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Mathematics

2 answers:

N76 [4]3 years ago

8 0

I think the answer is December

ddd [48]3 years ago

8 0

Answer:

Step-by-step explanation:

First Rhonda makes a 20% down payment. 20% of $85 is:

0.20 × $85 = $17

So what's left is:

$85 - $17 = $68

So the number of monthly payments at $8 per month is:

$68 / 8 = 8.5

Rounding up, it will take 9 months to make all the payments. If her father's birthday is on the third Sunday of June, she needs to make her ninth and final payment on June 1. So her first monthly payment needs to be 8 months before that, or October 1.

Answer is D.

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Prove that: (Sec A- cosec A)(1+ tan A+cot A) = tan A× sec A - cot A × cosec A

mash [69]

Answer:

Step-by-step explanation:

$(sec A-cosecA)(1+tanA+cotA)=tanAsecA-cotAcosecA$

we take the LHS so here goes,

$(sec A-cosecA)(1+tanA+cotA)=tanAsecA-cotAcosecA\\(\frac{1}{cosA} -\frac{1}{sinA})(1+\frac{sinA}{cosA}+\frac{cosA}{sinA})\\\\(\frac{sinA-cosA}{sinAcosA})(\frac{sinAcosA+sin^2A+cos^2A}{sinAcosA})\\$

since , $sin^2A+cos^2A=1$

the identity becomes,

$(\frac{sinA-cosA}{sinAcosA})(\frac{1+sinAcosA}{sinAcosA})\\\\(\frac{sinA+sin^2AcosA-cosA-cos^2AsinA}{sin^2Acos^2A})\\\\$

now, we know,

$sin^2A=1-cos^2A$ and $cos^2A=1-sin^2A$

the identity becomes,

$(\frac{sinA+(1-cos^2A)cosA-cosA-(1-sin^2A)sinA}{sin^2Acos^2A} )\\\\$

$(\frac{sinA+cosA-cos^3A-cosA-sinA+sin^3A}{sin^2Acos^2A})$

sin A and cos A cancel out it becomes zero

$\frac{sin^3A-cos^3A}{sin^2Acos^2A} \\\\$

Splitting the denominator the identity becomes

$\frac{sin^3A}{sin^2Acos^2A}-\frac{cos^3A}{sin^2Acos^2A} \\\\\frac{sinA}{cos^2A} - \frac{cosA}{sin^2A} \\\\\frac{sinA}{cosA}(\frac{1}{cosA})-\frac{cosA}{sinA}(\frac{1}{sinA})\\\\$