A purse contains $2.45 in dimes and

Mathematics

1 answer:

dimulka [17.4K]3 years ago

8 0

Answer:

dimes = 12, quarters =5

Step-by-step explanation:

Let x stand for quarters. So if there are two more than twice as many dimes than quarters than it would look something like

d = 2x+2

q = x

Now multiply them by the value of the coin, to get them into cents. So you'd multiply dimes by twn (2x+2*10) which is equal to 20x+20. Multiply quarters by 25. x*25=25x

Now make an equation. 20x+20+25x=245

45x=225

x=5

q = 5

d = 12

You might be interested in

Please help -2/5 + 4/5 on a numberline

4vir4ik [10]

Answer:

it will be 2/5

then it should be between 0 and 1

make 5 parts b/w 0 and 1 then the 2 nd one is 2/5.

I think so this is the ans.

8 0

4 years ago

Y = 6x - 14<br> y = - 8x

cupoosta [38]

Answer:

x = 1 and y = -8x

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

PLEASE PLEASE HELP IF U KNOW IT!!

larisa86 [58]

I believe the flagpole is 24ft

5 0

4 years ago

Find an equation of the tangent line to the hyperbola x2 a2 − y2 b2 = 1 at the point (x0, y0).

nalin [4]

The equation of the tangent line is $y = \frac{b^{2} x_{0}x }{a^{2} y_{0} } - \frac{b^{2} x_{0}^{2} }{a^{2} y_{0} } + y_{0}$

To find the tangent to the hyperbola $\frac{x^{2} }{a^{2} } - \frac{y^{2} }{b^{2} }$ at (x₀, y₀), we differentiate the equation implicitly to find the equation of the tangent at (x₀, y₀).

So, $\frac{d}{dx} (\frac{x^{2} }{a^{2} } - \frac{y^{2} }{b^{2} }) = \frac{d0}{dx}\\\frac{d}{dx} \frac{x^{2} }{a^{2} } - \frac{d}{dx}\frac{y^{2} }{b^{2} }= 0\\\frac{2x }{a^{2} } - \frac{dy}{dx}\frac{2y }{b^{2} } = 0\\\frac{2x }{a^{2} } = \frac{dy}{dx}\frac{2y }{b^{2} } \\\frac{dy}{dx} = \frac{b^{2}x }{a^{2}y }$

So, at (x₀, y₀)

$\frac{dy}{dx} = \frac{b^{2} x_{0} }{a^{2} y_{0} }$

So, the equation of the tangent line is gotten from the standard equation of a line in point-slope form

So, $\frac{y - y_{0} }{x - x_{0} } = \frac{b^{2} x_{0} }{a^{2} y_{0} } \\y - y_{0} = \frac{b^{2} x_{0} }{a^{2} y_{0} }(x - x_{0}) \\y = \frac{b^{2} x_{0} }{a^{2} y_{0} }(x - x_{0}) + y_{0} \\y = \frac{b^{2} x_{0}x }{a^{2} y_{0} } - \frac{b^{2} x_{0}^{2} }{a^{2} y_{0} } + y_{0}$