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nexus9112 [7]
3 years ago
15

Is there such thing as a triangle with two right angles? Explain your thinking

Mathematics
2 answers:
Darina [25.2K]3 years ago
8 0

Answer:

no there can be no such thing to lose legs or those 2 angles added up already are 180 degrees and the 180 has to be split up between 3 angles and if 2 of the angles were 90 degrees the other 2 angles also would be creating a square

Step-by-step explanation:

muminat3 years ago
6 0

Answer:

no

Step-by-step explanation:

right angles are like Ls for example L that’s a right angle so u can’t to that but u can L\ for a right triangle (hope this makes sense)

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elixir [45]
ANSWER:

i think it’s c
5 0
3 years ago
The volume of a triangular prism with base dimensions of 4 cm height and 10 cm base and a prism height of 15 cm is??
jonny [76]
I think you can do this way:
Volume = B*h where B is the area of the base.
B= 4 * 10 = 40
V= 40 * 15
V= 600 cm^3
5 0
3 years ago
Read 2 more answers
Show work please.<br><br> solve system of equations using matrices.
nadya68 [22]

Answer:

(t, t -1, t)

Step-by-step explanation:

You have three unknowns but only 2 equations, so you can't really SOLVE this...you can get a solution with a variable still in it (I forget what this is called.  I think it refers to infinite many solutions).  Here's how it works:

Set up your matrix:

\left[\begin{array}{ccc}1&-2&1\\2&-1&-1\\\end{array}\right] \left[\begin{array}{ccc}2\\1\\\end{array}\right]

You want to change the number in position 21 (the 2 in the scond row) to a 0 so you have y and z left.  Do this by multiplying the top row by -2 then adding it to the second row to get that 2 to become a 0.  Multiplying in a -2 to the top row gives you:

\left[\begin{array}{ccc}-2&4&-2\\2&-1&-1\\\end{array}\right]\left[\begin{array}{ccc}-4\\1\\\end{array}\right]

Then add, keeping the first row the same and changing the second to reflect the addition:

\left[\begin{array}{ccc}-2&4&-2\\0&3&-3\\\end{array}\right] \left[\begin{array}{ccc}-4\\-3\\\end{array}\right]

The second equation is this now:

3y - 3z = -3.  Solving for y gives you y = z - 1.  Let's let z = t (some random real number that will make the system true.  Any number will work.  I'll show you at the end.  Just bear with me...)

lf z = t, and if y = z - 1, then y = t - 1.  So far we have that y = t - 1 and z = t.  Now we solve for x:

From the first equation in the original system,

x - 2y + z = 2.  Subbing in t - 1 for y and t for z:

x - 2(t - 1) + t = 2.  Simplify to get

x - 2t + 2 + t = 2  and  x - t = 0, and x = t.  So the solution set is (t, t - 1, t).  Picking a random value for t of, let's say 2, sub that in and make sure it works.  If:

x - 2y + z = 2, then t - 2(t - 1) + t = 2 becomes t - 2t + 2 + t = 2, and with t = 2, 2 - 2(2) + 2 + 2 = 2.    Check it:  2 - 4 + 4 = 2 and 2 = 2.  You could pick any value for t and it will work.

6 0
3 years ago
Please help quickly im on a test<br> if f(x)=x+4, and g(x)=2x+4, what is f(2)+g(2)?
JulsSmile [24]

Answer:

14

Step-by-step explanation:

f(2)+g(2)

(2+4)+(2×2+4)

6+8

14

5 0
2 years ago
Please help meeeee.....
dimulka [17.4K]
The answer should be 4
5 0
3 years ago
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