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2 years ago

A circle whose center is (2,0) and passes through the point (-5,0).

Mathematics

2 answers:

Nana76 [90]2 years ago

5 0

2179 because of da measurements

vampirchik [111]2 years ago

4 0

Answer:

2791

Step-by-step explanation:

Yea

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Model the two step equation by drawing algebra tiles 3t+10=16

dedylja [7]

3t+10=16
-10. -10
3t=6
/3. /3
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3 years ago

Convert a 30% markup percent on cost to markup percent on selling price. (Round to the nearest hundredth percent.)

Art [367]

Answer:

Option C

23.08% markup on selling price.

Step-by-step explanation:

Given in the question,

markup percentage on cost price = 30%

To find,

markup percentage on selling price

Markup is the ratio between the cost of a good or service and its selling price.

Let suppose that cost price percentage = 100%

As we know that,

<h3>cost% + markup% = selling%</h3><h3>100% + 30% = 130%</h3>

So percent markup selling price = 30 / 130 x 100

= 23.0769

Hence, 30% markup on cost price = 23.0769% markup on selling price.

4 0

2 years ago

What is 9+(5x10)+1-(15x2)

solmaris [256]

Step-by-step explanation:

Answer:30. hope this helps

5 0

2 years ago

Read 2 more answers

A parabola had a vertex at (3,4) and passes through the point (5,-4) what is the standard form equation for the parabola.. help

8_murik_8 [283]

Answer:

i need help on this one aswell

Step-by-step explanation:

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4 0

2 years ago

Use separation of variables to solve dy dx − tan x = y2 tan x with y(0) = √3. Find the value of c in radians, not degrees

a_sh-v [17]

Answer:

$y(x)=tan(-log(cos(x))+\frac{\pi }{3} )$

Step-by-step explanation:

Rewrite the equation as:

$\frac{dy(x)}{dx}-tan(x)=y(x)^{2} *tan(x)$

Isolating $\frac{dy}{dx}$

$\frac{dy}{dx} =tan(x)+tan(x)*y^{2}$

Factor:

$\frac{dy}{dx} =tan(x)*(1+y^{2} )$

Dividing both sides by $(1+y^{2} )$ and multiplying them by $dx$

$\frac{dy}{1+y^{2} } =tan(x)dx$

Integrate both sides:

$\int\ \frac{dy}{1+y^{2} } = \int\ tan(x) dx$

Evaluate the integrals:

$arctan(y)=-log(cos(x))+C_1$

Solving for y:

$y(x)=tan(-log(cos(x))+C_1)$

Evaluating the initial condition:

$y(0)=\sqrt{3} =tan(-log(cos(0))+C_1)=tan(-log(1)+C_1)=tan(0+C_1)$

$\sqrt{3} =tan(C_1)\\arctan(\sqrt{3} )=C_1\\60=C_1$

Converting 60 degrees to radians:

$60degrees*\frac{\pi }{180degrees} =\frac{\pi }{3}$

Replacing $C_1$ in the diferential equation solution:

$y(x)=tan(-log(cos(x))+\frac{\pi }{3} )$

3 0

2 years ago