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Alex_Xolod [135]
2 years ago
8

A line passes through point A(12,18). A second point on the line has an x-value that is 125% of the x-value of point A and a y-v

alue that is 75% of the y-value of point A. Use point A to write an equation of the line in point-slope form.
Mathematics
1 answer:
Reil [10]2 years ago
7 0

Answer:

(y - 18) = -3/2 (x - 12)

Step-by-step explanation:

125% of 12 equals 15

75% of 18 equals 13 1/2

so the slope equals 4.5/-3 or -3/2

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A dog is standing 5 feet from the base of a tree, looking up at a cat that has climbed 16 feet up the tree. What is the angle of
Katena32 [7]

Answer:

x=72.6

Step-by-step explanation:

use inverse trig because your finding a missing angle. use tan because you haved opp/adj

tan-1=(16/5)

x=tan-1(16/5)

*use a calculator

x=72.6

5 0
3 years ago
In each of the following pairs of numbers, how much larger is the first number than the second?
lapo4ka [179]

Answer: a. 154 is larger than 27 by 127.

b. 25 is larger than 12 by 13.

c.135 is larger than 127 by 8.

d.46 is larger than 24 by 22.


Step-by-step explanation:

Subtract second number from the first number.

a. 154 and 27

Difference=154-27=127

Hence, 154 is larger than 27 by 127.

b. 25 and 12

Difference=25-12=13

Hence, 25 is larger than 12 by 13.

c. 135 and 127

Difference=135-127=8

Hence, 135 is larger than 127 by 8.

d. 46 and 24

Difference=46-24=22

Hence, 46 is larger than 24 by 22.

4 0
3 years ago
Read 2 more answers
Polygon ABCD is plotted on a coordinate plane and then rotated 90° clockwise about point C to form polygon A′B′C′D′. Match each
horrorfan [7]

The vertices of ABCD after 90 degrees clockwise rotation about point C are: A' (0,6), B' (3,7), C' (4,6) and D' (4,3)

<h3>How to match the vertices of the polygon?</h3>

The image of the polygon ABCD is not given; however, the question can still be answered because the coordinates are known.

The vertices of polygon ABCD are given as:

A = (4, 6)

B = (5, 3)

C = (4, 2)

D = (1, 2)

The rule of rotation about point C is:

(x,y) = (a + b - y, x + b - a)

Where:

(a, b) = (4, 2) --- the point of rotation.

So, we have:

(x,y) = (4 + 2 - y, x + 4 - 2)

(x,y) = (6 - y, x + 2)

The above means that:

A' = (6 - 6, 4 + 2) = (0,6)

B' = (6 - 3, 5 + 2) = (3,7)

C' = (6 - 2, 4 + 2) = (4,6)

D' = (6 - 2, 1 + 2) = (4,3)

Hence, the image of the rotation and their vertices (i.e. coordinates) are:

A' = (0,6)

B' =  (3,7)

C' = (4,6)

D' =  (4,3)

Read more about rotation at:

brainly.com/question/4289712

#SPJ1

8 0
2 years ago
Last 2 questions please help
Jlenok [28]
4.
Minimum: 20
First Quartile: 44
Median: 51
Third Quartile: 57
Maximum: 86

Ascending order : 20 41 44 44 49 53 55 57 80 86
6 0
2 years ago
Find the derivative of the function at P 0 in the direction of A. ​f(x,y,z) = 3 e^x cos(yz)​, P0 (0, 0, 0), A = - i + 2 j + 3k
Alik [6]

The derivative of f(x,y,z) at a point p_0=(x_0,y_0,z_0) in the direction of a vector \vec a=a_x\,\vec\imath+a_y\,\vec\jmath+a_z\,\vec k is

\nabla f(x_0,y_0,z_0)\cdot\dfrac{\vec a}{\|\vec a\|}

We have

f(x,y,z)=3e^x\cos(yz)\implies\nabla f(x,y,z)=3e^x\cos(yz)\,\vec\imath-3ze^x\sin(yz)\,\vec\jmath-3ye^x\sin(yz)\,\vec k

and

\vec a=-\vec\imath+2\,\vec\jmath+3\,\vec k\implies\|\vec a\|=\sqrt{(-1)^2+2^2+3^2}=\sqrt{14}

Then the derivative at p_0 in the direction of \vec a is

3\,\vec\imath\cdot\dfrac{-\vec\imath+2\,\vec\jmath+3\,\vec k}{\sqrt{14}}=-\dfrac3{\sqrt{14}}

3 0
3 years ago
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