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2 years ago

Chord AB is congruent to chord CD. What is the measure of arc CD?

Mathematics

1 answer:

tatiyna2 years ago

8 0

Is there measurements for AB AND CD

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The graph h = −16t^2 + 25t + 15 models the height and time of a ball that was thrown off of a building where h is the height in

Olenka [21]

Answer:

Step-by-step explanation:

If you graph the equation f(x) = -16x^2+25x+15. You would need to pay attention to the x value which is somewhere in between 0.75-0.9.

Another way is to convert the equation into vertex form.

6 0

2 years ago

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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

Terrell wrote that 5-4.2=8. Which statement about this answer is true?

Nuetrik [128]

Answer:

No.

Step-by-step explanation:

No. 5 - 4.2 = 0.8

7 0

2 years ago

6/15 divided by 1/7

mylen [45]

Question-
6/15 divided by 1/7

Answer-
Find the reciprocal of the divisor
Reciprocal of 1/7: 7/1Now, multiply it with the dividend
So, 6/15 ÷ 1/7 = 6/15 × 7/1= (6 × 7)/(15 × 1) = 42/15After reducing the fraction, the answer is 14/5In mixed form: 2 4/5

6 0

3 years ago

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during a sale , a shop allows discount of the marked price of clothing . What will a costumer pay for a dress with a marked pric

ollegr [7]

Answer:

35 d

Step-by-step explanation:

3 0

2 years ago

Read 2 more answers