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Ksenya-84 [330]
3 years ago
10

Identify the graphed linear equation.

Mathematics
2 answers:
erastova [34]3 years ago
6 0
D
It is because the slope is negative and the y intercept is -2
alisha [4.7K]3 years ago
6 0

The y-intercept is -2, which gives you the b of the slope-intercept form (y=mx + b). This would rule out choices A and C. It rises 5 and goes to the left 1 in the negative direction. This gives you the m part (rise/run). This would give you the answer of:

D) y = -5x - 2


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In parallelogram ABCD, AD = 4x - 10 and BC = 3x + 2. Find AD.
Helga [31]

Answer:12


Step-by-step explanation:

4x - 10= 3x+2 add 10= 4x = 3x+12 subtract 3x=1x 1x=12 divide 12 by 1=12 should be 12 sorry if I got it wrong...

8 0
3 years ago
The volume of a right cylinder is 27π cubic centimeters, and the high is 3 centimeters (cm). What is the radius of the cylinder?
Dmitrij [34]

Answer:

B

Step-by-step explanation:

The volume is the height x area of base, 27pi / 3 = 9pi

Area of base is radius square x pi, 9 pi = 3^2 x pi

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2 years ago
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Solve for x using basic trigonometric ratios.
yuradex [85]
Sine 35 degrees = x / 20

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x = .57358 * 20

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Please help me with this answer!
Delvig [45]

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Step-by-step explanation:

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Let u = (1, k) and v = (2, 1). Find k such that The distance between u and v is 1 u and v are orthogonal The angle between u and
SVETLANKA909090 [29]

Answer:k=1,k=-2,k=8\pm 5\sqrt{3}

Step-by-step explanation:

Given two vectors

u=1\hat{i}+k\hat{j}

v=2\hat{i}+1\hat{j}

\left ( i\right )Distance between them is given by

|u-v|=\sqrt{\left ( 2-1\right )^2+\left ( 1-k\right )^2}=1

squaring both side

1^{2}+\left ( 1-k\right )^2=1

k^2-2k+1=0

\left ( k-1\right )^2=0

k=1

\left ( ii\right )

angle between u and v is 90 i.e. orthogonal

u\dot v=0

\left ( 1\hat{i}+k\hat{j}\right )\dot \left ( 2\hat{i}+1\hat{j}\right )=0

2+k=0

k=-2

\left ( iii\right )

angle between u & v is \frac{\pi }{3}

u\dot v=|u||v|cos\left (\frac{\pi }{3}\right )

|u|=\sqrt{1^2+k^2}

|v|=\sqrt{2^2+1^2}

2+k=\left ( \sqrt{1+k^2}\right )\left ( \sqrt{5}\right )cos\left ( \frac{\pi }{3}\right )

\left ( 4+2\right )^2=\left ( 1+k^2\right )5

k^2-16k-11=0

k=8\pm 5\sqrt{3}

7 0
3 years ago
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