Trigonometry Identities help please

Mathematics

1 answer:

AfilCa [17]3 years ago

6 0

$\sin(x) = - \frac{ \sqrt{2} }{2}$
$\tan(x) = - \frac{ \sqrt{2} }{ \sqrt{2} }$

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Write a list of steps that are needed to find the measure of ∠B.

andrew11 [14]

Answer:

The measure of angle B is 41°.

Step-by-step explanation:

180 - 113 = 67.

67 + 72 = 139.

180 - 139 = 41.

I hope this helped! If you need more details about it, ask me!

3 0

3 years ago

A company that makes hard candy have a standard bag of hard candy with 150 pieces. The hard candy has three distinct colors red,

anzhelika [568]

Answer:

The degrees of freedom, Df = The number of bags produced on Monday - 1

Step-by-step explanation:

The number of degrees of freedom is the limiting number of values that are logically not influenced by other values such that they are capable of having variation

The degrees of freedom = The sample size - 1 = N - 1

Therefore, the degrees of freedom, Df = The number of bags produced on Monday - 1

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3 years ago

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valina [46]

Answer:

Step-by-step explanation:

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3 years ago

17+4h+2=1-5h what does the variable equal ?

olga2289 [7]

1. 17+4h+2=1-5h
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3. 18+4h=-5h
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3 years ago

Solve the system of equations using matrices. Use the Gauss- Jordan elimination method And find a solution set

KATRIN_1 [288]

$\begin{gathered} x+y+z=4 \\ x-y-z=0 \\ x-y+z=8 \\ \text{The system using matrix is} \\ \begin{bmatrix}{1} & {1} & {1}, & {4} \\ {1} & {-1} & {-1,} & {0} \\ {1} & {-1} & {1,} & {8}{}{}\end{bmatrix}\rightarrow F2=F2-F1=\begin{bmatrix}{1} & {1} & {1}, & {4} \\ {0} & {-2} & {-2,} & {-4} \\ {1} & {-1} & {1,} & {8}{}{}\end{bmatrix} \\ \rightarrow F3=F3-F1=\begin{bmatrix}{1} & {1} & {1}, & {4} \\ {0} & {-2} & {-2,} & {-4} \\ {0} & {-2} & {0,} & {4}{}{}\end{bmatrix}\rightarrow F2=-\frac{1}{2}F2 \\ =\begin{bmatrix}{1} & {1} & {1}, & {4} \\ {0} & {1} & {1,} & {2} \\ {0} & {-2} & {0,} & {4}{}{}\end{bmatrix}\rightarrow F3=F3+2F2=\begin{bmatrix}{1} & {1} & {1}, & {4} \\ {0} & {1} & {1,} & {2} \\ {0} & {0} & {2,} & {8}{}{}\end{bmatrix}\rightarrow F3=\frac{1}{2}F3 \\ =\begin{bmatrix}{1} & {1} & {1}, & {4} \\ {0} & {1} & {1,} & {2} \\ {0} & {0} & {1,} & {4}{}{}\end{bmatrix}\rightarrow F2=F2-F3=\begin{bmatrix}{1} & {1} & {1}, & {4} \\ {0} & {1} & {0,} & {-2} \\ {0} & {0} & {1,} & {4}{}{}\end{bmatrix} \\ \rightarrow F1=F1-F3=\begin{bmatrix}{1} & {1} & {0}, & {0} \\ {0} & {1} & {0,} & {-2} \\ {0} & {0} & {1,} & {4}{}{}\end{bmatrix}\rightarrow F1=F1-F2 \\ =\begin{bmatrix}{1} & {0} & {0}, & {2} \\ {0} & {1} & {0,} & {-2} \\ {0} & {0} & {1,} & {4}{}{}\end{bmatrix} \\ \text{Therefore, the solution is }x=2,\text{ y=-2 and z=4} \end{gathered}$

6 0

1 year ago