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3 years ago

7

What can you buy with a credit card ?

1 answer:

telo118 [61]3 years ago

8 0

Answer:

Most goods and services

Step-by-step explanation:

A credit card can be used to purchase most goods and services, whether they are legal or not.

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The area of a parallelogram is 96 cm squared the height is 4 cm less than the length what is

andrey2020 [161]

Area of paralellogram=legnth times height

l=legnth
h=height
area=96

lh=96

h is 4 less than l
h=-4+l
sub
l(-4+l)=96
-4l+l^2=96
minus 96 both sides
l^2-4l-96=0
factor
(l-12)(x+8)=0
set to zero
l-12=0
l=12

l+8=0
l=-8, imossible, legnth cannot be negative, disreagard this soultion

legnth=12cm

8 0

2 years ago

Read 2 more answers

Find all the solutions for the equation:

Contact [7]

$2y^2\,\mathrm dx-(x+y)^2\,\mathrm dy=0$

Divide both sides by $x^2\,\mathrm dx$ to get

$2\left(\dfrac yx\right)^2-\left(1+\dfrac yx\right)^2\dfrac{\mathrm dy}{\mathrm dx}=0$

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{2\left(\frac yx\right)^2}{\left(1+\frac yx\right)^2}$

Substitute $v(x)=\dfrac{y(x)}x$ , so that $\dfrac{\mathrm dv(x)}{\mathrm dx}=\dfrac{x\frac{\mathrm dy(x)}{\mathrm dx}-y(x)}{x^2}$ . Then

$x\dfrac{\mathrm dv}{\mathrm dx}+v=\dfrac{2v^2}{(1+v)^2}$

$x\dfrac{\mathrm dv}{\mathrm dx}=\dfrac{2v^2-v(1+v)^2}{(1+v)^2}$

$x\dfrac{\mathrm dv}{\mathrm dx}=-\dfrac{v(1+v^2)}{(1+v)^2}$

The remaining ODE is separable. Separating the variables gives

$\dfrac{(1+v)^2}{v(1+v^2)}\,\mathrm dv=-\dfrac{\mathrm dx}x$

Integrate both sides. On the left, split up the integrand into partial fractions.

$\dfrac{(1+v)^2}{v(1+v^2)}=\dfrac{v^2+2v+1}{v(v^2+1)}=\dfrac av+\dfrac{bv+c}{v^2+1}$

$\implies v^2+2v+1=a(v^2+1)+(bv+c)v$

$\implies v^2+2v+1=(a+b)v^2+cv+a$

$\implies a=1,b=0,c=2$

Then

$\displaystyle\int\frac{(1+v)^2}{v(1+v^2)}\,\mathrm dv=\int\left(\frac1v+\frac2{v^2+1}\right)\,\mathrm dv=\ln|v|+2\tan^{-1}v$

On the right, we have

$\displaystyle-\int\frac{\mathrm dx}x=-\ln|x|+C$

Solving for $v(x)$ explicitly is unlikely to succeed, so we leave the solution in implicit form,

$\ln|v(x)|+2\tan^{-1}v(x)=-\ln|x|+C$

and finally solve in terms of $y(x)$ by replacing $v(x)=\dfrac{y(x)}x$ :

$\ln\left|\frac{y(x)}x\right|+2\tan^{-1}\dfrac{y(x)}x=-\ln|x|+C$

$\ln|y(x)|-\ln|x|+2\tan^{-1}\dfrac{y(x)}x=-\ln|x|+C$

$\boxed{\ln|y(x)|+2\tan^{-1}\dfrac{y(x)}x=C}$

7 0

3 years ago

Did anyone take the Linear Functions Unit Test for CCA grade 10?? I could really use the help!!

gulaghasi [49]

Yeah post the questions I'll see if I can answer them

6 0

3 years ago

Evaluate 10.2x + 9.4y when x = 2 and y = 3.

IrinaK [193]

Answer:

48.6

Step-by-step explanation:

Substitute 2 for x and 3 for y

Multiply

Simplify by rounding then add to arrive at

Answer

7 0

3 years ago

A ball is thrown into the air with an upward velocity of 48 ft/s. Its height h in feet after t seconds is given by the function

stiv31 [10]

The answer to this, your question is h = -16t • 2 + 48t + 6
h = -32t + 48t + 6
h = 16t + 6 HOPE I HELPED IN A WAY! ~Marcey<3

7 0

3 years ago

Read 2 more answers