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2 years ago

Please help! (giving brainliest)

Mathematics

1 answer:

emmasim [6.3K]2 years ago

6 0

luca: $5\cdot \frac{1}{3}=\frac{5\cdot \:1}{3} =\frac{5}{3}$ 1 remainder 2

bailey: $6\cdot \frac{4}{5}=\frac{6\cdot \:4}{5} =\frac{24}{5}$ = 4 remainder 4

corin: $3\frac{1}{4}\cdot \:2\frac{1}{3} =\frac{13}{4}\cdot \:2\frac{1}{3} =\frac{13}{4}\cdot \frac{7}{3} =\frac{13\cdot \:7}{4\cdot \:3} =\frac{91}{4\cdot \:3} =\frac{91}{12} =7\frac{7}{12}$

sabra: $1\frac{4}{5}\cdot \frac{1}{2} =\frac{9}{5}\cdot \frac{1}{2} =\frac{9\cdot \:1}{5\cdot \:2} =\frac{9}{5\cdot \:2} =\frac{9}{10}$ 0.9

<em>So D. is the right answer.</em>

<h3><em>Hope I helped you!</em></h3><h3><em>Success!</em></h3>

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Answer:

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Part 2) $tan(A)=4\frac{\sqrt{65}}{65}$

Part 3) $cot(A)=\frac{\sqrt{65}}{4}$

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Step-by-step explanation:

Let

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we have that

$sin(A)=\frac{4}{9}$

step 1

Find the cos(A)

we know that

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substitute the value of sin(A)

$cos^{2}(A)+(\frac{4}{9})^{2}=1$

$cos^{2}(A)=1-(\frac{4}{9})^{2}$

$cos^{2}(A)=1-(\frac{16}{81})$

$cos^{2}(A)=(\frac{65}{81})$

$cos(A)=\frac{\sqrt{65}}{9}$

step 2

Find the tan(A)

we know that

$tan(A)=\frac{sin(A)}{cos(A)}$

substitute the values

$tan(A)=\frac{\frac{4}{9}}{\frac{\sqrt{65}}{9}}$

$tan(A)=\frac{4}{\sqrt{65}}$

$tan(A)=4\frac{\sqrt{65}}{65}$

step 3

Find the cot(A)

we know that

$cot(A)=\frac{1}{tan(A)}$

we have

$tan(A)=\frac{4}{\sqrt{65}}$

substitute

$cot(A)=\frac{1}{\frac{4}{\sqrt{65}}}$

$cot(A)=\frac{\sqrt{65}}{4}$

step 4

Find the sec(A)

we know that

$sec(A)=\frac{1}{cos(A)}$

we have

$cos(A)=\frac{\sqrt{65}}{9}$

substitute

$sec(A)=\frac{1}{\frac{\sqrt{65}}{9}}$

$sec(A)=\frac{9}{\sqrt{65}}$

$sec(A)=9\frac{\sqrt{65}}{65}$

step 5

Find the csc(A)

we know that

$csc(A)=\frac{1}{sin(A)}$

we have

$sin(A)=\frac{4}{9}$

substitute the value

$csc(A)=\frac{1}{\frac{4}{9}}$

$csc(A)=\frac{9}{4}$

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