The area of a square is expressed as the length of the side to the power of two, A = s^2. We were given the area of the enlarged photo which is 256 in^2. Also, it was stated that the length of the enlarged photo is the length of the original photo plus ten inches. So, from these statements we can make an equation to solve for x which represents the length of the original photo.
A = s^2
where s = (x+10)
A = (x+10)^2 = 256
Solving for x,
x= 6 in.
The dimensions of the original photo is 6 x 6.
Answer:
c- shifted 3 units right and 4 units up
Step-by-step explanation:
In this problem, we have a quadrilateral named as ABCD. Recall that a quadrilateral is a two-dimensional shape having four sides. So, we need to identify what transformation has been performed to get A'B'C'D', which is the same quadrilateral shifted certain units right and up. So take one point, say, B, so how do we need to do to obtain point B'? well, we need to move that point 3 units right and 4 units up, but how can we know this? just count the number of squares you need to move from B to B' horizontally and vertically, which is in fact 3 units right and 4 units up.
Answer:
7
Step-by-step explanation:
An equilateral triangle has a semiperimeter of 6 meters.
Heron’s formula: Area = StartRoot s (s minus a) (s minus b) (s minus c) EndRoot
An equilateral triangle has a semiperimeter of 6 meters. What is the area of the triangle? Round to the nearest square meter.
2 square meters
7 square meters
20 square meters
78 square meters
Answer:
A and C
Step-by-step explanation:
To determine which events are equal, we explicitly define the elements in each set builder.
For event A
A={1.3}
for event B
B={x|x is a number on a die}
The possible numbers on a die are 1,2,3,4,5 and 6. Hence event B is computed as
B={1,2,3,4,5,6}
for event C
![C=[x|x^{2}-4x+3]\\solving x^{2}-4x+3\\x^{2}-4x+3=0\\x^{2}-3x-x+3=0\\x(x-3)-1(x-3)=0\\x=3 or x=1](https://tex.z-dn.net/?f=C%3D%5Bx%7Cx%5E%7B2%7D-4x%2B3%5D%5C%5Csolving%20%20x%5E%7B2%7D-4x%2B3%5C%5Cx%5E%7B2%7D-4x%2B3%3D0%5C%5Cx%5E%7B2%7D-3x-x%2B3%3D0%5C%5Cx%28x-3%29-1%28x-3%29%3D0%5C%5Cx%3D3%20or%20x%3D1)
Hence the set c is C={1,3}
and for the set D {x| x is the number of heads when six coins re tossed }
In the tossing a six coins it is possible not to have any head and it is possible to have head ranging from 1 to 6
Hence the set D can be expressed as
D={0,1,2,3,4,5,6}
In conclusion, when all the set are compared only set A and set C are equal
Thanks for posting your question here. The answer to the above problem is x = <span>48.125. Below is the solution:
</span>
x+x/7+1/11(x+x/7)=60
x = x/1 = x • 7/7
x <span>• 7 + x/ 7 = 8x/7 - 60 = 0
</span>x + x/7 + 1/11 <span>• 8x/7 - 60 = 0
</span>8x <span>• 11 + 8x/ 77 = 96x/ 77
</span>96x - 4620 = 12 <span>• (8x-385)
</span>8x - 385 = 0
x = 48.125
