4725-2250= 2475
2475/2250 x100= 110
It increased by 110%
Answer:
yes it is possible to get a common denominator grater than 12 and still get the right answer.
Step-by-step explanation:
This is because the numerator is always directly corelated to de denominator therefore the fraction wouldn't change because when simplified it is the same fraction as it would be with 12 as the denominator
The equation of line T is 2x - y = 7 ⇒ (6x - 3y = 21)
Step-by-step explanation:
A dilation is a transformation that produces an image that is the same shape as the original, but is a different size
- If the equation of a line is ax + by = k is dilated, with center origin and scale factor k, then the equation of the image of the line is kax + kby = kc
- The line and its image are parallel
- The coordinates of a general point on the image is (kx , ky)
Line L is mapped onto the line T by a dilation centered at the origin and a scale factor of 3.
That means lint T is the image of line L after dilation
∵ The equation of line L is 2x - y = 7
∵ Line L is dilated by scale factor 3 and centered at origin
- That means multiply the equation of line L by 3 to find the
equation of line t
∵ Line T is the image of line L after dilation
∴ The equation of line T is (3)(2x) - (3)(y) = (3)(7)
∴ The equation of line T is 6x - 3y = 21
<em>Very important note:</em>
The equation of line T is the same with equation of line L but multiplied by the scale factor 3 ⇒ L and T are coincide lines (same line)
That means the equation of lines T and L is 2x - y = 7
The equation of line T is 2x - y = 7 ⇒ (6x - 3y = 21)
Learn more:
You can learn more about dilation in brainly.com/question/2480897
#LearnwithBrainly
First term ,a=4 , common difference =4-7=-3, n =50
sum of first 50terms= (50/2)[2×4+(50-1)(-3)]
=25×[8+49]×-3
=25×57×-3
=25× -171
= -42925
derivation of the formula for the sum of n terms
Progression, S
S=a1+a2+a3+a4+...+an
S=a1+(a1+d)+(a1+2d)+(a1+3d)+...+[a1+(n−1)d] → Equation (1)
S=an+an−1+an−2+an−3+...+a1
S=an+(an−d)+(an−2d)+(an−3d)+...+[an−(n−1)d] → Equation (2)
Add Equations (1) and (2)
2S=(a1+an)+(a1+an)+(a1+an)+(a1+an)+...+(a1+an)
2S=n(a1+an)
S=n/2(a1+an)
Substitute an = a1 + (n - 1)d to the above equation, we have
S=n/2{a1+[a1+(n−1)d]}
S=n/2[2a1+(n−1)d]
<u>First let's calculate the exponents.</u>
<u>Now we should multiply and divide.</u>
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<u>Now we should add and subtract.</u>
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<u>Convert the improper fractions into mixed numbers.</u>
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<u>Answer : </u>
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