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Triss [41]
3 years ago
11

In the figure, side AB is given by the expression 5x+5x+3, and side BC is 3x+92x−4. The simplified expression for the area of re

ctangle ABCD is ________ and the restriction on x is_____
Mathematics
1 answer:
kherson [118]3 years ago
3 0
We are asked to solve for the simplified expression of rectangle ABCD and the restriction on x. The solution is shown below:
L = 3X + 92X -4 = 95X -4
W =5X + 5X +3 =10X +3

The formula is Area = L*W
Area = (95x-4) * (10x +3)
Area = 950x² + 285x - 40x -12
 Area = 950x² + 245x -12
 x should be greater than zero    
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Write in logarithmic form.<br> 729 = 9^3
kolezko [41]

Answer:

Step-by-step explanation:

729 = 9³

by definition

a = b^{c} can be written as log_{b} a = c

by the same reasoning

729 = 9^{3} can be written as log_{9} 729 = 3

4 0
3 years ago
The ellipse with foci at (4, 0) and (-4, 0); y-intercepts (0, 3) and (0, -3).
Nesterboy [21]
Distance of Foci = c

Then,

c = 4

As c^2 + b^2 = a^2

And, b = 3

Then us will have:

a^2 = 4^2 + 3^2

a^2 = 16 + 9

a^2 = 25

a = 5


The equation of ellipse to this question is:

x^2 / a^2 + y^2/b^2 = 1

Then,

x^2 / 25 + y^2 / 9 = 1
4 0
3 years ago
Jose and Sue are playing a board game. If they land on a red or blue square, they lose 1 point. The inequality statement compare
Nina [5.8K]
The inequality statement shows that both Jose and Sue landed on red or blue squares and hence both lost some points. The points they lost were more than the points they gained, which resulted in an overall score having a negative value for each player.

Jose had a final score of -2 and Sue had a final score of -5. This means, Sue scored less and probably lost more points as compared to Jose. There is a difference of 3 points between the scores of both players. 


3 0
3 years ago
Read 2 more answers
Find Dy/dX when x^y×y^x=1​
victus00 [196]

by dy/dx, we'll be assuming that "y" is encapsulating a function, a function in terms of "x".

x^y\cdot y^x=1\implies \stackrel{\textit{\large product rule}}{\stackrel{\textit{chain rule}}{yx^{y-1}(1)}\cdot y^x+x^y\cdot \stackrel{\textit{chain rule}}{xy^{x-1}\cdot \frac{dy}{dx}}}~~ = ~~0 \\\\\\ y^{1+x}x^{y-1}+x^{y+1}y^{x-1}\cdot \cfrac{dy}{dx}=0\implies \cfrac{dy}{dx}=\cfrac{-y^{1+x}x^{y-1}}{x^{y+1}y^{x-1}} \\\\\\ \cfrac{dy}{dx}=-~~ y^{(1+x)-(x-1)}~~x^{(y-1)-(y+1)} \\\\\\ \cfrac{dy}{dx}=-~~ y^2x^{-2}\implies \cfrac{dy}{dx}=\cfrac{-y^2}{x^2}

7 0
3 years ago
Find the values of x, y, and z.<br> X=?<br> Y=?<br> Z=?
astraxan [27]
I haven’t taken geometry in a couple years, so I hope this is right. I’m so sorry if it’s not!!

X= 91

Y= 62

Z= 45
4 0
3 years ago
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