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Yuki888 [10]
3 years ago
12

An insecticible contains 85 centigrams of inert ingredient for every 1 centigram of active ingredient. if a quantity of the weig

hs 688 centigrams, how much of each type of ingredient does it contain?
Mathematics
1 answer:
11111nata11111 [884]3 years ago
7 0
(85/86)*688 cg = 680 cg . . . inert ingredient
.. 688 cg -680 cg = 8 cg . . . .active ingredient
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If the length of segment RQ is 8, what is the length of segment PQ?
artcher [175]
Half of 8 is 4 so it’s b
7 0
3 years ago
What is the area of a regular hexagon with a distance from its center to a vertex of 1 cm? (Hint: A regular hexagon can be divid
devlian [24]
<h3>Answer:</h3><h3>Exact area = \frac{3}{2}\sqrt{3} square cm</h3><h3>Approximate area = 2.598 square cm</h3>

=================================================

Work Shown:

s = side length of equilateral triangle = 1 cm

A = area of equilateral triangle with side length 's'

A = \frac{\sqrt{3}}{4}*s^2

A = \frac{\sqrt{3}}{4}*1^2

A = \frac{\sqrt{3}}{4}

This is just one of the 6 equilateral triangles (see diagram below)

Multiply by 6 to get the area of all 6 equilateral triangles, or the entire hexagonal area

6*A = 6*\frac{\sqrt{3}}{4}

6A = \frac{3\sqrt{3}}{2}

6A \approx 2.598

4 0
3 years ago
How do you do this algerbra problem 2/3x-8=9
Zigmanuir [339]
Write the division as a fraction
2/3x - 8 = 9
Multiply both sides of the equation by 3
2x-24=27
Move the constant to the right and change the sign
2x=27+24
Add the numbers
2x=51
Divide both sides by 2
2x/2=51/2
X= 25.5
4 0
3 years ago
When a house agent sells a house, she gets 8% commission on the sales price. if a house is sold for 450 000, what will the commi
Wittaler [7]

Answer:

36,000 dollars

Step-by-step explanation:

5 0
3 years ago
John inherited $25,000 and invested part of it in a money market account, part in municipal bonds, and part in a mutual fund. Af
Lena [83]

Answer:

Step-by-step explanation:

Understand

Let :

x = The amount of money invested in the money market account.  

y = The amount of money invested in municipal bonds.  

z = The amount of money invested in a mutual fund.  

The total investment is represented as:

x + y + z = $25,000

The interest earned can be represented as:

0.06x + 0.07y + 0.08z = $1,620

The amount invested in municipal bonds and mutual funds can be represented as:

y - z = $6,000

Process

The system of three equations can now be solved either by substitution, elimination, or with matrices.  Any process appropriately applied will supply a correct answer.  For this example, let's solve the problem by using matrices.

Solving with Matrices

The process of using matrices is essentially a shortcut for the process of elimination. Each row of the matrix represents an equation, and each column represents coefficients of one of the variables.  

Step 1:        Create a three-row by four-column matrix using coefficients and the constant of each equation.

The vertical lines in the matrix stands for the equal signs between both sides of each equation. The first column contains the coefficients of x; the second column contains the coefficients of y; the third column contains the coefficients of z; the last column contains the constants.  

We want to convert the original matrix to the following matrix.

because you can then read the matrix as x = a, y = b, and z = c.

Step 2:        We work with column 1 first. The number 1 is already in cell 11(Row 1- Col 1).  

Add -0.06 times Row 1 to Row 2 to form a new Row 2.

-0.06 [Row 1] + [Row 2] = [New Row 2]

Step 3:        We will now work with column 1. We want 1 in Cell 22, and we achieve this by multiplying Row 2 by 100.

100[Row 2] = [New Row 2]

Step 4:        Let's now manipulate the matrix so that there are zeros in Cell 12 and Cell 32. We do this by adding Row 2 to Row 1 and Row 3 to Row 2. That gives us a new Row 1 and a new Row 3.

- [Row 2] + [Row 1] = [New Row 1]

- [Row 2] + [Row 3] = [New Row 3]

Step 5:        Let's now manipulate the matrix so that there is a 1 in Cell 33. We do this by multiplying Row 3 by -1.

-1 [Row 3] = [New Row 3]

Step 6:        Let's now manipulate the matrix so that there are zeros in Cell 13 and Cell 23. We do this by adding Row 3 to Row 1 for a new Row 1 and adding -2 times Row 3 to Row 2 for a new Row 3.  

1[Row 3] + [Row 1] = [New Row 1]

-2[Row 3] + [Row 2] = [New Row 2]

Solution

You can now read the answers off the matrix:

x = $15,000 in the money market

y = $8,000 in municipal bonds

z = $2,000 in a mutual fund

6 0
3 years ago
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