<span>w= 33.75, the weight of the larger sphere
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Answer: She has $1022 left.
Step-by-step explanation:
The total amount that Amee received an the end of year as bonus at work is $1550.
She went on a shopping spree, spending $225 at the department store, $275 at the home furnishing store, and $28 at the card shop. Therefore, the total amount that she spent at the department store, the home furnishing store, and the card shop is
225 + 275 + 28 = $528
Therefore, the total amount of her bonus that she has left is
1550 - 528 = $1022
(i) speed = distance / time
so time = distance / speed
here we have
time t = 1080/x hours
(ii) return flight time = 1080 / (x + 30) hours
(a) 1080/x - 1080/(x + 30) = 1/2
Multiplying through by the LCD 2x(x + 30) we get:-
1080*2(x + 30) - 2x*1080 = x(x+30)
2160x + 64800 - 2160x = x^2 + 30x
x^2 + 30x - 64800 = 0
(b) factoring; -64800 = 270 * -240 ans 270-240 = 30 so we have
(x + 270)(x - 240) = 0 so x = 240 ( we ignore the negative -270)
So the speed for outward journey is 240 km/hr
(c) time ffor outward flight = 1080 / 240 = 4 1/2 hours
(d) average speed for whole flight = distance / time
Time for outward journey = 4.5 hours and time for return journey = d / v
= 1080 / (240+30) = 4 hours
Therefore the average speed for whole journey = 2160 / 8.5 = 254.1 km/hr
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.
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