All categories

3 years ago

If y varies directly as x and y is 12 when x is 1.2, what is the constant of variation for this relation?.

Mathematics

2 answers:

11111nata11111 [884]3 years ago

7 0

Hello,

y=k*x ; if x=1.2,y=12
==>12=k*1.2==>k=12/1.2==>k=10

Alex787 [66]3 years ago

4 0

From the question, we now that y is 12 when x is 1.2

in order to find out the constant of variation for this relation, we just need to substitute it to :

y = k x

12 = k 1.2

k = 12/1.2

k = 10

hope this helps

You might be interested in

2) (3 pt) Write the function from the table and graph.хf(x)-10004122130.52) f(x) =

DaniilM [7]

(x - h)^2 = 4p(y - k)

(-1 - 3)^2 = 4p(8 - 0.5)

(-4)^2 = 4p(7.5)

16 = 30p

p = 16/30

p = 8/15

(x - 3)^2 = 16/15(y - 0.5)

15(x^2 - 6x + 9) = 16y - 8

15x^2 - 90x + 135 = 16y - 8

16y = 15x^2 - 90x + 135 + 8

y = 15/16 x^2 - 90/16 x + 143/16

f(x) = 15/16 x^2 - 90/16x + 143/16

4 0

9 months ago

F(x) = 3x + x3

____ [38]

Answer:

Please check the explanation.

Step-by-step explanation:

Given

f(x) = 3x + x³

Taking differentiate

$\frac{d}{dx}\left(3x+x^3\right)$

$\mathrm{Apply\:the\:Sum/Difference\:Rule}:\quad \left(f\pm g\right)'=f\:'\pm g'$

$=\frac{d}{dx}\left(3x\right)+\frac{d}{dx}\left(x^3\right)$

solving

$\frac{d}{dx}\left(3x\right)$

$\mathrm{Take\:the\:constant\:out}:\quad \left(a\cdot f\right)'=a\cdot f\:'$

$=3\frac{d}{dx}\left(x\right)$

$\mathrm{Apply\:the\:common\:derivative}:\quad \frac{d}{dx}\left(x\right)=1$

$=3\cdot \:1$

$=3$

now solving

$\frac{d}{dx}\left(x^3\right)$

$\mathrm{Apply\:the\:Power\:Rule}:\quad \frac{d}{dx}\left(x^a\right)=a\cdot x^{a-1}$

$=3x^{3-1}$

$=3x^2$

Thus, the expression becomes

$\frac{d}{dx}\left(3x+x^3\right)=\frac{d}{dx}\left(3x\right)+\frac{d}{dx}\left(x^3\right)$

$=3+3x^2$

Thus,

f'(x) = 3 + 3x²

Given that f'(x) = 15

substituting the value f'(x) = 15 in f'(x) = 3 + 3x²

f'(x) = 3 + 3x²

15 = 3 + 3x²

switch sides

3 + 3x² = 15

3x² = 15-3

3x² = 12

Divide both sides by 3

x² = 4

$\mathrm{For\:}x^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}$

$x=\sqrt{4},\:x=-\sqrt{4}$

$x=2,\:x=-2$

Thus, the value of x will be:

$x=2,\:x=-2$

5 0

2 years ago

Factor -1/4 out of -1/2-5/4y

In-s [12.5K]

When we factor out something, we divide the factor by the equation. For example if we were to factor 3 out of 6, It will be 3(2). We got this answer by dividing 3 and 6. Using this concept:

$\frac{-1}{4}[( \frac{-1}{2} / \frac{-1}{4}) - ( \frac{5}{4} y / \frac{-1}{4})]$

Simplifying:

$\frac{-1}{4}(2 + 5y)$

Hope this helps!

3 0

3 years ago

6. Use the Rational Root (Zero) Theorem to list all the possible real roots (zeros).

maks197457 [2]

The left side
−
52
-
52
does not equal to the right side
0
0
, which means that the given statement is false.
False

8 0

2 years ago

Help??<br> anyone?<br> 7th grade math <br> tysm

prohojiy [21]

Answer:

if it isn't compound interest, i think it is $3029.9447

Step-by-step explanation:

4 0

2 years ago