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3 years ago

How to graph X = (-1/6) y

Mathematics

1 answer:

jenyasd209 [6]3 years ago

8 0

Answer:

X=(-1/6)y

Step-by-step explanation:

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Write an algebraic equation for this tape diagram. Then determine the value of x. You may use a calculator.

tresset_1 [31]

Answer:

3× + 16 = 22.75

3× = 22.75 - 16

3x = 6.75

3 3

x = 2.25

7 0

2 years ago

System of equations, special case infinitely many solutions?????

Mashcka [7]

One possible system is

1x + 3y = 4

2x + 6y = 8

Note how 2 is twice as large as 1, 6 is twice as large as 3, and 8 is twice as large as 4.

In other words, the second equation is the result of multiplying both sides of the first equation by 2.

1x+3y = 4

2*(1x+3y) = 2*4

2x+6y = 8

Effectively the two equations in bold are the same which produces the same line. The two lines overlap perfectly to intersect infinitely many times. An intersection is a solution.

4 0

2 years ago

Es 18 de junio de 1815, las tropas napoleónicas se encuentran justo adelante, tu eres el ingeniero en balística y el general Wel

Airida [17]

Answer:

89.1° or -1.4°

Step-by-step explanation:

1. Location:

You are on the Mont-Saint-Jean escarpment, near the Belgian town of Waterloo.

The French troops are about 50 m below you and 1.2 km distant.

2. Finding the firing angle

Data:

R = 1200 m

u = 600 m/s

h = -50 m (the height of the target)

a = 9.8 m/s²

We have two conditions.

Horizontal distance

(1) 1200 = 600t cosθ

Vertical distance

(2) -50 = 600t sinθ - 4.9t²

Divide each side of (1) by 600cosθ.

$(3) \, t =\dfrac{2}{\cos \theta}$

Substitute (3) into (2)

$-50 = 600t \sin \theta - 4.9t^{2} = 600 \left( \dfrac{2}{\cos \theta} \right ) \sin \theta - 4.9 \left( \dfrac{2}{\cos \theta} \right )^{2}\\\\(4) \, -50 = 1200 \tan \theta - \dfrac{19.6}{\cos^{2} \theta}$

Recall that

(5) sec²θ = 1/cos²θ = tan²θ + 1

Substitute (5) into (4)

$-50 = 1200 \tan \theta - 19.6 \left(\tan^{2} \theta}+ 1\right )$

Set up a quadratic equation

$\begin{array}{rcl}-50 & = & 1200 \tan \theta - 19.6\tan^{2} \theta -19.6 \\0 & = & 1200 \tan \theta - 19.6\tan^{2} \theta + 30.4\\0 & =&19.6\tan^{2} \theta - 1200 \tan \theta - 30.4\\0 & =&\tan^{2} \theta - 61.224 \tan \theta - 1.551\\\end{array}$