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3 years ago

Where do i put parantheses into 140/2+12-4x2=2

Mathematics

1 answer:

mixer [17]3 years ago

8 0

(140/2)+12-(4 x 2)= 2

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Which is an equation of a direct proportion

Over [174]

K = y/x OR y = kx

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3 years ago

-4x+3(5x+6)=7(2x+1)-3x

m_a_m_a [10]

-4x+15x+18=14x+7-3x

-4x+15x=11x

11x+18=14x-3x+7

14x-3x=11x

11x+7

11x+18=11x+7

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4 years ago

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A bacteria culture starts with 100 bacteria and doubles in size every half hour. (a) how many bacteria are there after 4 hours?

Savatey [412]

19200 all you have to do to figure this out is multiply 100 times 8 since the is 8 halfs in 4 hours.
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5 0

4 years ago

What does X equal? 14x+24=18x-32

exis [7]

Answer:

x =14

Step-by-step explanation:

14x+24=18x-32

Subtract 14x from each side

14x-14x+24=18x-14x-32

24 = 4x -32

Add 32 from each side

24+32=4x-32+32

56 = 4x

Divide each side by 4

56/4 = 4x/4

14=x

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3 years ago

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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