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nalin [4]
2 years ago
12

Help please !!! will give brainliest

Mathematics
1 answer:
vodka [1.7K]2 years ago
6 0

Answer:

x=-1

Step-by-step explanation:

(15-6+12+x ) / 4 = 5

20 = 27-6+x

20=21+x

x=-1

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Answers <br> A. 2<br> B. -2 <br> C. 10<br> D. -10
Mashcka [7]

Answer:b

10 to 20 is 10 and 80 to 60 is -20 so it’s -20/10 which is -2

7 0
2 years ago
Math, yes I don't know if I got it right
olga nikolaevna [1]

Answer:

4*(n)^2

Step-by-step explanation:


8 0
3 years ago
Read 2 more answers
as a gift, you fill the calendar with packets of chocolate candy. Each packet has a volume of 2 cubic inches. Find the maximum n
qaws [65]
The picture of the calendar is shown in the attached image.

Now, first we will get the volume of the calendar itself, we can note the calendar has the shape of a triangular prism.
Volume of triangular prism = area of base * depth
The area of base = area of triangle = 1/2 * base * depth 
Therefore:
Volume of prism = 1/2 * base * height * depth
where:
base = 4 in
height is he height of the base = 6 in
depth is the depth of the calendar = 8 in
Therefore:
Volume of calendar = 1/2 * 4 * 6 * 8 = 96 in^3

Now, we are given that the volume of each candy is 2 in^3, this means that:
number of candies to fill the calendar = volume of calendar / volume of candy
                                                            = 96/2 
                                                            = 48 candies

Hope this helps :)

4 0
3 years ago
Solve 73 make sure to also define the limits in the parts a and b
Aleks04 [339]

73.

f(x)=\frac{3x^4+3x^3-36x^2}{x^4-25x^2+144}

a)

\lim_{x\to\infty}f(x)=\lim_{x\to\infty}(\frac{3+\frac{3}{x}-\frac{36}{x^2}}{1-\frac{25}{x^2}+\frac{144}{x^4}})=3\lim_{x\to-\infty}f(x)=\lim_{x\to-\infty}(\frac{3+\frac{3}{x}-\frac{36}{x^2}}{1-\frac{25}{x^2}+\frac{144}{x^4}})=3\cdot\frac{1}{2}=3

b)

Since we can't divide by zero, we need to find when:

x^4-2x^2+144=0

But before, we can factor the numerator and the denominator:

\begin{gathered} \frac{3x^2(x^2+x-12)}{x^4-25x^2+144}=\frac{3x^2((x+4)(x-3))}{(x-3)(x-3)(x+4)(x+4)} \\ so: \\ \frac{3x^2}{(x+3)(x-4)} \end{gathered}

Now, we can conclude that the vertical asymptotes are located at:

\begin{gathered} (x+3)(x-4)=0 \\ so: \\ x=-3 \\ x=4 \end{gathered}

so, for x = -3:

\lim_{x\to-3^-}f(x)=\lim_{x\to-3^-}-\frac{162}{x^4-25x^2+144}=-162(-\infty)=\infty\lim_{x\to-3^+}f(x)=\lim_{x\to-3^+}-\frac{162}{x^4-25x^2+144}=-162(\infty)=-\infty

For x = 4:

\lim_{x\to4^-}f(x)=\lim_{n\to4^-}\frac{384}{x^4-25x^2+144}=384(-\infty)=-\infty\lim_{x\to4^-}f(x)=\lim_{n\to4^-}\frac{384}{x^4-25x^2+144}=384(-\infty)=-\infty

4 0
10 months ago
Help me plzzz with my math
Kipish [7]

Answer:

The answer is A.

Step-by-step explanation:

Hope this helped! :)

5 0
3 years ago
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