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nikklg [1K]
3 years ago
12

I need some help with these two questions please help!

Mathematics
1 answer:
antoniya [11.8K]3 years ago
6 0

Answer:

Step-by-step explanation:

-10.6*0.5=m+11.7

-5.3=m+11.7

-5.3-11.7=m

-17=m

14.2=2(-5.8+t)

14.2=-11.6+2t

14.2+11.6=2t

25.8=2t

25.8/2=t

12.9=t

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The area of an circle with 2.2 radius
lana66690 [7]
The area of a circle is
Лx r²
 so you  would take 3.14 times 2.2²
 3.14 x 4.84
so your answer would be 15.19. 
4 0
3 years ago
you invest $4000 in an account that pays simple interest of 4% for 10 years. how much will you have in 10 years?
bulgar [2K]
Annual interest - 4%

interest in 1 year 
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160

interest in 10 years
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4 0
3 years ago
Read 2 more answers
Part 3 - Discussion/Explanation Question
SpyIntel [72]

Step-by-step explanation:

Vertical asymptote can be Identites if there is a factor only in the denominator. This means that the function will be infinitely discounted at that point.

For example,

\frac{1}{x - 5}

Set the expression in the denominator equal to 0, because you can't divide by 0.

x - 5 = 0

x = 5

So the vertical asymptote is x=5.

Disclaimer if you see something like this

\frac{(x - 5)(x + 3)}{(x - 5)}

x=5 won't be a vertical asymptote, it will be a hole because it in the numerator and denominator.

Horizontal:

If we have a function like this

\frac{1}{x}

We can determine what happens to the y values as x gets bigger, as x gets bigger, we will get smaller answers for y values. The y values will get closer to 0 but never reach it.

Remember a constant can be represent by

a \times  {x}^{0}

For example,

1 = 1 \times  {x}^{0}

2 =  2 \times {x}^{0}

And so on,

and

x =  {x}^{1}

So our equation is basically

\frac{1 \times  {x}^{0} }{ {x}^{1} }

Look at the degrees, since the numerator has a smaller degree than the denominator, the denominator will grow larger than the numerator as x gets larger, so since the larger number is the denominator, our y values will approach 0.

So anytime, the degree of the numerator < denominator, the horizontal asymptote is x=0.

Consider the function

\frac{3 {x}^{2} }{ {x}^{2}  + 1}

As x get larger, the only thing that will matter will be the leading coefficient of the leading degree term. So as x approach infinity and negative infinity, the horizontal asymptote will the numerator of the leading coefficient/ the leading coefficient of the denominator

So in this case,

x =  \frac{3}{1}

Finally, if the numerator has a greater degree than denominator, the value of horizontal asymptote will be larger and larger such there would be no horizontal asymptote instead of a oblique asymptote.

8 0
2 years ago
Round nearest hundredth​
maks197457 [2]

Answer:.

Step-by-step explanation:nearest hundredth the answer for this question is joe mama

4 0
3 years ago
Ill give brainliest i really need help on this
Sever21 [200]
I think the answer is

B.
8 0
2 years ago
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