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2 years ago

Evaluate 6j+23+kl−3, when j=3, k=7, and l=33.

Mathematics

1 answer:

Dmitry [639]2 years ago

3 0

Answer:

269

Step-by-step explanation:

6(3)+23+(7)(33)-3

\/ \/

18 + 23 + 231 - 3 = 269

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Let $f(x) = 12x^9 + x$. Find $f(3) + f(-3)$.

BlackZzzverrR [31]

Answer:

Ok, here goes: f(3) means take f(x) and plug in 3 wherever we see x. So

f(3) = 12*39 + 3 = 236196 + 3

f(3) = 236199

Likewise, f(-3) means take f(x) and plug in (-3) wherever we see x. So

f(-3) = 12*(-3)9 + (-3) = -236196 - 3

f(-3) = -236199

Putting those two things together, we get

f(3) + f(-3) = 236199 - 236199 = 0

Step-by-step explanation:

7 0

1 year ago

Cos (90-theta) • cosec90-theta) =tano. How?

denis-greek [22]

Answer:

______________________________________________________

TRIGONOMETRY IDENTITIES TO BE USED IN THE QUESTION :-

For any right angled triangle with one angle α ,

$\cos (90 - \alpha ) = \sin \alpha$ or $\sin(90 - \alpha ) = \cos\alpha$

$cosec \: (90 - \alpha ) = \sec\alpha$ or $\sec(90 - \alpha ) = cosec\:\alpha$

SOME GENERAL TRIGNOMETRIC FORMULAS :-

$\sin \alpha = \frac{1}{cosec \: \alpha }$ or $cosec \: \alpha = \frac{1}{\sin \alpha }$

$\cos \alpha = \frac{1}{\sec \alpha }$ or $\sec \alpha = \frac{1}{\cos \alpha }$

______________________________________________________

Now , lets come to the question.

In a right angled triangle , let one angle be α (in place of theta) .

So , lets solve L.H.S.

$\cos (90 - \alpha ) \times cosec(90 - \alpha )$

$=> sin\alpha \times \sec\alpha$

$=> \sin\alpha \times \frac{1}{\cos\alpha }$

$=> \frac{\sin\alpha }{\cos\alpha }$

$=> \tan\alpha$ = R.H.S.

∴ L.H.S. = R.H.S. (Proved)

3 0

3 years ago

7.980 divided on 0.19

Gnom [1K]

7.980 divide by 0.19 is 42.

8 0

3 years ago

What is 13/169 in simplest form?

PolarNik [594]

The answer is 1/13, because 13 goes into 169 13 times

3 0

2 years ago

I need to solve these two equations using either substitution or elimination

vitfil [10]

Answer:

y = -15

x = -10

Step-by-step explanation:

1/4x - 1/2 = 5 (1)

3x + 2/3y = -40 (2)

Multiply (1) by 4

1/4x - 1/2 = 5

1/4x * 4 - 1/2y * 4 = 5 * 4

4/4x - 4/2y = 20

x - 2y = 20 (3)

Multiply (2) by 3

3x + 2/3y = -40 (2)

3x * 3 + 2/3y * 3 = -40 * 3

9x + 6/3y = -120

9x + 2y = -120 (4)

x - 2y = 20 (3)

9x + 2y = -120 (4)

Using elimination method, Add (3) and (4)

9x + x = -120 + 20

10x = -100

x = -100/10

x = -10

Substitute x = -10 into (3)

x - 2y = 20 (3)

-10 - 2y = 20

-2y = 20 + 10

-2y = 30

y = 30/-2

y = -15

x = -10

8 0

3 years ago