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irakobra [83]
4 years ago
10

Jake cut out 3 small circles and 5 big circles from a piece of cardboard. The radius of each small circle was x inches and the r

adius of each big circle was y inches. The expression below shows the total number of square inches of cardboard that Jake used for cutting out the circles. 3πx2 + 5πy2 What does the term 5πy2 in the expression represent?
Mathematics
1 answer:
Harlamova29_29 [7]4 years ago
7 0
The term above represents the radius of both the circles
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Answer:

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Step-by-step explanation:

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3 years ago
Explain why the graph of quadratic function could not contain both a minimum vertex and a maximum vertex at the same time.
KonstantinChe [14]
A quadratic function is a function of the form f(x)=ax^2+bx+c. The vertex, (h,k) of a quadratic function is determined by the formula: h= \frac{-b}{2a} and k=f(h); where h is the x-coordinate of the vertex and k is the y-coordinate of the vertex. The value of a determines if the <span>parabola opens upward or downward; if</span>a is positive, the parabola<span> opens upward and the vertex is the minimum value, but if </span>a is negative <span>the graph opens downward and the vertex is the maximum value. Since the quadratic function only has one vertex, it </span><span>could not contain both a minimum vertex and a maximum vertex at the same time.</span>
3 0
3 years ago
Please help with the this give tiny explanation
Mila [183]

Answer:

21

first we begin from 1 to 100 we find 11 numbers contain 1 then we count 1s from 10 to 19 we get 9 numbers but there is an exciption in number 11 as it cotains 2 1s so thenumber of 1s is 21

5 0
3 years ago
How would I find the integral of <img src="https://tex.z-dn.net/?f=%5Cint%5Cfrac%7Btdt%7D%7Bt%5E4%2B2%7D" id="TexFormula1" title
kotegsom [21]
Let t=\sqrt y, so that t^2=y, t^4=y^2, and \mathrm dt=\dfrac{\mathrm dy}{2\sqrt y}. Then

\displaystyle\int\frac t{t^4+2}\,\mathrm dt=\int\frac{\sqrt y}{2\sqrt y(y^2+2)}\,\mathrm dy=\frac12\int\frac{\mathrm dy}{y^2+2}

Now let y=\sqrt2\tan z, so that \mathrm dy=\sqrt2\sec^2z\,\mathrm dz. Then

\displaystyle\frac12\int\frac{\mathrm dy}{y^2+2}=\frac12\int\frac{\sqrt2\sec^2z}{(\sqrt2\tan z)+2}\,\mathrm dz=\frac{\sqrt2}4\int\frac{\sec^2z}{\tan^2z+1}\,\mathrm dz=\frac1{2\sqrt2}\int\mathrm dz=\dfrac1{2\sqrt2}z+C

Transform back to y to get

\dfrac1{2\sqrt2}\arctan\left(\dfrac y{\sqrt2}\right)+C

and again to get back a result in terms of t.

\dfrac1{2\sqrt2}\arctan\left(\dfrac{t^2}{\sqrt2}\right)+C
3 0
3 years ago
Hiiii help??????? Please and thanks
timofeeve [1]

Answer:

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