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2 years ago

Write y=3/5x+7 in standard form

Mathematics

2 answers:

andriy [413]2 years ago

6 0

Answer:

3x-5y=-35

Step-by-step explanation:

I have no clue what the step by step explanation i looked it up

larisa [96]2 years ago

5 0

I think the answer is 3x−5y=−35

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Plz solve this problem of trigonometry<br>i am an aakashian

makkiz [27]

Step-by-step explanation:

$\bf L.H.S = \tt \dfrac{sec\: \theta + tan \: \theta - 1}{tan \: \theta - sec \: \theta + 1} \\ \\$

$: \implies \tt \dfrac{\frac{1}{cos \: \theta} + \frac{sin \: \theta}{cos \: \theta} - 1}{ \frac{sin \: \theta}{cos \: \theta} - \frac{1}{cos \: \theta} + 1 } \: = \dfrac{1 + sin \: \theta - cos \: \theta}{sin \: \theta + cos \: \theta} \\ \\$

$: \implies \tt\dfrac{ sin \: \theta - (cos \: \theta - 1)}{sin \: \theta + (cos \: \theta - 1)} \: \times \: \dfrac{ sin \: \theta - (cos \: \theta - 1)}{sin \: \theta - (cos \: \theta - 1)} \\ \\$

$: \implies \tt\dfrac{ sin^{2} \: \theta + cos^{2} \: \theta + 1 - 2 \: cos \: \theta - 2 \: sin \: \theta \: (cos \: \theta - 1)}{sin^{2} \: \theta - (cos \: \theta - 1)^{2} } \\ \\$

$: \implies \tt\dfrac{1 + 1 - 2 \: cos \: \theta - 2 \: sin \: \theta \: cos \: \theta + 2 \: sin \: \theta}{sin^{2} \: \theta + cos^{2} \: \theta - 1 + 2 \: cos \: \theta } \\ \\$

$: \implies \tt\dfrac{2 - 2 \: cos \: \theta - 2 \: sin \: \theta \: cos \: \theta + 2 \: sin \: \theta}{sin^{2} \: \theta + cos^{2} \: \theta - sin^{2} \: \theta - cos^{2} \: \theta + 2 \: cos \: \theta } \\ \\$

$: \implies \tt\dfrac{2 (1 - \: cos \: \theta )- 2 \: sin \: \theta (1 - \: cos \: \theta)}{ 2 \: cos \: \theta - 2 \: cos^{2} \: \theta} \\ \\$

$: \implies \tt\dfrac{(2 + 2 \: sin \: \theta) \: \cancel{(1 - cos\: \theta)}}{2 \: cos \: \theta \: \cancel{(1 - cos \: \theta)}} \: = \: \dfrac{1 + sin \: \theta}{cos \: \theta} \\ \\$

$: \implies\tt\dfrac{1 + sin \: \theta}{cos \: \theta} \: \times \: \dfrac{1 - sin \: \theta}{1 - sin \: \theta} \\ \\$

$: \implies\tt\dfrac{1 + sin^{2} \: \theta}{cos \: (1 - sin \: \theta)} \\ \\$

$: \implies\tt\dfrac{cos^{2} \: \theta}{cos \: \theta (1 - sin \: \theta)} \\ \\$

$: \implies\tt\dfrac{cos \: \theta}{1 - sin \: \theta} \: = \: \bf{ R.H.S}\\ \\$

$\huge\bigstar \:\underline{\red{\sf Hence, Proved}} \: \bigstar \\$

6 0

4 years ago

Rewrite 2/20 : 1/40 as a unit rate

Usimov [2.4K]

Answer:

yes

Step-by-step explanation:

3 0

3 years ago

Help please asap please

olga55 [171]

Answer:

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Step-by-step explanation:

3 0

4 years ago

Simplify or solve for x in (X+9)/2=3

Tems11 [23]

Answer: -3, write the division as a fraction, multiply both sides by 2, move constants to the right, then you're left with x=6-9 which is -3

6 0

4 years ago

Which choice has the same equation in both forms?

Soloha48 [4]

Answer:

8x + 4y = 12 & y = -2x + 3 are the same equations.

Step-by-step explanation:

To know which choice has the same equation in both forms, we need to transform the standard form into the slope intercept form.

Then, solving for y in the first equation, we get:

$8x + 4y = 12\\8x + 4y - 8x = 12 -8x\\4y=12-8x\\\frac{4y}{4}=\frac{12-8x}{4} \\y=3 - 2x\\y=-2x+3$

It means that 8x + 4y = 12 & y = -2x + 3 are the same equations.

At the same way, we get that:

6x + 3y = 18 & y = 2x + 6 are not the same equation:

$6x + 3y = 18\\6x + 3y - 6x = 18-6x\\3y=18-6x\\\frac{3y}{3}=\frac{18-6x}{3} \\y=6 - 2x\\y=-2x+6$

2x + 4y = 8 & y = -2x + 2 are not the same equation:

$2x + 4y = 8\\2x + 4y -2x= 8-2x\\4y=8-2x\\\frac{4y}{4}=\frac{8-2x}{4}\\ y=2-\frac{1}{2}x$

3x + 4y = 12 & y = ¾x + 3 are not the same equation:

$3x + 4y = 12\\3x + 4y -3x= 12-3x\\4y=12-3x\\\frac{4y}{4}=\frac{12-3x}{4}\\ y=3-\frac{3}{4}x$

5 0

3 years ago