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2 years ago

How do I solve and find x?

Mathematics

1 answer:

oee [108]2 years ago

5 0

Answer:

x=60

Step-by-step explanation:

First get x all alone on 1 side of the equation. Add 14.5 to both sides of the equation to cancel out the negative. You will then be left with

x/5= 12

Multiply both sides of the equation by 5 to cancel out the division and you'll get x=60

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The area of a rectangle is (4x2 − 49y2) square units. Determine the dimensions of the rectangle by factoring the area expression

Pani-rosa [81]

Answer:

Step-by-step explanation:

$(4x^{2} - 49y^{2} ) = (2x+7y)(2x-7y)$

4 0

3 years ago

HELP FAST find the value of x.

sergeinik [125]

Answer:

81=6x+3(vertically opposite angle are equal)

81-3=6x

78=6x

x=13°

hope it helps.

6 0

2 years ago

<img src="https://tex.z-dn.net/?f=4%20%5Ctimes%208%20%5E%7B2x%20%2B%201%3F%7D%20%20%3D%2032%20%5E%7Bx%20%2B%202%7D%20" id="TexFo

svet-max [94.6K]

Answer:

x = 5

Step-by-step explanation:

__________________________________________________________

FACTS TO KNOW BEFORE SOLVING :-

$a^x \times a^y = a^{x+y}$

In an equation , if the bases are same in both L.H.S. & R.H.S. then , the power of the bases on both the sides of equation should be equal. For e.g. : $a^x = a^y$ ⇒ $x = y$ [∵ Bases are equal on both the sides]

__________________________________________________________

$4 \times 8^{2x+1} = 32^{x+2}$

Lets express it in terms of 2.

$=> 2^2 \times 2^{3 (2x+1)} = 2^{5(x+2)}$

$=> 2^2 \times 2^{6x+3} = 2^{5x+10}$

$=> 2^{2 + 6x+3} = 2^{5x+10}$

$=> 2^{6x+5} = 2^{5x+10}$

Here the bases on both the sides are equal. Hence ,

$=> 6x + 5 = 5x + 10$

$=> 6x - 5x = 10 - 5$

$=> x = 5$

6 0

2 years ago

3c-4 (a+b) if a =-1, b=2 and c = 1/3

Ksju [112]

Plug in each variable.

3(1/3)-4(-1+2)

Simplify

1-4(-1+2)

-3(1)

-3

6 0

3 years ago

Find y' by implicit differentiation: xy + 2x + 3x^2 = 4

NeX [460]

(xy)' + (2x)' + (3x^2)' = (4)'

y + xy' + 2 + 6x = 0

xy' = -y -2 -6x

y' = [-y -2 -6x] / x

Now solve y from the original equation and substitue

xy + 2x + 3x^2 = 4 => y = [-2x - 3x^2 + 4] / x

y' = [(-2x - 3x^2 +4) / x - 2 - 6x ] / x

y' = [-2x - 3x^2 + 4 -2x -6x^2 ] x^2 = [ -4x - 9x^2 + 4] / x^2 =

= [-9x^2 - 4x + 4] / x^2

3 0

3 years ago