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2 years ago

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Tucker ran 3/4 of a mile, took a break, ran 3/8 of a mile, and then took a second break. After the second break, Tucker ran some

more. If he ran a combined total of 2 miles, how far did Tucker run after his second break?

2 answers:

Nataly_w [17]2 years ago

6 0

Answer:

7/8 of a mile

Step-by-step explanation:

hope this helps

I am Lyosha [343]2 years ago

6 0

7/8 the guys right, supper helpful ty

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Find the equation of all tangent lines having slope of -1 that are tangent to the curve y=(9)/(x+1)

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Answer:

$f(x)=\frac9{x+1}\\ f'(x)=-\frac9{(x+1)^2}\\ f'(x)=-1\ \iff\ -\frac9{(x+1)^2}=-1\ \to \ \frac9{(x+1)^2}=1\ \to \ (x+1)^2=9\\ |x+1|=3\ \to \ x+1=3\ \vee\ x+1=-3\\ x_1=2\ \vee\ x_2=-4\\ f(x_1)=f(2)=\frac9{2+1}=3\\ f(x_2)=f(-4)=\frac9{-4+1}=-3$

First tangent line:

$y=f'(x_1)\cdot (x-x_1)+f(x_1)\ \to \ y=-1(x-2)+3\ \to \ y=-x+5$

Second tangent line:

$y=f'(x_2)\cdot (x-x_2)+f(x_2)\ \to \ y=-1(x+4)-3\ \to \ y=-x-7$

Notice: slope of -1 means that both $f'(x_1), \ f'(x_2)$ are equal to -1, so $f'(x_1)=-1 \ and \ f'(x_2)=-1$

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1. Given points A(3, -5) and B(19, -1), find the coordinates of point C that sit 3/8 of the way along line AB, closer to A than

zaharov [31]

1. C(x, y) = (7.3, –3.9)

2. C(x, y) = (17, –1.5)

Solution:

Question 1:

Let the points are A(3, –5) and B(19, –1).

C is the point that on the segment AB in the fraction $\frac{3}{8}$ .

Point divides segment in the ratio formula:

$$C(x, y)=\left(\frac{mx_2+nx_1}{m+n} , \frac{my_2+ny_1}{m+n}\right)$

Here, $x_1=3, y_1=-5, x_2=19, y_2=-1$ and m = 3, n = 8

$$C(x, y)=\left(\frac{3\times19+8\times3}{3+8} , \frac{3\times(-1)+8\times(-5)}{3+8}\right)$

$$=\left(\frac{57+24}{11} , \frac{-3-40}{11}\right)$

$$=\left(\frac{81}{11} , \frac{-43}{11}\right)$

C(x, y) = (7.3, –3.9)

Question 2:

Let the points are A(3, –5) and B(19, –1).

C is the point that on the segment AB in the fraction $\frac{3}{8}$ .

Point divides segment in the ratio formula:

$$C(x, y)=\left(\frac{mx_2+nx_1}{m+n} , \frac{my_2+ny_1}{m+n}\right)$

Here, $x_1=3, y_1=-5, x_2=19, y_2=-1$ and m = 7, n = 1

$$C(x, y)=\left(\frac{7\times19+1\times3}{7+1} , \frac{7\times(-1)+1\times(-5)}{7+1}\right)$

$$=\left(\frac{133+3}{8} , \frac{-7-5}{8}\right)$

$$=\left(\frac{136}{8} , \frac{-12}{8}\right)$

C(x, y) = (17, –1.5)

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