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lukranit [14]
3 years ago
11

What is the solution to the system of equations? Use the substitution method. {y=−2x+5 {3x−4y=2 Enter your answer in the boxes.

Mathematics
1 answer:
Nataly_w [17]3 years ago
7 0
1. You have the following equations:
<span>
 y=−2x+5  (i)
 3x−4y=2  (ii)

 2. You must substitute the equation (i) into the equation (ii), as below:

 </span>3x−4y=2
 3x-4(−2x+5)=2
 3x+8x-20=2
 11x-20=2

 3. Now, you must clear the variable "x":

 11x=2+20
 x=22/11
 x=2

 4. You have the value of "x", so you can find the value of "y". When you substitute x=2 into the equation (i), you obtain:

 y=−2x+5
 y=-2(2)+5
 y=-4+5
 y=1

 5. Therefore, the answer is:

 x=2
 y=1
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Step-by-step explanation:

Using the cofunction identity

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sin30° = cos(90 - 30)° = cos60°

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List the partical products of 35 times 7
White raven [17]
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3 0
2 years ago
Janeel has a 10-inch by 12-inch photograph. She wants to scan the photograph, then reduce the result
Vaselesa [24]

<u>Complete Question:</u>

Janeel has a 10 inch by 12 inch photograph. She wants to scan the photograph, then reduce the results by the same amount in each dimension to post on her Web site. Janeel wants the area of the image to be one eight of the original photograph. Write an equation to represent the area of the reduced image. Find the dimensions of the reduced image.

<u>Correct Answer:</u>

A) (10-x)(12-x)=15

B) Dimensions are : Length = 10-x = 3 inch , Breadth = 12-x = 5 inch

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

a. Write an equation to represent the area of the reduced image.

Let the reduced dimensions is by x , So the new dimensions are

length=10-x\\breadth=12-x

According to question , Area of new image is :

⇒ Area = \frac{1}{8}Length(breadth)

⇒ Area = \frac{1}{8}(10)(12)

⇒ Area = 15

So the equation will be :

⇒ (10-x)(12-x)=15

b. Find the dimensions of the reduced image

Let's solve :  (10-x)(12-x)=15

⇒ 120-10x-12x+x^2=15

⇒ 120-22x+x^2=15

⇒ x^2-22x+105=0

By Quadratic formula :

⇒ x = \frac{-b \pm \sqrt{b^2-4ac} }{2a}

⇒ x = \frac{22 \pm8 }{2}

⇒ x = \frac{22 +8 }{2} , x = \frac{22 -8 }{2}

⇒ x = 15 , x =7

x = 15 is rejected ! as 15 > 10 ! Side can't be negative

⇒ x =7

Therefore, Dimensions are : Length = 10-x = 3 inch , Breadth = 12-x = 5 inch

5 0
2 years ago
Point A(–4, 2) is reflected over the line x = 3 to create the point A'. What are the coordinates of A'?
iragen [17]
A ( - 4, 2)
Reflected over x = 3.
x = 3 is a vertical line, the point (-4, 2) is 7 units on the left side of the vertical line. When you reflect across the line to the right side we need to be 7 units away from the vertical line on the right side. 3 + 7 = 10
The x value would be 10. (10, 2)
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8 0
3 years ago
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