How to rearrange an equation to find the missing variab;e?

1 answer:

5 0

You figure out the problem with the numbers you have then find the variable it’s kinda like a puzzle you have to break down the process

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8 0

3 years ago

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Which is another way to check the sum of 104+34+228+877?

Alex787 [66]

The answer is digit sum method.

Digit sum method is method used to check the sum of sum numbers. If the sum of all of the digits of numbers is equal to the sum of all of the digits of the total sum, then the arithmetic process was correct.

We need to check the sum of 104+34+228+877:
104 + 34 + 228 + 877 = 1243

Let's first summarize the digits of individual numbers:
104 *** 1 + 0 + 4 = 5
34 *** 3 + 4 = 7
228 *** 2 + 2 + 8 = 12 *** 1 + 2 = 3
877 *** 8 + 7 + 7 = 22 *** 2 + 2 = 4

Now, let's sum these sums:
5 + 7 + 3 + 4 = 19 *** 1 + 9 = 10 *** 1 + 0 = 1

Then, let's summarize the digits of the total sum:
1243 *** 1 + 2 + 4 + 3 = 10 *** 1 + 0 = 1

Since the sums of the digits on the both sides of equation is 1, than the arithmetic process was correct and the sum of 104 + 34 + 228 + 877 is really 1243.

7 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.

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

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