A student looked at the diagram below and concluded x=y.

Mathematics

1 answer:

8090 [49]2 years ago

8 0

Answer:

$y = 85$

$x = 95$

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7.9-3.47=please show me how to rewrite it vertically?

Lemur [1.5K]

Answer:

4.43

Step-by-step explanation:

7.90

- 3.47

__________

4. 4 3

I couldn't regroup but of you cross out 0 and put 10, 10-7= 3

and you have to borrow one from 9 and make it 8 and 8-4=4

4 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:

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