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3 years ago

How many triangles exist with the given angle measures? 60 60 60

1 answer:

6 0

The sum of the angles in a triangle is 180°, so the angles do form a triangle. Since two of the three angles are congruent, the triangle is an isosceles triangle. An infinite number of isosceles triangles exist with different side lengths. Therefore, more than one triangle exists with the given angle measures.

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Which is equivalent to (2-5i)(-2+5i)

Nadusha1986 [10]

= -4 + 10i + 10i -25i^2 = -4 + 20i + 25 = 21 + 20i, because i^2 = -1.

6 0

3 years ago

Read 2 more answers

The graph of g(x) is shown. Determine the transformations from the parent function f(x)=x2 to the new function g(x)=15x2+5

Viktor [21]

Answer:

The graph of g is translated down 5 units from the graph of f and stretched horizontally

Step-by-step explanation:

was correct on test :)

pls brainliest XD

.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.

7 0

2 years ago

Help! It’s timed and I’m not sure

tensa zangetsu [6.8K]

Answer:

x < 0

Hope this helps!

7 0

3 years ago

IL COSL al we creeree viall

kakasveta [241]

Answer:

In the next year, Anthony worked 2,084 hours

Step-by-step explanation:

Anthony worked 1,697 hours in 2010.

We also know Anthony worked 22.8% more hours than in 2010.

The problem requires to calculate how much did Anthony work in the next year.

It can be calculated as follows:

Take 22.8% of 1,697:

$22.8 * 1,697 / 100 = 386.916\approx 387\ hours$

Now calculate by adding it to the original number of hours:

1,697 + 387 = 2,084 hours

In the next year, Anthony worked 2,084 hours

8 0

3 years ago

Suppose that a die-hardly-ever battery has an exponential time-to-failure distribution with a mean of 48 months. at 60 months, t

mel-nik [20]

Since the time to failure distribution is exponential, the distribution is F (n) = 1 – e ^ (-x/48)

The probability of failure between months 60 and 72 is the same as the probability that the battery will fail in 12 months, since the exponential distribution is memoryless F (12) = 0.221

7 0

4 years ago