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3 years ago

DB is a diameter of circle C. KAis a tangent line to circle C at point A.

Mathematics

1 answer:

Arte-miy333 [17]3 years ago

5 0

Answer:

BAD is 67°

ABC is 90°

Step-by-step explanation:

ABC is 90 due to the diameter rule in circle theorem

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Solve for y: ax+yb=c

Ganezh [65]

Answer:

y = $\frac{c-ax}{b}$

Step-by-step explanation:

ax + yb = c ( subtract ax from both sides )

yb = c - ax ( isolate y by dividing both sides by b )

y = $\frac{c-ax}{b}$

4 0

3 years ago

find the coordinates of the image of point Z (5,-1) when it is translated 1 unit to the left and 2 units down. pls help!!

DaniilM [7]

Answer:

$Z^{'} (5 - 1, -1 - 2) = (4, -3)$

Step-by-step explanation:

The formula of the translation can be written as

$x_2 = x_1 - 1$ , because a shift to the left is along the x - axis and it's negative because it's on the left of the graph.

$y_2 = y_1 - 2$ , because a shift up or down is along the y axis.

To compute the new point Z:

5 0

3 years ago

Read 2 more answers

If GHJ is equal to LMK, with a scale factor of 5:6, find the perimeter of GHJ.

xxTIMURxx [149]

Answer: Weeeell, 35

Step-by-step explanation:

1. By definition, the perimeter of a triangle is the sum of the lenght of each side.

2. Then, the perimeter of the triangle LMK is:

P_{LMK}=14+17+11\\P_{LMK}=42

3. If both triangles are congruent and the scale factor is 5:6 (Which you can rewrite as a fraction: 5/6), you must multiply the perimeter of the triangle LMK by this scale factor.

4. Then, you have that the perimeter of the triangle GHJ is:

P_{GHJ}=42*\frac{5}{6}=35

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4 0

3 years ago

Consider the curve defined by the equation y=6x2+14x. Set up an integral that represents the length of curve from the point (−2,

torisob [31]

Answer:

32.66 units

Step-by-step explanation:

We are given that

$y=6x^2+14x$

Point A=(-2,-4) and point B=(1,20)

Differentiate w.r. t x

$\frac{dy}{dx}=12x+14$

We know that length of curve

$s=\int_{a}^{b}\sqrt{1+(\frac{dy}{dx})^2}dx$

We have a=-2 and b=1

Using the formula

Length of curve= $s=\int_{-2}^{1}\sqrt{1+(12x+14)^2}dx$

Using substitution method

Substitute t=12x+14

Differentiate w.r t. x

$dt=12dx$

$dx=\frac{1}{12}dt$

Length of curve= $s=\frac{1}{12}\int_{-2}^{1}\sqrt{1+t^2}dt$

We know that

$\sqrt{x^2+a^2}dx=\frac{x\sqrt {x^2+a^2}}{2}+\frac{1}{2}\ln(x+\sqrt {x^2+a^2})+C$

By using the formula

Length of curve= $s=\frac{1}{12}[\frac{t}{2}\sqrt{1+t^2}+\frac{1}{2}ln(t+\sqrt{1+t^2})]^{1}_{-2}$

Length of curve= $s=\frac{1}{12}[\frac{12x+14}{2}\sqrt{1+(12x+14)^2}+\frac{1}{2}ln(12x+14+\sqrt{1+(12x+14)^2})]^{1}_{-2}$

Length of curve= $s=\frac{1}{12}(\frac{(12+14)\sqrt{1+(26)^2}}{2}+\frac{1}{2}ln(26+\sqrt{1+(26)^2})-\frac{12(-2)+14}{2}\sqrt{1+(-10)^2}-\frac{1}{2}ln(-10+\sqrt{1+(-10)^2})$