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Maru [420]
3 years ago
9

I need help with my math

Mathematics
1 answer:
Nat2105 [25]3 years ago
3 0

Answer:

Part A. C=9

Part B. (w+3)² =139

Part C. w = 8.8 inch

Step-by-step explanation:

Given from the question length of the the picture = (2w+12) inches

Width of the picture = w inches

Area of the picture = 260 inch²

Part A. Area of the picture with the given dimensions= w×(2w+12)

Or w(2w+12) = 260

2w²+12w = 260

2(w²+6w) = 2×(130)

w²+6w = 130

Or w²+6w +9 = 130+9 ⇒ which is in the form of w²+6w+c = 130+c

Therefore for c = 9 we will get a perfect square trinomial.

Part B. As we have seen the equation in part A.

As required equation will be (w+3)²=139

Part C. Since (w+3)² = 139

Then by taking under root on both the sides of the equation

(w+3) =√139 = 11.8

(w+3)-3=11.8-3

w = 8.8 inch

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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
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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
-8+4y-3y=11

Combining like terms, we have:
-8+y=11

Add 8 to each side:
-8+y+8=11+8
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This is <u>not the y-coordinate</u> of the answer we're looking for, so <u>B is not correct</u>.

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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:
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Using the distributive property, we have:
-2(-11)-2(-2y)-y=-13
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Combining like terms:
22+3y=-13

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Divide both sides by 3:
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This is <u>not the y-coordinate</u> of the answer we're looking for, so <u>C is not correct</u>.  

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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:

\left \{ {{2x-y=7} \atop {-(2x+7y=31)}} \right. &#10;\\&#10;\\-8y=-24

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:
2x-y=7
2x-3=7

Add 3 to each side:
2x-3+3=7+3
2x=10

Divide each side by 2:
2x/2=10/2
x=5

This gives us the x- and y-coordinate we need, so <u>D is the correct answer</u>.
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3 years ago
Evaluate f[g(x)] when f(x)=7x-5 and g(x)=x+2
Alex

Answer:

7x+9

Step-by-step explanation:

f(x)=7x-5

g(x)=x+2

f[g(x)]=  Substitute g(x) in for x in the function f(x)

           =7(g(x))-5

          = 7( x+2) - 5

Distribute

           = 7x+14 - 5

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Step-by-step explanation:

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