All categories

3 years ago

I'm giving brainliest to best answer

Mathematics

1 answer:

Natali5045456 [20]3 years ago

3 0

Answer:

2) 62.5 wpm

3) 12.7 miles per gallon

4)18 commercials per hour

5) 22.9 ounces per dollar

Step-by-step explanation:

2) 500 words / 8 minutes = 62.5 words per 1 minute

3) 216 miles / 17 gallons = 12.7 miles per gallon

4) 36 commercials / 2 hours = 18 commercials per hour

5) 64 oz / $2.79 = 22.9 oz per $1

You might be interested in

Simplify the expression below. (-5 × -4)÷(-3+5)

ohaa [14]

Answer:

-10

Step-by-step explanation:

7 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

a plane averaged 400 mph on a trip going east, but only 280 mph on the return trip. The total flying time in both directions was

Liula [17]

Answer:

1400 miles

Step-by-step explanation:

Speed is the time rate of change of distance. It is the ratio of distance to time and is given by the equation:

$Speed(S) = \frac{distance(d)}{time(t)}\\ S=\frac{d}{t}$

plane averaged 400 mph on a trip going east, but only 280 mph on the return trip.

The time spent (t₁) in going east is given by:

$S=\frac{d}{t_1}\\ t_1=\frac{d}{S} =\frac{d}{400mph}$

The time spent (t₂) in going east is given by:

$S=\frac{d}{t_2}\\ t_2=\frac{d}{S} =\frac{d}{280mph}$

The total time (t) = t₁ + t₂

t = t₁ + t₂ = 8.5 hours

$\frac{d}{400}+\frac{d}{280} =8.5\\280d+400d=952000\\680d=952000\\d=1400miles\\$

Therefore the one way distance is 1400 miles

7 0

3 years ago

How do u write 1.2 as a fraction

Ivahew [28]

Split the number into its whole number component and decimal component.

for the decimal component, 0.2 is the same as 2/10

Find the GCF of both 2 and 10 and that would be 2

divide both the numerator and denominator by the GCF

simplify and you get 1/5

Lastly combine the whole number component with the fraction and your answer would be 1 1/5.

6 0

3 years ago

Wich term best describes a figure formed by three segments connecting three noncollinear points ?

Nat2105 [25]

A triangle is the answer.

4 0

3 years ago

Read 2 more answers