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3 years ago

Examples of a scientific law at least one

Mathematics

2 answers:

vladimir1956 [14]3 years ago

8 0

Newton's law!! ^^

Hope that helped

Tatiana [17]3 years ago

7 0

mathematical descriptions

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△ABC ≅ △XYZ, which side corresponds to side CA?

Rus_ich [418]

Zx because letters always have to line up woth each other c is place z and a is place x (i loved this unit )

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3 years ago

How many 5/16 foot pieces of wood can you cut from a board that is 1 19/16 feet long?

lapo4ka [179]

Answer:

5 pieces

Step-by-step explanation:

To find the number of pieces divide 1 7/8 by 3/8

Change 1 7/8 to an improper fraction

substitute this value into the equation

to solve this complex fraction remove the bottom fraction by multiplying both the top and bottom fractions by the inverse of the bottom fraction.

Multiplying by the inverse results in a product of 1 so the result is

dividing by 1 gives just the top fraction.

Dividing out common factors leaves

8 0

3 years ago

Write the improper fractions for these mixed numbers. 6 1/3 8 1/2 4 3/5

Juliette [100K]

6 1/3 is 19/3
8 1/2 is 17/2
4 3/5 is 23/5
Pls mark brainiest

3 0

3 years ago

Read 2 more answers

If a scalene triangle has its measures 4 m, 11 m and 8 m, find the largest angle.

soldier1979 [14.2K]

Answer:

129.8 approximately

Step-by-step explanation:

So this sounds like a problem for the Law of Cosines. The largest angle is always opposite the largest side in a triangle.

So 11 is the largest side so the angle opposite to it is what we are trying to find. Let's call that angle, X.

My math is case sensitive.

X is the angle opposite to the side x.

Law of cosines formula is:

$x^2=a^2+b^2-2ab \cos(X)$

So we are looking for X.

We know x=11, a=4, and b=8 (it didn't matter if you called b=4 and a=8).

$11^2=4^2+8^2-2(4)(8)\cos(X)$

$121=16+64-64\cos(X)$

$121=80-64\cos(X)$

Subtract 80 on both sides:

$121-80=-64\cos(X)$

$41=-64\cos(X)$

Divide both sides by -64:

$\frac{41}{-64}=\cos(X)$

Now do the inverse of cosine of both sides or just arccos( )

[these are same thing]

$\arccos(\frac{-41}{64})=X$

Time for the calculator:

X=129.8 approximately

3 0

3 years ago

Can anyone figure this out?

Verizon [17]

$\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ N(\stackrel{x_1}{-3}~,~\stackrel{y_1}{10})\qquad A(\stackrel{x_2}{6}~,~\stackrel{y_2}{3})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ NA=\sqrt{(6+3)^2+(3-10)^2}\implies NA=\sqrt{130} \\\\[-0.35em] ~\dotfill\\\\ A(\stackrel{x_2}{6}~,~\stackrel{y_2}{3})\qquad D(\stackrel{x_1}{6}~,~\stackrel{y_1}{-1}) \\\\\\ AD=\sqrt{(6-6)^2+(-1-3)^2}\implies AD=4 \\\\[-0.35em] ~\dotfill$

$\bf D(\stackrel{x_1}{6}~,~\stackrel{y_1}{-1})\qquad N(\stackrel{x_1}{-3}~,~\stackrel{y_1}{10}) \\\\\\ DN=\sqrt{(-3-6)^2+(10+1)^2}\implies DN=\sqrt{202}$

now that we know how long each one is, let's plug those in Heron's Area formula.

$\bf \qquad \textit{Heron's area formula} \\\\ A=\sqrt{s(s-a)(s-b)(s-c)}\qquad \begin{cases} s=\frac{a+b+c}{2}\\[-0.5em] \hrulefill\\ a=\sqrt{130}\\ b=4\\ c=\sqrt{202}\\[1em] s=\frac{\sqrt{130}+4+\sqrt{202}}{2}\\[1em] s\approx 14.81 \end{cases} \\\\\\ A=\sqrt{14.81(14.81-\sqrt{130})(14.81-4)(14.81-\sqrt{202})} \\\\\\ A=\sqrt{324}\implies A=18$

5 0

3 years ago