Find the missing angle in the triangle below

Mathematics

1 answer:

Deffense [45]3 years ago

4 0

Answer:

125

Step-by-step explanation:

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F left parenthesis x right parenthesis equals 9 x cubed plus 2 x squared minus 5 x plus 4 and g left parenthesis x right parenth

Jet001 [13]

The answer is 2x(2x²+x+1).

When we subtract polynomials we combine like terms:
(9x³+2x²-5x+4)-(5x³-7x+4)

9x³-5x³=4x³
2x²- 0 = 2x²
-5x--7x=-5x+7x=2x
4-4=0

This gives us
4x³+2x²+2x

Each of these is divisible by 2, and each has an x, so we factor those out:
2x( )

4x³/2x = 2x²:
2x(2x² )

2x²/2x=x:
2x(2x²+x )

2x/2x = 1:
2x(2x²+x+1)

3 0

3 years ago

Increase 2850 by 18%

faust18 [17]

3363 it is. That is the answer. The problem would be "2850 + 18%" right. Because that's what I did.

4 0

3 years ago

Let f (x) = 3x − 1 and ε > 0. Find a δ > 0 such that 0 < ∣x − 5∣ < δ implies ∣f (x) − 14∣ < ε. (Find the largest

s344n2d4d5 [400]

Answer:

This proves that f is continous at x=5.

Step-by-step explanation:

Taking f(x) = 3x-1 and $\varepsilon>0$ , we want to find a $\delta$ such that $|f(x)-14|$

At first, we will assume that this delta exists and we will try to figure out its value.

Suppose that $|x-5|$ . Then

$|f(x)-14| = |3x-1-14| = |3x-15|=|3(x-5)| = 3|x-5|< 3\delta$ .

Then, if $|x-5|$ , then $|f(x)-14|$ . So, in this case, if $3\delta \leq \varepsilon$ we get that $|f(x)-14|$ . The maximum value of delta is $\frac{\varepsilon}{3}$ .