Convert the mixed number to an improper fraction

Mathematics

1 answer:

Goryan [66]3 years ago

5 0

The answer is b. Multiply 8 by 6 then add 7. Hope it helps.

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Determine the solutuon for x^2-3x-28>0

natita [175]

Answer:

x < -4 or x > 7.

Step-by-step explanation:

We first determine the critical points by solving x^2 - 3x - 28 = 0:

x^2 - 3x - 28 = 0

(x - 7)(x + 4) = 0

x = 7, - 4

so the critical points are -4 and 7.

Create a Table (pos = positive and neg = negative):

Value of x< - 4 -4 < x < 7 x > 7

---------------------|----------- |--------------------- |---------------------

x + 4 NEG POS POS

x - 7 NEG NEG POS

(x + 4)(x - 7) POS NEG POS

So the function is positive (>0) for x < -4 or x > 7.

You can also do this by drawing the graph of the function.

6 0

3 years ago

How do you this question (pre-cal)

Novay_Z [31]

The goal to proving identities is to transform one side into the other. We can only pick one side to transform while the other side stays the same the entire time. The general rule of thumb is to transform the more complicated side (though there may be exceptions to this guideline).

So I'll take the left hand side and try to turn it into $\csc^2( B )$

One way we can do that is through the following steps:

$\frac{\tan(B) + \cot(B)}{\tan(B)} = \csc^2(B)\\\\\frac{\tan(B)}{\tan(B)} + \frac{\cot(B)}{\tan(B)} = \csc^2(B)\\\\1 + \cot(B)*\frac{1}{\tan(B)} = \csc^2(B)\\\\1 + \cot(B)*\cot(B) = \csc^2(B)\\\\1 + \cot^2(B) = \csc^2(B)\\\\1 + \frac{cos^2(B)}{\sin^2(B)} = \csc^2(B)\\\\\frac{sin^2(B)}{\sin^2(B)}+\frac{cos^2(B)}{\sin^2(B)} = \csc^2(B)\\\\\frac{sin^2(B)+cos^2(B)}{\sin^2(B)} = \csc^2(B)\\\\\frac{1}{\sin^2(B)} = \csc^2(B)\\\\\csc^2(B)=\csc^2(B) \ \ {\Large \checkmark}\\\\$