Daily math problems for October 13, 2026 — Descriptive Statistics & Sampling, Euclidean Geometry, Calculus & more

Mean $\bar{x}={\displaystyle \frac{2+4+4+6}{4}}=4$.

Squared deviations: $(2-4{)}^2+(4-4{)}^2+(4-4{)}^2+(6-4{)}^2=4+0+0+4=8$.

Variance $={\displaystyle \frac{8}{4}}=2$.

$\frac{1}{2}\times 2\pi \times 10=10\pi$ cm.

${\displaystyle \frac{f(3+h)-f(3)}{h}}={\displaystyle \frac{(3+h{)}^2-9}{h}}={\displaystyle \frac{9+6h+h^2-9}{h}}={\displaystyle \frac{6h+h^2}{h}}=6+h$.

${\displaystyle \underset{h\to 0}{\lim }(6+h)=6}$.

By Vieta's formulas, the sum of the roots is $-\frac{b}{a}$ and the product is $\frac{c}{a}$: $$r+s=7$$ $$rs=10$$ Use the identity that rewrites $r^2+s^2$ in terms of the sum and product: $$r^2+s^2=(r+s{)}^2-2rs$$ Substitute the known values: $$r^2+s^2=7^2-2(10)=49-20=29$$

$\bullet$ Apply the chain rule, multiplying by the derivative of the inner function $3x^2-5$:

$$f^{\mathrm{\prime }}(x)=4(3x^2-5{)}^3\cdot 6x=24x(3x^2-5{)}^3$$

$\bullet$ Evaluate the inner expression at $x=1$:

$$3(1{)}^2-5=-2$$

$\bullet$ Substitute, using $(-2{)}^3=-8$:

$$f^{\mathrm{\prime }}(1)=24(1)(-2{)}^3=24\cdot (-8)=-192$$

$a_5={\displaystyle \frac{a_3+a_7}{2}}={\displaystyle \frac{11+27}{2}}=19$.

$c^2=4^2+5^2-2(4)(5)\cos {60}^{\circ }=16+25-40\cdot \frac{1}{2}=41-20=21$.

So $c=\sqrt{21}$.

$\bullet$ Build the pieces for $A(1)=\left(\begin{array}{cc}1& 2\\ 0& 1\end{array}\right)$:

$$A(1{)}^2=\left(\begin{array}{cc}1& 4\\ 0& 1\end{array}\right)$$

$$2A(1{)}^t=\left(\begin{array}{cc}2& 0\\ 4& 2\end{array}\right)$$

$$A(1{)}^{-1}=\left(\begin{array}{cc}1& -2\\ 0& 1\end{array}\right)$$

$\bullet$ Assemble $f(A(1))=A^2+2A^t-A^{-1}$ and take the determinant:

$$f(A(1))=\left(\begin{array}{cc}2& 6\\ 4& 2\end{array}\right),\det =4-24=-20$$

$\bullet$ For $A(-1)=\left(\begin{array}{cc}-1& 2\\ 0& 1\end{array}\right)$ we have $A(-1{)}^2=I$ and:

$$2A(-1{)}^t=\left(\begin{array}{cc}-2& 0\\ 4& 2\end{array}\right),A(-1{)}^{-1}=\left(\begin{array}{cc}-1& 2\\ 0& 1\end{array}\right)$$

$\bullet$ Assemble $f(A(-1))$:

$$f(A(-1))=\left(\begin{array}{cc}0& -2\\ 4& 2\end{array}\right),\det =0-(-8)=8$$

$\bullet$ Multiply the two determinants:

$$\det \text{\hspace{0.17em}⁣}(f(A(1))\cdot f(A(-1)))=(-20)(8)=-160$$

$\int \text{\hspace{0.17em}⁣}({\displaystyle \frac{1}{x}}-{\displaystyle \frac{1}{x+1}})dx=\ln \mathrm{\mid }x\mathrm{\mid }-\ln \mathrm{\mid }x+1\mathrm{\mid }+C=\ln \text{\hspace{0.17em}⁣}\mid {\displaystyle \frac{x}{x+1}}\mid +C$.

${10}^3-10=990$, and $990\mathrm{/}6=165$. The induction proof factors $n^3-n=(n-1)n(n+1)$, which is the product of three consecutive integers and is therefore divisible by both $2$ and $3$, hence by $6$.