HOME >
Mathematics > Hi there i just have about 4 questions i couldn't solve in math
Hi there i just have about 4 questions i couldn't solve in math
[From: ] [author: ] [Date: 11-05-20] [Hit: ]
which means length=15Diag^2=8^2+15^2diag=172.Sides a b and hypotenuse is ca=3b+3c=4b-3By pythagoras theorem(4b-3)^2=b^2+(3b+3)^216b^2-24b+9=10b^2+18b+96b^2-42b=0b^2-7b=0b=0 or b=7B cannot =0 as it is a length so b=7perimeter:b+3b+3+4b-3=563.b=2h+20.5bh=300.5(2h+2)h=302h^2+2h-60=0h^2+h-30=0(h+6)(h-5)h=-6 or h =5, because h is a length,......
= 12 units
4)
-2(4x - 1)^2 + 18 = 0
(4x - 1)^2 - 9 = 0
(16x^2 - 8x + 1) - 9 = 0
16x^2 - 8x - 8 = 0
2x^2 - x - 1 = 0
(2x + 1)(x - 1) = 0
x = -1/2 or x = 1
-
1.
l=2w-1
l*w=120
(2w-1)w=120
2w^2-w-120=0
w=8 or w=-7.5
w cannot =-7.5 as a length cannot be negative, so w=8, which means length=15
Diag^2=8^2+15^2
diag=17
2.
Sides a b and hypotenuse is c
a=3b+3
c=4b-3
By pythagoras theorem
(4b-3)^2=b^2+(3b+3)^2
16b^2-24b+9=10b^2+18b+9
6b^2-42b=0
b^2-7b=0
b=0 or b=7
B cannot =0 as it is a length so b=7
perimeter:b+3b+3+4b-3
=56
3.
b=2h+2
0.5bh=30
0.5(2h+2)h=30
2h^2+2h-60=0
h^2+h-30=0
(h+6)(h-5)
h=-6 or h =5, because h is a length, h=5
b=2(5)+2
b=12
4.
-2(4x-1)^2+18=0
(4x-1)^2=9
4x-1=3 or -3
x=1 or x=-0.5
keywords: just,039,Hi,solve,math,about,questions,couldn,have,in,there,Hi there i just have about 4 questions i couldn't solve in math
