1.)A picture 8 in. by 12 in. is placed in a frame which has a uniform width. if the area of the frame is equal to the area of the picture, find the dimension of the frame
(8+2x)(12+2x)-96 = 96
(8+2x)(12+2x)=96+96
96+4x(2)=192
4x(2)=192-96
4x(2)=96
x(2)=96/4
x/2=24/2
x=12
is it correct that the answer is 12? i'm not confident with my answer,please help!
Arman has 1160 annual income from bonds bearing 3% and 5 % interest. then he added 25% more of the 3% bonds and 40% more of the 5 % bonds, thereby increasing his annual income by 410. find his initial investment in each type
of bond
x=original 3% investment
y=original 5% investment
.03x+.05y=1160
.03(1.25x)+.05(1.4y)=1570
3.75x+7y = 1570
i don't know what to do next,please help!
thank you
Not even close. Not only is the algebra wrong, but x is not the answer to the question.
(8+2x)(12+2x)=96+96
96+40x+4x^2 = 96+96
4x^2+40x-96 = 0
x^2+10x-24 = 0
(x-2)(x+12) = 0
x = 2 or -12
Naturally, -12 is not a possibility, so x=2
The frame is 2" wide, making the dimensions of the framed picture 12x16
Note that 12x16 = 96+96
on the 2nd,
first, note that .05x1.4 is NOT 7. It is 0.07
.03x + .05y = 1160
.0375x + .07y=1570
7(1st)-5(2nd):
.21x + .35y = 8120
.1875x + .35y = 7850
.0225x = 270
x = 12000
y = 16000