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3 years ago

Speed of a 747 airplane is 9,944 inches per second. What is the nearest foot per second

Mathematics

1 answer:

laiz [17]3 years ago

3 0

Answer:

12 inches = 1 foot

9,944 inches = 9,944/12 = 826.67 ft

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Rewrite the following in the form log (c). 4 log(5)

Murljashka [212]

Answer:

Step-by-step explanation:4log(5) = log(5^4) = log(625).

This problem involves using one of the properties of logs, where a coefficient (in this case the "4") for a logarithm equals the "inside of a logarithm" raised to power of whatever number the coefficient is.

The property in mathematical terms is: Alog(B) = log(B^A).

So, 4log(5)= log(5^4) = log(625)

5 0

3 years ago

What’s the geometric formula

Anni [7]

I think the answer would be be C.

8 0

3 years ago

Read 2 more answers

Prove that: (b²-c²/a)CosA+(c²-a²/b)CosB+(a²-b²/c)CosC = 0

IRISSAK [1]

Prove that:

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

Proof: 

We know that, by Law of Cosines,

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

Taking LHS

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

Substituting the value of cos A, cos B and cos C,

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

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

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

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

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

On combining the fractions,

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

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

Regrouping the terms,

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

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

$\longmapsto\dfrac{0}{2abc}$

$\longmapsto\bf 0=RHS$

LHS = RHS proved.

7 0

2 years ago

The width of a rectangle is 5 cm less than the length. The perimeter is 38 cm. Find the dimensions of the rectangle

VikaD [51]

Answer:

Step-by-step explanation:

38=2(5)+2(x ) find X

2x5-10

38-10=28

28/2=14

14/2=7

x =7

so you length is 7

8 0

3 years ago

Find an equation of a line that has a slope 7 and passes through the point 3,11

choli [55]

You’re given that m=7 (slope is 7), so setup the slope-intercept form with that info: y = 7x + b.

Now you need to find b. Plug in (3,11) for x and y, x=3 and y=11.

11 = 7*3 + b

Notice the only variable left is b, so you’ll find b this way.

11 = 21 + b and then subtract 21 from both sides:

11 - 21 = b

-10 = b

Plug that back into the equation and you have y = 7x -10

5 0

3 years ago