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3 years ago

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you have a 5-inch by 7-inch picture that you framed. When you hung the picture, you found that the entire area is 69.56 in. What

is the width of the frame?

2 answers:

riadik2000 [5.3K]3 years ago

8 0

Answer: d) 2.4 inches

<u>Step-by-step explanation:</u>

Let x represent the width of the frame

Then the area of the frame is:

(x + 5)(x + 7) = 69.56

Rewrite it in Standard form of a quadratic equation:

x² + 12x + 35 = 69.56

x² + 12x - 34.56 = 0

Use Quadratic Formula to solve for x:

$x=\dfrac{-b+\sqrt{b^2-4ac}}{2a}\\\\\\x=\dfrac{-12+\sqrt{12^2-4(1)(-34.56)}}{2(1)}\\\\\\x=\dfrac{-12+\sqrt{282.24}}{2}\\\\\\x=\dfrac{-12+16.6}{2}\\\\\\x=\dfrac{4.8}{2}\\\\\\x=\large\boxed{2.4}$

Montano1993 [528]3 years ago

4 0

Answer:

d. 2.4 in

Step-by-step explanation:

So initially w/o frame the area was 5 x 7 = 35 sq in.

Now we have to add the width of the frame to both dimensions.

(5+w)(7+w) = 69.56 sq in

35 + 7w + 5w + w² = 69.56

w² + 12w + 35 = 69.56

w² + 12w - 34.56 = 0

I use the quadratic formula to solve this (x = -b±√b²-4ac / 2a). I cheat by using a quadratic program on the calculator :,)

w = 2.4

w = -14.4

Since we can't have a negative width, the answer must be <u>w = 2.4 inches.</u>

You can also just plug the answer choices one-by-one into the calculator with guess-and-check because this is multiple choice.

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3 years ago

Read 2 more answers

Use the Polynomial Identity below to help you create a list of 10 Pythagorean Triples:

Stolb23 [73]

Answer:

(3,4,5)

(6,8,10)

(5,12,13)

(8,15,17)

(12,16,20)

(7,24,25)

(10,24,26)

(20,21,29)

(16,30,34)

(9,40,41)

Just choose 2 numbers from {1,2,3,4,5,6,7,8,...} and make sure the one you input for x is larger.

Post the three in the comments and I will check them for you.

Step-by-step explanation:

We need to choose 2 positive integers for x and y where x>y.

Positive integers are {1,2,3,4,5,6,7,.....}.

I'm going to start with (x,y)=(2,1).

x=2 and y=1.

$(2^2+1^2)^2=(2^2-1^2)^2+(2\cdot2\cdot1)^2$

$(4+1)^2=(4-1)^2+(4)^2$

$(5)^2=(3)^2+(4)^2$

So one Pythagorean Triple is (3,4,5).

I'm going to choose (x,y)=(3,1).

x=3 and y=1.

$(3^2+1^2)^2=(3^2-1^2)^2+(2\cdot3\cdot1)^2$

$(9+1)^2=(9-1)^2+(6)^2$

$(10)^2=(8)^2+(6)^2$

So another Pythagorean Triple is (6,8,10).

I'm going to choose (x,y)=(3,2).

x=3 and y=2.

$(3^2+2^2)^2=(3^2-2^2)^2+(2\cdot3\cdot2)^2$

$(9+4)^2=(9-4)^2+(12)^2$

$(13)^2=(5)^2+(12)^2$

So another is (5,12,13).

I'm going to choose (x,y)=(4,1).

$(4^2+1^2)^2=(4^2-1^2)^2+(2\cdot4\cdot1)^2$

$(16+1)^2=(16-1)^2+(8)^2$

$(17)^2=(15)^2+(8)^2$

Another is (8,15,17).

I'm going to choose (x,y)=(4,2).

$(4^2+2^2)^2=(4^2-2^2)^2+(2\cdot4\cdot2)^2$

$(16+4)^2=(16-4)^2+(16)^2$

$(20)^2=(12)^2+(16)^2$

We have another which is (12,16,20).

I'm going to choose (x,y)=(4,3).

$(4^2+3^2)^2=(4^2-3^2)^2+(2\cdot4\cdot3)^2$

$(16+9)^2=(16-9)^2+(24)^2$

$(25)^2=(7)^2+(24)^2$

We have another is (7,24,25).

You are just choosing numbers from the positive integer set {1,2,3,4,... } and making sure the number you plug in for x is higher than the number for y.

I will do one more.

Let's choose (x,y)=(5,1).

$(5^2+1^2)^2=(5^2-1^2)^2+(2\cdot5\cdot1)^2$

$(25+1)^2=(25-1)^2+(10)^2$

$(26)^2=(24)^2+(10)^2$

So (10,24,26) is another.

Let (x,y)=(5,2).

$(5^2+2^2)^2=(5^2-2^2)^2+(2\cdot5\cdot2)^2$

$(25+4)^2=(25-4)^2+(20)^2$

$(29)^2=(21)^2+(20)^2$

So another Pythagorean Triple is (20,21,29).

Choose (x,y)=(5,3).

$(5^2+3^2)^2=(5^2-3^2)^2+(2\cdot5\cdot3)^2$

$(25+9)^2=(25-9)^2+(30)^2$

$(34)^2=(16)^2+(30)^2$

Another Pythagorean Triple is (16,30,34).

Let (x,y)=(5,4)

$(5^2+4^2)^2=(5^2-4^2)^2+(2\cdot5\cdot4)^2$

$(25+16)^2=(25-16)^2+(40)^2$

$(41)^2=(9)^2+(40)^2$

Another is (9,40,41).

5 0

3 years ago

A drawing of a man is 3 inches high. The actual man is 75

oksano4ka [1.4K]

Answer:

1 to 25

Step-by-step explanation:

the actual person is 25 times the drawing

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