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3 years ago

A store sold 6 fruit trees five of the trees were apple trees what fraction of the trees were Apple trees?

Mathematics

2 answers:

Dovator [93]3 years ago

6 0

5/6 of the trees were apple trees. (5 over 6)

never [62]3 years ago

5 0

The fraction would be 5/6 because there are 6 apple trees in all and 5 of those fruit tress are apple

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What is true about the figure? In the figure, angles C and D are a Therefore angle C must measure

Nuetrik [128]

Answer

In the figure, angles C and D are a

✔ linear pair

Therefore, angle C must measure

✔ 56°

Step-by-step explanation:

3 0

4 years ago

two cars travel at same speed to different destinations. car A reaches its destination in 24 minutes. car B reaches its destinat

Harlamova29_29 [7]

Answer:

The speeds of the cars is: 0.625 miles/minute

Step-by-step explanation:

We use systems of equations in two variables to solve this problem.

Recall that the definition of speed (v) is the quotient of the distance traveled divided the time it took : $v=\frac{distance}{time}$ . Notice as well that the speed of both cars is the same, but their times are different because they covered different distances. So if we find the distances they covered, we can easily find what their speed was.

Writing the velocity equation for car A (which reached its destination in 24 minutes) is:

$v\,*\,24\,min=d_A$

Now we write a similar equation for car B which travels 5 miles further than car A and does it in 32 minutes:

$v\,*\,32\,min=d_A+5\,miles$

Now we solve for $d_A$ in this last equation and make the substitution in the equation for car A:

$v\,*\,32\,min=d_A+5\,miles\\d_A=v\,*\,32\,min-5\,miles\\\\v\,*\,24\,min=v\,*\,32\,min-5\,miles\\v\,(24\,min-32\,min)=-5\,miles\\v\,(-8\,min)=-5\, miles\\v=\frac{-5}{-8} \frac{miles}{min} \\v=0.625\,\frac{miles}{min}$

So this is the speed of both cars: 0.625 miles/minute

3 0

4 years ago

Please help me with this very hard question. Determine if each pair of expressions is equivalent. Mark all equivalent expression

Shalnov [3]

It’s the 3rd one hopefully that helps you

7 0

2 years 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,

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

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

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

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