<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) 
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

According to question , Area of new image is :
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So the equation will be :
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b. Find the dimensions of the reduced image
Let's solve : 
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By Quadratic formula :
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x = 15 is rejected ! as 15 > 10 ! Side can't be negative
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Therefore, Dimensions are : Length = 10-x = 3 inch , Breadth = 12-x = 5 inch
Answer:
14 6/7 is the circumference
Step-by-step explanation:
you first of all look for the radius using the formulae multiply both sides by the reciprocal of 22/7
7/22 ×22/7r=64×7/22=224/11{2 4/11)
2 4/11 is the radius
cir =2×22/7×2 4/11
2×22/7× 26/11=104/7
==== =14 6/7
all i can do but am not sure sorry
<h3>Answer: 1.15</h3>
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Work Shown:
QR = 73
QP = 55
PR = x
Pythagorean Theorem
a^2 + b^2 = c^2
(QP)^2 + (PR)^2 = (QR)^2
(55)^2 + (x)^2 = (73)^2
3025 + x^2 = 5329
x^2 = 5329-3025
x^2 = 2304
x = sqrt(2304)
x = 48
PR = 48
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tan(angle) = opposite/adjacent
tan(R) = QP/PR
tan(R) = 55/48
tan(R) = 1.1458333 approximately
tan(R) = 1.15
X=2 here’s why 90-32 is 58. 29x=58 then x is 2
Answer:
2/5
Step-by-step explanation:
From the two points we have marked, they are clearly 10 units apart from -5 to 5, and the change in the y-axis is 4. So the change between the two points is 4/10 (rise over run).
4/10 simplifies to 2/5.