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# Victoria's Responses in WyzAnt Answers

#### Solving 5x^2+13x-6?

How do you solve the equation 5x2+13x-6 by factoring completely?

Thank you!

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Asked by Courtnee from Mitchell, IN
10

Courtnee,

First we need to determine which factors of 5 and 6 subtract to 13:

Factors of 5: 5 and 1

Factors of 6: 3 and 2, 6 and 1

Since we are trying to get to 13, I am focusing on the first pair that can get me greater than 13:

5*3=15, which leaves me with 1*2=2

15-2=13!

Now I need to set this into parenthetical units:

(5x     )(x     ) I start by placing my x terms

(5x-2)(x+3) Since I need a negative 2x, I place the negative two where it multiplies by the isolated x, and this allows me to place my last factor in the remaining spot.

If you want to check your work, you can always foil out the answer, to make sure that it is the original equation.

#### 2[4-2(3-x)]-1=4[2(4x-3)+7]-25

2[4-2(3-x)]-1=4[2(4x-3)+7]-25

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Asked by Mildred from Owensboro, KY
10

This is a question mostly of order of operations. I solve from the center out:

2[4-2(3-x)]-1=4[2(4x-3)+7]-25

2[4-6+2x]-1=4[8x-6+7]-25

2(2x-2)-1=4(8x+1)-25

4x-4-1=24x+4-25

4x-5=24x-21 (add 21 to both sides)

4x+16=24x (subtract 4x from both sides)

16=20x (divide both sides by 20)

16/20=x (simplify)

3/4=x

#### Phyllis invested 57000 dollars,

Phyllis invested 57000 dollars, a portion earning a simple interest rate of 5 percent per year and the rest earning a rate of 7 percent per year. After one year the total interest earned on these investments was 3290 dollars. How much money did she invest at each rate?

At rate 5 percent :
At rate 7 percent :

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Asked by Treena from Fox Island, WA
00

Treena,

For this problem, we need to use the basic interest formula, I=Prt (Interest equals principal multiplied by rate multiplied by time). We are told that the investment is for 1 year (t) and that the interest is \$3,290 (I). The concern is that there are two rates and we do not know how much of our principal is applied to each rate. We must set our variables:

For our first rate of 5%, we will use a variable of x for our principal

For our second rate of 7%, our variable becomes 57,000-x (since the two variables must add to 57,000)

Now we can create our equation. We must note each investment separately in the equation;

I=Prt+Prt

3,290=x(.05)1 + (57,000-x).07(1)

3,290=.05x+3,990-.07x

3,290=3,990-.02x

-700=-.02x

35,000=x Remember that x is the investment at 5%.

So now we can solve for our second investment;

7% Investment: 57,000-x

57,000-35000=22,000

#### The measure of the third angle is 40 degrees more than that of the smallest angle.

One angle in a triangle has a measure that is three times as large as the smallest angle. The measure of the third angle is 40 degrees more than that of the smallest angle. Find the measure of the LARGEST angle.

Answer: The LARGEST angle has a measure of degrees.

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Asked by Treena from Fox Island, WA
00

Treena,

There are three variables here:

Smallest angle: x

Next angle: 3x

Final angle: x+40

All of the angles of a triangle add to 180 degrees, so we can create an equation:

180=x+3x+x+40

180=5x+40

140=5x

28=x

Now we plug this number into our two other angles, in order to solve for all measurements:

3x

3(28)

84 degrees

x+40

28+40

68 degrees

In order to check our answers, we will add all the measures together, to ensure that they are 180 degrees:

28+84+68=180 degrees!

The largest of the three angles is 84 degrees

#### A man flies a small airplane from Fargo to Bismarck, North Dakota ---

A man flies a small airplane from Fargo to Bismarck, North Dakota --- a distance of 180 miles. Because he is flying into a head wind, the trip takes him 2 hours. On the way back, the wind is still blowing at the same speed, so the return trip takes only 1 hour and 12 minutes. What is the plane's speed in still air, and how fast is the wind blowing?

his plane speed equals mph.
the wind speed equals mph.

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Asked by Treena from Fox Island, WA
00

Treena,

This is another problem related to distance, so we will utilize the distance formula (d=rt). The issue with this problem is that there are two variables in relation to the rate. The first variable is the plane speed (r) and the second variable is the wind speed (w). We will first set up our equations for the two flights, and then we will address how to solve for both variables.

Flight into the wind: In this flight, the plane speed is working against the wind, so our rate is set up as r-w, since the wind is decreasing our plane speed. This flight took 2 hours (t) and traveled 180 miles (d), so the equation is 180=(r-w)2

Second flight: In this flight, the plane speed is working with the wind, so our rate is set up as r+w, since the wind is increasing our plane speed. This flight took 1.2 hours (t) and traveled 180 miles (d), so the equation is 180=(r+w)1.2

Now, you have two choices, you can use elimination or substitution, in order to address the multiple variables. I will use substitution. Since both flights traveled the same distance, we can set the rt portions of our equations equal to each other and solve for one variable:

rt=rt

(r-w)2=(r+w)1.2

2r-2w=1.2r+1.2w

0.8r=3.2w

r=4w

Now, we can plug in 4w for r, in either equation:

180=(r+w)1.2

180=(4w+w)1.2

180=5w(1.2)

150=5w

30=w

Since we know that r=4w, we can plug in w, to solve for r:

r=4w

r=4(30)

r=120

Now we know our plane speed and wind speed. If we want, we can plug both into our second equation, in order to check the values:

180=(r-w)2

180=(120-30)2

180=90(2)

180=180 This is a true statement, so our values are valid!

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