$(sec x)+(cos x)=(tan x)*(sin x)\\identities\\(sec x)= 1/cos x\\(tan x)=sin x/cos x\\\\and\\(cos x)^{2}+(sin x)^{2}=1\\(cos x)^{2}=1-(sin x)^{2}\\\\(1/cos x )+(cos x) = (sin x)*(sin x)/(cos x)\\ [(1/cos x)+(cos x)]*(cos x)=(sin x)^{2}\\\\1+(cos x)^{2}=(sin x)^{2}\\1+1-(sin x)^{2}=(sin x)^{2}\\2=2*(sin x)^{2}\\2/2=(sin x)^{2}\\1=(sin x)^{2}\\\\This is not true ,so\\(sec x)+(cos x)\neq (tan x)*(sin x)$

6 0

3 years ago

Simplify each expression using the laws of exponents. No negative exponents should be in your final answer. (On the image attach

Cloud [144]

Answer:

Answer is given below with explanations.

Step-by-step explanation:

$by \: simplifying \: the \: given \: expression \: using \: \\ law \: of \: exponents \: we \: get \\ \frac{15 {x}^{15} {y}^{ - 3} {z}^{4} }{18 {x}^{18} { {y}^{4} {z}^{ - 4} }^{} } \\ = \frac{5 \: {x}^{ - 3} {y}^{ - 7} {z}^{8} }{6} \\ we \: may \: also \: write \: as \\ = \frac{5 {z}^{8} }{6 {x}^{3} {y}^{7} }$

$formulae \: used \ \: : \\ \frac{ {x}^{m} }{ { {x}^{n} }^{} } = {x}^{m - n} \: \: \: and \\ {x}^{ - m} = \frac{1}{ {x}^{m} }$

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5 0

3 years ago

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Sara says, "If you subtract 19 from my number and multiply the difference by - 6, the result is - 72." What is Sara's number?​

Sara says, "If you subtract 19 from my number and multiply the difference by - 6, the result is - 72." What is Sara's number?