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2 years ago

rashaads car used 2/3 of a gallon to travel 14 1/2 miles .How many miles can the car go on one gallon of gas

Mathematics

1 answer:

Brut [27]2 years ago

7 0

Answer:

21.75 or 21 $\frac{3}{4}$

Step-by-step explanation:

What I did was 14.5 / 2 to see how many miles were on a 1/3 gallon, then multiplied that answer (which was 7.25) by 3 to get the full 3/3 or 1 gallon.

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Which is a function? look at pic

-Dominant- [34]

Answer:

option 2 {(12, 3), (11,2), ...}

Step-by-step explanation:

For functions, multiple x-values can have the same y-value but each y-value must have a unique x-value. The second option matches this criterion.

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1 year ago

Please simplify 10{N + 3).

lana [24]

Answer:

Step-by-step explanation:

I must assume that you meant 10(N + 3). Let's not mix { with ).

10(N + 3) is as simple as this expression gets. You could, however, re-write it as 10N + 30.

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2 years ago

Lilia is saving money for dance lessons. She already has $10. Each week she is going to save $5. Write an equation, in slope-int

katrin [286]

The equation which represents the amount of money, y, that Lilia will save in x weeks is;

y =$5x + $10.

<h3>Slope-intercept form of linear equations</h3>

The slope-intercept form of linear equations usually takes the form;

y = mx + c

According to the information given;

Lilia already has $10.

Therefore, the y-intercept of the equation is; $10.

And since she saves $5 each week;

The slope of the equation is; $5.

By substitution, the required equation is;

y =$5x + $10.

Read more on slope-intercept form of equations;

brainly.com/question/1884491

4 0

1 year ago

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

Please Help with angles!

777dan777 [17]

Step-by-step explanation:

003:

To solve this, we'll be using the Pythagorean theorem, $a^2+b^2=c^2$

$a^2+8^2=15^2\\a^2+64=225\\a^2=161\\a = \sqrt{161}$

004:

For the next couple questions we'll be using a fun rule called SOHCAHTOA, which stands for Sin (Opposite/Hypotenuse) Cos (Adjacent/Hypotenuse) Tan (Opposite/Adjacent). From this we can see that Tan θ is 8/ $\sqrt{161}$ .

005:

From this rule again, we can tell that sin Ф (Opposite/Hypotenuse) is going to be $\sqrt{161}$ /15.

Answer:

003: $\sqrt{161}$

004: 8/ $\sqrt{161}$

005: $\sqrt{161}$ /15

7 0

2 years ago