All categories

3 years ago

Can someone understand this i know its substitution for sure if i dont get wrong

Mathematics

1 answer:

svlad2 [7]3 years ago

7 0

4x - y = 3

y = 2x + 13

Substitute the second equation into the first equation:

4x - y = 3 [since y = 2x + 13, you can substitute (2x + 13) for "y" in the equation]

4x - (2x + 13) = 3 Distribute/multiply - into (2x + 13)

4x - 2x - 13 = 3

2x - 13 = 3 Add 13 on both sides

2x = 16 Divide 2 on both sides

x = 8

Now that you know x = 8, plug it into one of the equations to find y

y = 2x + 13

y = 2(8) + 13

y = 16 + 13

y = 29

x = 8, y = 29 or (8, 29)

[proof]

4x - y = 3

4(8) - 29 = 3

32 - 29 = 3

3 = 3

y = 2x + 13

29 = 2(8) + 13

29 = 16 + 13

29 = 29

You might be interested in

From the top of an 95 meter cliff, Tim spots a hiker down below at an angle

kirza4 [7]

Step-by-step explanation:

jsvsbsbbssnbshsshvsvsv

7 0

3 years ago

Can someone help me on letters B and C?

skelet666 [1.2K]

Step-by-step explanation:

if he is just talking about the length than it is,

4(13x + 8)

of he is talking about overall length he swam than,

4(13x + 8) + 3(7x - 2)

if x = 12

than

perimeter = 2l + 2b

so, after inserting the values we got,

→ 2[13(12) + 8] + 2[7(12) - 2]

→ 2 (156 + 8) + 2(84 - 2)

→ 2 × 164 + 2 × 82

→ 328 + 164

→ 492 ft. ← perimeter

and yeah, according to me

a is wrong as,

formula of perimeter = 2l + 2b or 2(l+b)

so, the answer should be,

2(13x + 8) + 2(7x - 2) or 2 [ (13x + 8) + (7x - 2)

hope this answer helps you dear...take care and may u have a great day ahead!

4 0

2 years ago

BRAINLIEST PLZ JACOB HELP MEE

Viktor [21]

It’s 13 and 15 you felllmeeyebebenr

8 0

2 years ago

Let f(x) = 1/x^2 (a) Use the definition of the derivatve to find f'(x). (b) Find the equation of the tangent line at x=2

Verdich [7]

Answer:

(a) $f'(x)=-\frac{2}{x^3}$

(b) $y=-0.25x+0.75$

Step-by-step explanation:

The given function is

$f(x)=\frac{1}{x^2}$ .... (1)

According to the first principle of the derivative,

$f'(x)=lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$

$f'(x)=lim_{h\rightarrow 0}\frac{\frac{1}{(x+h)^2}-\frac{1}{x^2}}{h}$

$f'(x)=lim_{h\rightarrow 0}\frac{\frac{x^2-(x+h)^2}{x^2(x+h)^2}}{h}$

$f'(x)=lim_{h\rightarrow 0}\frac{x^2-x^2-2xh-h^2}{hx^2(x+h)^2}$

$f'(x)=lim_{h\rightarrow 0}\frac{-2xh-h^2}{hx^2(x+h)^2}$

$f'(x)=lim_{h\rightarrow 0}\frac{-h(2x+h)}{hx^2(x+h)^2}$

Cancel out common factors.

$f'(x)=lim_{h\rightarrow 0}\frac{-(2x+h)}{x^2(x+h)^2}$

By applying limit, we get

$f'(x)=\frac{-(2x+0)}{x^2(x+0)^2}$

$f'(x)=\frac{-2x)}{x^4}$

$f'(x)=\frac{-2)}{x^3}$ .... (2)

Therefore $f'(x)=-\frac{2}{x^3}$ .

(b)

Put x=2, to find the y-coordinate of point of tangency.

$f(x)=\frac{1}{2^2}=\frac{1}{4}=0.25$

The coordinates of point of tangency are (2,0.25).

The slope of tangent at x=2 is

$m=(\frac{dy}{dx})_{x=2}=f'(x)_{x=2}$

Substitute x=2 in equation 2.

$f'(2)=\frac{-2}{(2)^3}=\frac{-2}{8}=\frac{-1}{4}=-0.25$

The slope of the tangent line at x=2 is -0.25.

The slope of tangent is -0.25 and the tangent passes through the point (2,0.25).

Using point slope form the equation of tangent is

$y-y_1=m(x-x_1)$

$y-0.25=-0.25(x-2)$

$y-0.25=-0.25x+0.5$

$y=-0.25x+0.5+0.25$

$y=-0.25x+0.75$

Therefore the equation of the tangent line at x=2 is y=-0.25x+0.75.

5 0

3 years ago

one high speed internet provider has a $50 setup fee and costs $30 per month. another has no setup fee and costs $40 per month.

vlabodo [156]

Answer:

y=30x+50 and y=40

Step-by-step explanation:

7 0

3 years ago