PLSSSSS HELPPP

Grace bought 6.9 ounces of chocolate. 0.7 ounces of it was milk chocolate and the rest was dark chocolate. How much dark chocola

goblinko [34]

6.2 you subtract the 0.7 from 6.9 because 0.7 of it is milk chocolate

7 0

3 years ago

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Melissa has three different positive integers. she adds their reciprocals together and gets a sum of 1. what is the product of h

madreJ [45]

First let us assign the three positive integers to be x, y, and z.

From the given problem statement, we know that:

(1/x) + (1/y) + (1/z) = 1

Without loss of generality we can assume x < y < z.

We know that:

1 = (1/3) + (1/3) + (1/3)

Where x = y = z = 3 would be a solution

However this could not be true because x, y, and z must all be different integers. And x, y, and z cannot all be 3 or bigger than 3 because the sum would then be less than 1. So let us say that x is a denominator that is less than 3. So x = 2, and we have:

(1/2) + (1/y) + (1/z) = 1

Therefore

(1/y) + (1/z) = 1/2

We also know that:

(1/4) + (1/4) = (1/2)

and y = z = 4 would be a solution, however this is also not true because y and z must also be different. And y and z cannot be larger than 4, so y=3, therefore

(1/2) + (1/3) + (1/z) = 1

Now we are left by 1 variable so we calculate for z. Multiply both sides by 6z:

3z + 2z + 6 = 6z

z = 6

Therefore:

(1/2) + (1/3) + (1/6) = 1

so {x,y,z}={2,3,6}

8 0

3 years ago

The line 3x-8y=k cuts the x-axis and y- axis at the point A and B respectively. If the area of ∆AOB =12sq.units, find the value

Sladkaya [172]

Given:

The equation is:

$3x-8y=k$

It cuts the x-axis and y- axis at the point A and B respectively.

The area of ∆AOB =12 sq.units.

To find:

The value of k.

Solution:

We have,

$3x-8y=k$

Substituting $x=0$ to find the y-intercept.

$3(0)-8y=k$

$0-8y=k$

$y=\dfrac{k}{-8}$

$y=-\dfrac{k}{8}$

Substituting $y=0$ to find the x-intercept.

$3x-8(0)=k$

$3x-0=k$

$x=\dfrac{k}{3}$

Area of a triangle is:

$A=\dfrac{1}{2}\times base\times height$

The height of the ∆AOB is $OB=\dfrac{k}{8}$ because distance cannot be negative and the base of the ∆AOB is $OA=\dfrac{k}{3}$ . So, the area of the ∆AOB is: