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3 years ago

Help!! Stuck on math problem!

Mathematics

1 answer:

Brut [27]3 years ago

7 0

Vertical Asymptotes:
x=0
x=0
Horizontal Asymptotes:
y=0
y=0
No Oblique Asymptotes

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What is the value of y in the equation 6 + y = −3?

oee [108]

The value of y is -9

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3 years ago

2.05 times 10 to the power of 6 in standard notation

Natalka [10]

Answer:

2050000

Step-by-step explanation:

10 to the power of 6= 1000000

2.05 x 1000000 = 2,050,000

Hence, 2.05 × 10^6 = 2050000

Hope this helps!

Please mark brainliest if you think I helped! Would really appreciate!

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3 years ago

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I had a summer job cooking two items at a bbq place. I sell burgers for $9.00 and sell chicken sandwiches for $11.25. Because I

pishuonlain [190]

Answer:

He sold 180 chicken sandwiches and 150 burguers.

Step-by-step explanation:

Since 330 items were sold, then the amount of burguers and chicken sandwiches sold must be equal to this value. From this information we have:

burguers + sandwiches = 330

We know that each burguer cost $9 and each chicken sandwich cost $11.25 and that we made a total of $3,375. So we have:

9*burguers + 11.25*sandwiches = 3,375

From the first equation we have:

burguers = 330 - sandwiches

Applying it on the second equation:

9*(330 - sandwiches) + 11.25*sandwiches = 3,375

2,970 - 9*sandwiches + 11.25*sandwiches = 3,375

2.25*sandwiches = 3,375 - 2,970

2.25*sandwiches = 405

sandwiches = 405/2.25 = 180

So,

burguers = 330 - 180 = 150

He sold 180 chicken sandwiches and 150 burguers.

8 0

3 years ago

Which of the following are identities? Check all that apply

Natasha2012 [34]

Answer:

A, C

Step-by-step explanation:

Actually, those questions require us to develop those equations to derive into trigonometrical equations so that we can unveil them or not. Doing it only two alternatives, the other ones will not result in Trigonometrical Identities.

Examining

A) True

$\frac{1-tan^{2}x}{2tanx} =\frac{1}{tan2x} \\ \frac{1-tan^{2}x}{2tanx} =\frac{1}{\frac{2tanx}{1-tan^{2}x}}\\ tan2x=\frac{1-tan^{2}x}{2tanx}$

Double angle $tan2\alpha =\frac{1 -tan^{2}\alpha }{2tan\alpha}$

B) False,

No further development towards a Trig Identity

C) True

Double Angle Sine Formula $sin2\alpha =2sin\alpha *cos\alpha$

$sin(8x)=2sin(4x)cos(4x)\\2sin(4x)cos(4x)=2sin(4x)cos(4x)$

D) False No further development towards a Trig Identity

$[sin(x)-cos(x)]^{2} =1+sin(2x)\\ sin^{2} (x)-2sin(x)cos(x)+cos^{2}x=1+2sinxcosx\\ \\sin^{2} (x)+cos^{2}x=1+4sin(x)cos(x)$

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3 years ago

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Prime factorisation of 8232

Rufina [12.5K]

Answer:

2 x 2 x 2 x 3 x 7 x 7 x 7

Step-by-step explanation:

it is Wednesday my dudes

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3 years ago

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