Neither P, nor A are on the sketch
I guess P is the upper right corner of the rectangle
and A=(0,1)
P belongs to the line going through (1,0) and B(0,y)
<span>but we don't know the y-coordinate of B </span>
<span>the triangle is right and isosceles, so pythagoras a²+a²=2² ... 2a²=4 ... a²=2 ... a=sqrt2 </span>
now look at the right triangle BOA
<span>his hypotenuse is AB=sqrt2 and the <span>the kathete</span> OA is 1 </span>
so y²+1²=(sqrt2)² ... y²+1=2 ... y²=1.. y=1
so the coordinates of B are (0,1)
the line going through (1,0) and (0,1) is L(x)=-x+1
P belongs to this line, so the coordinates of P are P(x,-x+1) (0<x<1)
b) so if that's P, the height of the rectangle is -x+1 and the width=2x
<span>so its area A(x)=2x*(-x+1)= -2x²+2x
I hope my answer has come to your help. Thank you for posting your question here in Brainly.
</span>
Answer:
Step-by-step explanation:
We have to see what happens at each graph at the “end” of x axis This means look both to the left and right of the x axis
1) when x approaches -oo, f(x)approaches -oo. When x approaches +oo, f(x) approaches +oo
2) when x approaches -oo, f(x) approaches +oo. When x approaches +oo, f(x) approaches +oo
3) when x approaches -oo, f(x) approaches -oo. When x approaches +oo, f(x) approaches -oo
4) when x approaches -oo, f(x) approaches -oo. When x approaches +oo, f(x) approaches +oo
10+2+4=16 jehehehshhwbdhshbs hahahah
<h2>
Answer:</h2><h2>x=37.5° and angle ABC is 52.5°</h2><h2>
Step-by-step explanation:</h2>
We know that angle CBD is a right angle.
We also know that angle ABE is 180°.
Our unknown angle values are angle DBE and angle CBA.
angle CBD minus angle ABE will be 90°.
angle ABC plus angle DBE must equal 90°.
Solving for x:
(x+15)+(x)=90°
(x+15-15)+(x)=90°-15
(x)+(x)=75°
2x=75°
2x/2=75°/2
x=37.5°
Checking the work:
ABC=37.5°+15=52.5°
CBD=90°
DBE=37.5°
In conclusion, ABC plus CBD plus DBE equals 180°
<em>Hope this helped!</em>
Answer:
The original dimensions of the card were 9 inches length and 9 inches width
Step-by-step explanation:
Let x be the original length and width of the card in inches (remember that it was squared originally). The exercise says that the area of the new width, with dimensions x-4 and x-5 is 20 square inches, therefore
(x-4)*(x-5) = 20
x²-9x+20 = 20
x²-9x = 0
x*(x-9) = 0
x = 0 or x = 9
Since x must be positive, then it cant be 0, thus, x has to be 9. The original dimensions of the card were 9 inches length and 9 inches width.
