have 4 blackboards left, getting (4+4-1)!/4!(4-1)!=7!/(4!*3!)=35
Answer:
y= 9
Step-by-step explanation:
4y-y+11=38
3y+11=38
3y= 38-11
3y= 27
y= 27/3
y=9
hope it helps ,pls mark me as brainliest
Answer:
a) 625; 625; right triangle
b) 205; 256; obtuse triangle
Step-by-step explanation:
The squares are values found in your memory or using a calculator. It is straightforward addition to find their sum.
<h3>left side</h3>
7² +24² = 49 +576 = 625
25² = 625
The sum of the squares of the short sides is the square of the long side, so this is a right triangle.
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<h3>right side</h3>
6² +13² = 36 +169 = 205
16² = 256
The long side is longer than is needed to form a right triangle, so the largest angle is more than 90°. This is an <em>obtuse triangle</em> (as shown).
