<h3>Given</h3>
trapezoid PSTK with ∠P=90°, KS = 13, KP = 12, ST = 8
<h3>Find</h3>
the area of PSTK
<h3>Solution</h3>
It helps to draw a diagram.
∆ KPS is a right triangle with hypotenuse 13 and leg 12. Then the other leg (PS) is given by the Pythagorean theorem as
... KS² = PS² + KP²
... 13² = PS² + 12²
... PS = √(169 -144) = 5
This is the height of the trapezoid, which has bases 12 and 8. Then the area of the trapezoid is
... A = (1/2)(b1 +b2)h
... A = (1/2)(12 +8)·5
... A = 50
The area of trapezoid PSTK is 50 square units.
Answer:
( f h ) (x) = 6 x² - 1
Step-by-step explanation:
<u><em>Step(i)</em></u>:-
Given<em> f(x) = 3 x - 4</em>
g (x) = −x²+2 x−5
<em> h (x) = 2 x² + 1</em>
j (x) = 6 x + 2 - 8 x
K (x) = 3 x² - x + 7
<u><em>Step(ii)</em></u>:-
<em>( f h ) (x) = f ( h (x)) = f ( 2 x² + 1 )</em>
= 3 (2 x² + 1 ) - 4
= 3 ((2 x² ) + 3 - 4
= <em>6 x² - 1</em>
<u><em> Final answer:</em></u>-
∴ <em> ( f h ) (x) = 6 x² - 1</em>
Two rays sharing a common endpoint is the vertex.
Answer:
I disagree
'No it won;t
Step-by-step explanation:
Given the inequality functions;
-2(3 – x) > 2x – 6
Open the parenthesis
-2(3) -2(-x) > 2x - 6
-6 + 2x > 2x - 6
Collect like terms
2x-2x > -6 + 6
0x > 0
x > 0/0
<em>The value of x does not exist on any real number. I disagree with my friend since the value of x is an indeterminate function. If the inequality were ≥, it won't change anything as the value of x won't still exist on any real number </em>
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