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Leno4ka [110]
3 years ago
11

Look at this expression.

Mathematics
1 answer:
g100num [7]3 years ago
4 0
Multiply x<span> and </span>3

<span>Multiply x and 1</span>

<span>The x just gets copied along.</span>

<span>The answer is x</span>

x

<span>3*x evaluates to 3x</span>

Because of the minus sign

<span>3x becomes - 3x</span>

<span>The answer is -3x</span>

<span>Multiply y and 2</span>

<span>Multiply y and 1</span>

<span>The y just gets copied along.</span>

<span>The answer is y</span>

y

<span>2*y evaluates to 2y</span>

<span>-3*x-2*y evaluates to -3x-2y</span>

<span>The answer is -3x-2y-2</span>

<span>-3*x-2*y-2 evaluates to <span>-3x-2y-2</span></span>

<span><span>so the first one is right</span></span>

<span><span>
</span></span>

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Which equation represents “nine less than k is three”?
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k - 9 = 3

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5 0
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A box of Munchkins contains chocolate and glazed donut holes. If Jacob ate 2 chocolate
neonofarm [45]

Answer:

24 munchkins.

Step-by-step explanation:  

Let C be the number of chocolate and D be number of glazed donut holes in the original box.

We are told if Jacob ate 2 chocolate  munchkins, then 1/11 of the remaining Munchkins would be chocolate. We can represent this information as:

C-2=\frac{1}{11}*(C+D-2)...(1)

We are also told if he instead added 4  glazed Munchkins to the original box, 1/7 of the Munchkins would be chocolate. We can represent this information as:

C=\frac{1}{7}*(C+D+4)...(2)

Upon substituting C's value from equation (2) in equation (1) we will get,

\frac{1}{7}*(C+D+4)-2=\frac{1}{11}*(C+D-2)

Let us have a common denominator on right side of equation.

\frac{1}{7}*(C+D+4)-\frac{7*2}{7}=\frac{1}{11}*(C+D-2)

\frac{C+D+4-14}{7}=\frac{1}{11}*(C+D-2)

Multiplying both sides of our equation by 7, we will get,

7*\frac{C+D-10}{7}=7*\frac{1}{11}*(C+D-2)

C+D-10=\frac{7}{11}*(C+D-2)  

Multiplying both sides of our equation by 11, we will get,

11*(C+D-10)=11*\frac{7}{11}*(C+D-2)  

11*(C+D-10)=7*(C+D-2)

11C+11D-110=7C+7D-14

11C-7C+11D-7D=-14+110  

4C+4D=96

4(C+D)=96  

(C+D)=\frac{96}{4}

(C+D)=24

Therefore, the total number of Munchkins in original box is 24.

3 0
3 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.

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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

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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}

Solve for θ

Use the quadratic formula.

tanθ = 61.249 or -0.025

θ = arctan(61.249) = 89.1° or

θ = arctan(-0.025) = -1.4°

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Answer:

it is 12

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