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sasho [114]
2 years ago
14

I need help I need help

Mathematics
1 answer:
Cerrena [4.2K]2 years ago
5 0

Answer:

{ \boxed{ \bf{c}}} \:   \: \frac{3.8}{4}  = s

Step-by-step explanation:

{ \tt{4 \: pies \:  =   \: 3.8 \: ounces}} \\ { \tt{1 \: pie \:  =  \: ( \frac{1}{4}  \times 3.8) \: ounces}} \\ { \tt{ = 0.95 \: ounces}}

{ \underline{ \sf{ \blue{christ \: † \: alone}}}}

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Can you help me please? thank you <3
olganol [36]

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7 0
3 years ago
Determine the horizontal vertical and slant asymptote y=x^2+2x-3/x-7
lilavasa [31]

Answer:

<h2>A.Vertical:x=7</h2><h2>Slant:y=x+9</h2>

Step-by-step explanation:

f(x)=\dfrac{x^2+2x-3}{x-7}\\\\vertical\ asymptote:\\\\x-7=0\qquad\text{add 7 to both sides}\\\\\boxed{x=7}\\\\horizontal\ asymptote:\\\\\lim\limits_{x\to\pm\infty}\dfrac{x^2+2x-3}{x-7}=\lim\limits_{x\to\pm\infty}\dfrac{x^2\left(1+\frac{2}{x}-\frac{3}{x^2}\right)}{x\left(1-\frac{7}{x}\right)}=\lim\limits_{x\to\pm\infty}\dfrac{x\left(1+\frac{2}{x}-\frac{3}{x^2}\right)}{1-\frac{7}{x}}=\pm\infty\\\\\boxed{not\ exist}

slant\ asymptote:\\\\y=ax+b\\\\a=\lim\limits_{x\to\pm\infty}\dfrac{f(x)}{x}\\\\b=\lim\limits_{x\to\pm\infty}(f(x)-ax)\\\\a=\lim\limits_{x\to\pm\infty}\dfrac{\frac{x^2+2x-3}{x-7}}{x}=\lim\limits_{x\to\pm\infty}\dfrac{x^2+2x-3}{x(x-7)}=\lim\limits_{x\to\pm\infty}\dfrac{x^2+2x-3}{x^2-7x}\\\\=\lim\limits_{x\to\pm\infty}\dfrac{x^2\left(1+\frac{2}{x}-\frac{3}{x^2}\right)}{x^2\left(1-\frac{7}{x}\right)}=\lim\limits_{x\to\pm\infty}\dfrac{1+\frac{2}{x}-\frac{3}{x^2}}{1-\frac{7}{x}}=\dfrac{1}{1}=1

b=\lim\limits_{x\to\pm\infty}\left(\dfrac{x^2+2x-3}{x-7}-1x\right)=\lim\limits_{x\to\pm\infty}\left(\dfrac{x^2+2x-3}{x-7}-\dfrac{x(x-7)}{x-7}\right)\\\\=\lim\limits_{x\to\pm\infty}\left(\dfrac{x^2+2x-3}{x-7}-\dfrac{x^2-7x}{x-7}\right)=\lim\limits_{x\to\pm\infty}\dfrac{x^2+2x-3-(x^2-7x)}{x-7}\\\\=\lim\limits_{x\to\pm\infty}\dfrac{x^2+2x-3-x^2+7x}{x-7}=\lim\limits_{x\to\pm\infty}\dfrac{9x-3}{x-7}=\lim\limits_{x\to\pm\infty}\dfrac{x\left(9-\frac{3}{x}\right)}{x\left(1-\frac{7}{x}\right)}

=\lim\limits_{x\to\pm\infty}\dfrac{9-\frac{3}{x}}{1-\frac{7}{x}}=\dfrac{9}{1}=9\\\\\boxed{y=1x+9}

8 0
3 years ago
Quadrilateral RUST has a vertex at R(2,3). What are the coordinates of R' after dilation by a scale factor of 3, centered at the
coldgirl [10]
Concept of this answer is (3,5) in the y(x)=-3
6 0
2 years ago
Read 2 more answers
Plz help me plz I need your I am struggling so badly I need your help plz help me I’m begging you plz plz help me
mars1129 [50]
<h3>♫ - - - - - - - - - - - - - - - ~Hello There!~ - - - - - - - - - - - - - - - ♫</h3>

➷ Just substitute the given value into anywhere you see the term x:

2 x 4^2 + 12

Follow the PEMDAS rule:

(calculate the exponential first)

4^2 = 16

Now we have:

2 x 16 + 12

(now the multiplication needs to be calculated)

16 x 2 = 32

Now we have:

32 + 12

Add them together to get the answer:

44 <== this is the answer

<h3><u>✽</u></h3>

➶ Hope This Helps You!

➶ Good Luck (:

➶ Have A Great Day ^-^

↬ ʜᴀɴɴᴀʜ ♡

3 0
3 years ago
Read 2 more answers
Please help please please
Grace [21]
\Omega=\{1;\ 2;\ 3;\ 4;\ 5;\ 6;\ 7;\ 8;\ 9;\ 10\}\\\\|\Omega|=10\\\\A=\{1;\ 3;\ 6;\ 7\};\ B=\{2;\ 3\}\\\\A\ \cup\ B=\{1;\ 2;\ 3;\ 6;\ 7\}\\\\|A\ \cup\ B|=5\\\\P(A\ \cup\ B)=\dfrac{|A\ \cup\ B|}{|\Omega|}\to P(A\ \cup\ B)=\dfrac{5}{10}=\dfrac{1}{2}=0.5=0.50
4 0
2 years ago
Read 2 more answers
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