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Kryger [21]
3 years ago
10

The world's smallest plane has a length of 12 feet and the world's longest plane has a length of 302 feet. Write an absolute val

ue equation that represents the minimum and maximum lengths. Use x to represent the lengths.
Mathematics
1 answer:
inessss [21]3 years ago
3 0

Answer:

| 157 +/- 145 |

Step-by-step explanation:

302-12 = 290 total range

290/2 = 145 absolute value range

| 157 +/- 145 |

The two answers to that are 12 and 301

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Y=Mx+b and to find the slope
Grace [21]
M is the slope of the equation.

Hope this helps!
5 0
3 years ago
Prove that: (b²-c²/a)CosA+(c²-a²/b)CosB+(a²-b²/c)CosC = 0​
IRISSAK [1]

<u>Prove that:</u>

\:\:\sf\:\:\left(\dfrac{b^2-c^2}{a}\right)\cos A+\left(\dfrac{c^2-a^2}{b}\right)\cos B +\left(\dfrac{a^2-b^2}{c}\right)\cos C=0

<u>Proof: </u>

We know that, by Law of Cosines,

  • \sf \cos A=\dfrac{b^2+c^2-a^2}{2bc}
  • \sf \cos B=\dfrac{c^2+a^2-b^2}{2ca}
  • \sf \cos C=\dfrac{a^2+b^2-c^2}{2ab}

<u>Taking</u><u> </u><u>LHS</u>

\left(\dfrac{b^2-c^2}{a}\right)\cos A+\left(\dfrac{c^2-a^2}{b}\right)\cos B +\left(\dfrac{a^2-b^2}{c}\right)\cos C

<em>Substituting</em> the value of <em>cos A, cos B and cos C,</em>

\longmapsto\left(\dfrac{b^2-c^2}{a}\right)\left(\dfrac{b^2+c^2-a^2}{2bc}\right)+\left(\dfrac{c^2-a^2}{b}\right)\left(\dfrac{c^2+a^2-b^2}{2ca}\right)+\left(\dfrac{a^2-b^2}{c}\right)\left(\dfrac{a^2+b^2-c^2}{2ab}\right)

\longmapsto\left(\dfrac{(b^2-c^2)(b^2+c^2-a^2)}{2abc}\right)+\left(\dfrac{(c^2-a^2)(c^2+a^2-b^2)}{2abc}\right)+\left(\dfrac{(a^2-b^2)(a^2+b^2-c^2)}{2abc}\right)

\longmapsto\left(\dfrac{(b^2-c^2)(b^2+c^2)-(b^2-c^2)(a^2)}{2abc}\right)+\left(\dfrac{(c^2-a^2)(c^2+a^2)-(c^2-a^2)(b^2)}{2abc}\right)+\left(\dfrac{(a^2-b^2)(a^2+b^2)-(a^2-b^2)(c^2)}{2abc}\right)

\longmapsto\left(\dfrac{(b^4-c^4)-(a^2b^2-a^2c^2)}{2abc}\right)+\left(\dfrac{(c^4-a^4)-(b^2c^2-a^2b^2)}{2abc}\right)+\left(\dfrac{(a^4-b^4)-(a^2c^2-b^2c^2)}{2abc}\right)

\longmapsto\dfrac{b^4-c^4-a^2b^2+a^2c^2}{2abc}+\dfrac{c^4-a^4-b^2c^2+a^2b^2}{2abc}+\dfrac{a^4-b^4-a^2c^2+b^2c^2}{2abc}

<em>On combining the fractions,</em>

\longmapsto\dfrac{(b^4-c^4-a^2b^2+a^2c^2)+(c^4-a^4-b^2c^2+a^2b^2)+(a^4-b^4-a^2c^2+b^2c^2)}{2abc}

\longmapsto\dfrac{b^4-c^4-a^2b^2+a^2c^2+c^4-a^4-b^2c^2+a^2b^2+a^4-b^4-a^2c^2+b^2c^2}{2abc}

<em>Regrouping the terms,</em>

\longmapsto\dfrac{(a^4-a^4)+(b^4-b^4)+(c^4-c^4)+(a^2b^2-a^2b^2)+(b^2c^2-b^2c^2)+(a^2c^2-a^2c^2)}{2abc}

\longmapsto\dfrac{(0)+(0)+(0)+(0)+(0)+(0)}{2abc}

\longmapsto\dfrac{0}{2abc}

\longmapsto\bf 0=RHS

LHS = RHS proved.

7 0
2 years ago
For a chemistry lab, you used 2 ounces of calcium sulfate, 2.5 pounds of copper, and 14 lb of nickel. What was the combined weig
Flauer [41]
That would be 16.625 pounds and/or 266 pounds
7 0
3 years ago
Help please...Thanks
Maru [420]
There are 14 boys, therefore 16 girls.
Ratio of boys:girls is 14:16
14/16 = 7/8 so answer is the 3rd option.
4 0
3 years ago
Can anybody help me with questions 6-9?
Marina CMI [18]

6.  z=2a-p

7.  w= A/L

8.  w= v/Lh

9.  t= (A-P)/PR

3 0
3 years ago
Read 2 more answers
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