3 years ago

PLEASE HELP ME !! ASAP !! I NEED HELP !!

Mathematics

1 answer:

kvv77 [185]3 years ago

7 0

Answer:

c. 40

a. 3.5

Step-by-step explanation:

question 2:

2x+3x+4x=180

9x=180

x=180/9

x=20

2x=2(20)=40

question 3:

x/15=4/17

17x=60

x=60/17

x=3.5

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Sum of positive roots of the equation (x2 - 12x + 35) (x2 + 10x + 24) = 504 is

Digiron [165]

Answer:

(3)11

Step-by-step explanation:

We are given that

$(x^2-12x+35)(x^2+10x+24)=504$

We have to find the sum of positive roots of the equation.

$x^2(x^2+10x+24)-12x(x^2+10x+24)+35(x^2+10x+24)=504$

$x^4+10x^3+24x^2-12x^3-120x^2-288x+35x^2+350x+840=504$

$x^4-2x^3-61x^2+62x+840-504=0$

$x^4-2x^3-61x^2+62x+336=0$

Factor of 336

2,3,4,6,8,7,

Let x=2

$2^4-2^4-61(2^2)+62(2)+336\neq0$

x=2 is not the root of equation

x=-2

$(-2)^4-2(-2)^3-61(-2)^2+62(-2)+336=0$

Hence x=-2 is the root of equation.

x+2 is a factor of equation.

x=3

$3^4-2(3^3)-61(3^2)+62(3)+336=0$

Therefore, x=3 is the root of equation.

$(x+2)(x-3)(x^2-x-56)=0$

$(x+2)(x-3)(x^2-8x+7x-56)=0$

$(x+2)(x-3)(x(x-8)+7(x-8))=0$

$(x+2)(x-3)(x-8)(x+7)=0$

$x-8=0\implies x=8$

$x+7=0\implies x=-7$

Positive roots are 3 and 8

Sum of positive roots=3+8=11

Option (3) is true.

4 0

3 years ago

Help solve #2 please

borishaifa [10]

its pithagoreon so

c^2= a^2+ b^2

c^2= 32^2 +24^2

c^2= 1,024+576

c^2= 1600

c= $\sqrt{1600\\$ = 40

5 0

3 years ago