How to solve which fraction is larger 10/13 or 11/14 or 14/15 or 17/18

Answers

Answer 1

Answer: To solve these type of problem you have to make the denominator the same or use the claculator.

let us do this problem by making the denominator the same.

$(10)/(13) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (11)/(14) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (14)/(15) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (17)/(18) \n\n (10)/(13)* (3780)/(3780) \ \ \ \ \ \ \ \ \ (11)/(14) * (3510)/(3510) \ \ \ \ \ \ \ \ \ (14)/(15) * (3276)/(3276) \ \ \ \ \ \ \ \ \ (17)/(18) * (2730)/(2730)$

$(37800)/(49140) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (38610)/(49140) \ \ \ \ \ \ \ \ \ \ \ \ \ \ (45864)/(49140) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (46410)/(49140)$

So, now. the one with the largest numerator is the largest fraction

In this case, $(46410)/(49140)$ is the largest which is equal to $(17)/(18)$ .

So, the largest fraction is $(17)/(18)$ .

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What is 2 plus 2 on study island

Answers

Answer: 2 plus two is Four

Which two numbers add up to 5 and multiply to -45 ?

Answers

we have x+y=5
xy=-45
solve
x+y=5
subtract x from both sides
y=-x+5
subsiute -x+5 for y in second equaiton
x(-x+5)=-45
-x^2+5x=-45
add 2x^2-5x to both sides
0=x^2-5x-45
use quadratic equation which is
if you have 0=ax^2+bx+c then x= $\frac{-b+ \sqrt{b^(2)-4ac} }{2a}$ or $\frac{-b- \sqrt{b^(2)-4ac} }{2a}$
sbsitute
1x^2-5x-45
a=1
b=-5
c=-45
subsitute
$\frac{-(-5)+ \sqrt{-5^(2)-4(1)(-45)} }{2(1)}$
$(5+ √(25-(-180)) )/(2)$
$(5+ √(25+180) )/(2)$
$(5+ √(205) )/(2)$

or $\frac{5- \sqrt{} }{2}$ or aprox x=-4.65891 or 9.65891, so the 2 numbers would be those answers for x

How do you solve x^2 plus 16x plus 63=0?

Answers

$x^2+ 16x + 63=0 \n \na=1, \ b=16 , \ c=63 \n \n \Delta =b^2-4ac =16^2 -4\cdot1\cdot 63 = 256 -252 = 4 \n \nx_(1)=(-b-\Delta )/(2a)=(-16-√(4))/(2 )=(-16-2)/(2)=-(-18)/(2)=-9 \n \nx_(2)=(-b+\Delta )/(2a)=(-16+√(4))/(2 )=(-16+2)/(2)=-(-14)/(2)=-7 \n \n \nAnswer : \ x= -7 \ \ or \ \ x= -9$

Find the perimeter of triangle ABC, with coordinates A(-3, 0), B(0, 4), and C(3, 0)

Answers

$Perimeter\ = AB+BC+AC\n\n AB=√((x_B-x_A)^2+(y_B-y_A)^2)=√((0+3)^2+(4-0)^2)=\n√(3^2+4^2)=√(9+16)=√(25)=5\n\n BC=√((x_C-x_B)^2+(y_C-y_B)^2)=√((3-0)^2+(0-4)^2)=\n√(3^2+(-4)^2)=√(9+16)=√(25)=5\n\n\ AC=√((x_C-x_A)^2+(y_C-y_A)^2)=√((3+3)^2+(0-0)^2)=\n√(6^2+0^2)=√(36+0)=√(36)=6\n\nPerimeter=5+5+6=16$