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skelet666 [1.2K]
3 years ago
13

Kirk's average driving speed is 7 kilometers per hour faster than Thor's. In the same length of time it takes Kirk to drive 450

kilometers, Thor drives only 408 kilometers. What is Kirk's average speed?
Mathematics
1 answer:
alexandr402 [8]3 years ago
3 0

Answer:

x = 816 kilometers per hour.

Step-by-step explanation:

Given the following data;

Distance traveled by Kirk = 450km

Distance traveled by Thor = 408 km

Let the average driving speed of Thor be x.

Translating the word problem into an algebraic equation;

k = x + 6

We know that the time taken to cover a distance by an object with speed is given by the formula;

Time = distance/speed

Substituting into the equation, we have;

408/x = 405/x + 6

Cross-multiplying, we have;

408*(x + 6) = 405*x

408x + 2448 = 405x

408x - 405x = 2448

3x = 2448

x = 2448/3

x = 816 kilometers per hour.

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Donata bought 4 lemons and 4 plums at the local supermarket for a total of $ 21.60. Dan bought 11 lemons and 6 plums at the same
VashaNatasha [74]
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6 0
3 years ago
Can you please help me
max2010maxim [7]
Here’s the answer and each step I took to solve it. With the example I made on the side just continue to repeat that for each one to get the answers. You just replace that one number for the X to take the place for the next X value as you go down the table solving them.

7 0
3 years ago
Vanessa draws one side of equilateral ΔABC on the coordinate plane at points A(–2, 1) and B(4,1). What are the possible coordina
grandymaker [24]

Important question. No menu of choices given so we'll just have to figure it out.

It's not possible in two dimensions for all three vertices of an equilateral triangle to be lattice points, i.e. points with integer coordinates.  

Since A(-2,1) and B(4,1) have integer coordinates, our other point won't. There will be a √3 involved, the algebraic tell that there's a 30/60/90 triangle lurking somewhere in the problem.

Here the given side is parallel to the x axis which makes things easier.  

|AB| = 4 - -2 = 6

The x coordinate of the third vertex will be the same as that of the midpoint of AB, let's call it M,

M = ( (-2 + 4) / 2, (1 + 1)/2 ) = (1, 1)

We're making some progress.  Whatever our height h is our two candidates for the third vertex C are (1, 1 ± h).

Since |AB|=6, we get |AB|=|BC|=|AC|=6 because it's an equilateral triangle.

The altitude CM bisects the triangle into 2 30/60/90 triangles.  Let's take one of them, AMC.  Angle AMC is a right angle, so we have a right triangle with legs |AM|=3 and |CM|=h, and hypotenuse |AC|=6. The Pythagorean Theorem tells us:

3² + h² = 6²

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4 0
3 years ago
For which system of equations is (5, 3) the solution? A. 3x – 2y = 9 3x + 2y = 14 B. x – y = –2 4x – 3y = 11 C. –2x – y = –13 x
Alla [95]
The <u>correct answer</u> is:

D) \left \{ {{2x-y=7} \atop {2x+7y=31}} \right..

Explanation:

We solve each system to find the correct answer.

<u>For A:</u>
\left \{ {{3x-2y=9} \atop {3x+2y=14}} \right.

Since we have the coefficients of both variables the same, we will use <u>elimination </u>to solve this.  

Since the coefficients of y are -2 and 2, we can add the equations to solve, since -2+2=0 and cancels the y variable:
\left \{ {{3x-2y=9} \atop {+(3x+2y=14)}} \right. &#10;\\&#10;\\6x=23

Next we divide both sides by 6:
6x/6 = 23/6
x = 23/6

This is <u>not the x-coordinate</u> of the answer we are looking for, so <u>A is not correct</u>.

<u>For B</u>:
\left \{ {{x-y=-2} \atop {4x-3y=11}} \right.

For this equation, it will be easier to isolate a variable and use <u>substitution</u>, since the coefficient of both x and y in the first equation is 1:
x-y=-2

Add y to both sides:
x-y+y=-2+y
x=-2+y

We now substitute this in place of x in the second equation:
4x-3y=11
4(-2+y)-3y=11

Using the distributive property, we have:
4(-2)+4(y)-3y=11
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Combining like terms, we have:
-8+y=11

Add 8 to each side:
-8+y+8=11+8
y=19

This is <u>not the y-coordinate</u> of the answer we're looking for, so <u>B is not correct</u>.

<u>For C</u>:
Since the coefficient of x in the second equation is 1, we will use <u>substitution</u> again.

x+2y=-11

To isolate x, subtract 2y from each side:
x+2y-2y=-11-2y
x=-11-2y

Now substitute this in place of x in the first equation:
-2x-y=-13
-2(-11-2y)-y=-13

Using the distributive property, we have:
-2(-11)-2(-2y)-y=-13
22+4y-y=-13

Combining like terms:
22+3y=-13

Subtract 22 from each side:
22+3y-22=-13-22
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Divide both sides by 3:
3y/3 = -35/3
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Since the coefficients of x are the same in each equation, we will use <u>elimination</u>.  We have 2x in each equation; to eliminate this, we will subtract, since 2x-2x=0:

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Divide both sides by -8:
-8y/-8 = -24/-8
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The y-coordinate is correct; next we check the x-coordinate  Substitute the value for y into the first equation:
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Add 3 to each side:
2x-3+3=7+3
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Divide each side by 2:
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This gives us the x- and y-coordinate we need, so <u>D is the correct answer</u>.
7 0
3 years ago
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olasank [31]
X^2/(x- 9 = 81/(x - 9)
This is the equation for which you want the solution.
Multiplying both sides of the equation by (x - 9) we get
x^2(x - 9)/(x - 9) = 81(x - 9)/(x - 9)
So the (x - 9) goes out from both the denominator and the numerator and then the simplified equation becomes
x^2 = 81
x ^2 = (9)^2
x = 9
So the value of the unknown variable x comes out to be 9.
6 0
3 years ago
Read 2 more answers
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