How do you pass algebra class

Mathematics

1 answer:

Kamila [148]3 years ago

4 0

Take notes and pay attention since just slacking off for only 10 minutes can get you behind and struggling.

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A 15 ounce alloy containing 30% gold was mixed with an alloy containing 5% gold to get an alloy containing 20% gold. How many ou

Studentka2010 [4]

Answer:

4.5 % alloy

Step-by-step explanation:

Let x = amt of 30% alloy

The resulting total is to be 25 oz, therefore:

(25-x) = amt of 5% alloy:

A typical mixture equation:

.30x + .05(25-x) = .20(25)

.30x + 1.25 - .05x = 5 .30x - .05x = 5 - 1.25

.25x = 3.75

x = 3.75%2F.25

x = 15 oz of 30% alloy required

then

25-15 = 10 oz of 5% alloy:

Check solution

.30(15) + .05(10) = .20(25)

4.5 + .5 = 5

4 0

3 years ago

The given line segment has a midpoint at (3, 1).

Katena32 [7]

Answer:

$y=\frac{1}{3}x$

Step-by-step explanation:

The given line segment has a midpoint at (3, 1) and goes through (2, 4), (3, 1), and (4, -2). We can use any two of the three points to calculate the equation of the line. Let us use the points (2, 4) and (4, -2)

Therefore the line goes through (2, 4) and (4, -2). The equation of a line passing through $(x_1,y_1)\ and\ (x_2,y_2)$ is:

$\frac{y-y_1}{x-x_1}=\frac{y_2-y_1}{x_2-x_1}$ .

Therefore the line passing through (2, 4) and (4, -2) has an equation:

$\frac{y-y_1}{x-x_1}=\frac{y_2-y_1}{x_2-x_1}\\\frac{y-4}{x-2}=\frac{-2-4}{4-2}\\\frac{y-4}{x-2}=\frac{-6}{2}\\y-4=x-2(-3)\\y-4=-3x+6\\y=-3x+10$

Comparing with the general equation of line: y = mx + c, the slope (m) = -3 and the intercept on the y axis (c) = 10

Two lines are said to be perpendicular if the product of their slope is -1. If the slope of line one is m1 and the slope of line 2 = m2, then the two lines are perpendicular if:

$m_1m_2=-1$ .

Therefore The slope (m2) of the perpendicular bisector of y = -3x + 10 is:

$m_1m_2=-1\\-3m_2=-1\\m_2=\frac{1}{3}$

Since it is the perpendicular bisector of the given line segment, it passes through the midpoint (3, 1). The equation of the perpendicular bisector is:

$\frac{y-y_1}{x-x_1}=m\\\frac{y-1}{x-3}=\frac{1}{3}\\ y-1= \frac{1}{3}(x-3)\\ y-1=\frac{1}{3}x-1\\y=\frac{1}{3}x$