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sergey [27]
3 years ago
5

Add one third to 5 times d

Mathematics
1 answer:
DochEvi [55]3 years ago
5 0
As it is the problem is unsolvable because it doesn't have an =
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Triangle EFG is dilated by a scale factor of 3 centered at (0, 1) to create triangle E'F'G'. Which statement is true about the d
lesya [120]

Answer:

segment EH and segment E prime H prime both pass through the center of dilation.  

Step-by-step explanation:

Center of dilation is point (0.1), same as H. Both, E(0,5) and H(0,1) are placed over y-axis, then E' and H' are also located at y-axis.

After dilation, H' is placed at (0,1) because it coincides with the center of dilation

Distance between E and center of dilation is 4 units, then E' should be at 4*3=12 units from the center of dilation and over y-axis. Therefore, E' is placed at (0, 13)

So, segment E'H' goes from (0,1) to (0,13), and pass through the center of dilation, like segment EH.

4 0
3 years ago
College bookstore charge $60 for a yearly membership. Books will then only cost $7.60 each, tax included. Right in equation to r
Yuri [45]
Equation - 7.60b+60=m


7.60b+60=100
You’d be able to buy 5 books
4 0
4 years ago
How can i differentiate this equation?
Dmitry_Shevchenko [17]

\bf y=\cfrac{2x^2-10x}{\sqrt{x}}\implies y=\cfrac{2x^2-10x}{x^{\frac{1}{2}}} \\\\\\ \cfrac{dy}{dx}=\stackrel{\textit{quotient rule}}{\cfrac{(4x-10)(\sqrt{x})~~-~~(2x^2-10x)\left( \frac{1}{2}x^{-\frac{1}{2}} \right)}{\left( x^{\frac{1}{2}} \right)^2}} \\\\\\ \cfrac{dy}{dx}=\cfrac{(4x-10)(\sqrt{x})~~-~~(2x^2-10x)\left( \frac{1}{2\sqrt{x}} \right)}{\left( x^{\frac{1}{2}} \right)^2} \\\\\\ \cfrac{dy}{dx}=\cfrac{(4x-10)(\sqrt{x})~~-~~\left( \frac{2x^2-10x}{2\sqrt{x}} \right)}{x}


\bf\cfrac{dy}{dx}=\cfrac{(4x-10)(\sqrt{x})~~-~~\left( \frac{2x^2-10x}{2\sqrt{x}} \right)}{x} \\\\\\ \cfrac{dy}{dx}=\cfrac{ \frac{(4x-10)(\sqrt{x})(2\sqrt{x})~~-~~(2x^2-10x)}{2\sqrt{x}}}{x} \\\\\\ \cfrac{dy}{dx}=\cfrac{(4x-10)(\sqrt{x})(2\sqrt{x})~~-~~(2x^2-10x)}{2x\sqrt{x}}


\bf \cfrac{dy}{dx}=\cfrac{(4x-10)2x~~-~~(2x^2-10x)}{2x\sqrt{x}}\implies \cfrac{dy}{dx}=\cfrac{8x^2-20x~~-~~(2x^2-10x)}{2x\sqrt{x}} \\\\\\ \cfrac{dy}{dx}=\cfrac{8x^2-20x~~-~~2x^2+10x}{2x\sqrt{x}} \implies \cfrac{dy}{dx}=\cfrac{6x^2-10x}{2x\sqrt{x}} \\\\\\ \cfrac{dy}{dx}=\cfrac{2x(3x-5)}{2x\sqrt{x}}\implies \cfrac{dy}{dx}=\cfrac{3x-5}{\sqrt{x}}

8 0
3 years ago
What expression is equivalent to
Rasek [7]

= 18 1/5 + 22 2/5 + 40 1/5

Answer is D. 18 1/5 + 22 2/5 + 40 1/5


3 0
3 years ago
97 square inches to the nearest tenth
kupik [55]
97 Rounded to the nearest tenth would be 100
7 0
3 years ago
Read 2 more answers
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