Answer:
x=2
Step-by-step explanation:
i would set up a system of equation using the pythagorean theorem:
1^2+x^2=z^2
4^2+x^2=y^2
Add up these equations to get:
17+2x^2=y^2+z^2
But notice that
y^2+z^2=5^2
So you can substitute
17+2x^2=25
2x^2=8
x^2=4
x=2
:)
Answer:
C (√34)
Step-by-step explanation:
Pythagoras' theorem states that: a² + b² = c², where c is the hypotenuse and what we are trying to figure out.
So, 3² + 5² = c²
9 + 25 = 34 = c²
c = √34
:)
The area of rectangle ABCD can be expressed in simplified form as: 8y^2 - 10y.
<h3>How to Calculate the Area of a Rectangle?</h3>
The area of a rectangle is calculated using:
Area = (length)(height).
We are given the following:
Length of rectangle ABCD = side DC = 4y - 5
Height of rectangle ABCD = side AD = 2y
The area of rectangle ABCD = (length)(height) = (4y - 5)(2y).
Expand the expression
The area of rectangle ABCD = 2y(4y) - 2y(5) [distributive property of multiplication]
The area of rectangle ABCD = 8y^2 - 10y
Thus, the area of rectangle ABCD can be expressed in simplified form as: 8y^2 - 10y.
Learn more about area of rectangle on:
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1. 2x - (3x + 6).....negative before parentheses makes everything inside parentheses opposite
2. 2x - 3x - 6....2x - 3x = -1x
3. -1x - 6
Answer:
-1x - 6
<span>Put it in the form of y =mx +b, or in this instance, y> mx +b
move the (1/2) x to the right by adding it to both sides of the inequality
(1/3)y>(1/2)x +2
Multiply by 3 on both sides to get y by itself.
y>(3/2) x +6
This is a graph with y intercept of (0,6) and a moderate upward and to the right slope. Because it is > , the line on the graph will NOT be part of the solution.
The easiest way to find the side of the graph that the inequality satisfies is to use (0,0) and see if it works or doesn't work. In the original equation, 0-0>2 does NOT work, so the area where the inequality works is to the up and left of the graph, which should be a dotted line to show that the inequality is greater than only.
The point (6,-2) should work.
Test it. 6*(1/3)-(-2)*(1/2)>0 ; 2-(-1)=3, and 3>2 It does work.
The point (6,2) should not work
Test it. 6 *(1/3)-2(1/2)=2-1 ; 1 is NOT >2, so it does not work.
If the graph goes through the origin, then pick a point near the graph with a small x or y.</span>
