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3 years ago

I need this to be answered PLEASE!!!

Mathematics

1 answer:

STatiana [176]3 years ago

4 0

Since 45 minutes is 75% of an hour, you need to convert into a decimal and add 2. Which gives you 2.75. Then you have to divide 8,580 by 2.75 and you get a final answer of 3120 words. Hope this helps!

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If two angles are supplementary , they must

-BARSIC- [3]

If the two angles are supplementary, they must equal 180 degrees

7 0

4 years ago

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A dump truck brought 1⁄3 of a ton of rock on the first trip, 5⁄6 of a ton on the second trip, and 5⁄12 of a ton on the third tri

ludmilkaskok [199]

First convert the numbers to get the same denominator
1/3=4/12
5/6=10/12
5/12
Then add them all together
4/12+10/12+5/12= 19/12
Then reduce
1 7/12
Your answer is 1 ton and 7/12 of a ton

8 0

3 years ago

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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

3 years ago

Tereka and Emma each deposit money in savings accounts. The amount of money, in dollars, in Tereka’s account after x years is mo

Over [174]

Given:

Amount of money in Tereka’s account : $T(x)=472(1.04)^x$

Amount of money in Emma’s account : $E(x)=485(1.09)^x$

To find:

The difference between initial amounts Tereka and Emma deposit.

Solution:

Substitute x=0 in the given functions, to get the initial amount of money they deposit.

$T(0)=472(1.04)^0$

$T(0)=472(1)$

$T(0)=472$

So, the initial amount Tereka deposit is $472.

$E(0)=485(1.09)^0$

$E(0)=485(1)$

$E(0)=485$

So, the initial amount Emma deposit is $485.

Difference between the initial amounts Tereka and Emma deposit is

$485-472=13$

Therefore, the difference between initial amounts Tereka and Emma deposit is $13. Emma deposit $13 more than Tereka.

6 0

3 years ago

If there are 48 oranges then how many apples are there?

UkoKoshka [18]

If this is a actual question then you are missing parts of it! Please leave a comment with the full question and I would be glad to help you out! If looking at what is given then 0 apples as all that is stated is only 48 oranges and no apples.

5 0

3 years ago

Read 2 more answers