Daily math problems for October 17, 2026 — Fundamental Algebra, Calculus, Algebra & more

$\text{HCF}(6,9)=3$, so $6x+9=3(2x+3)$.

Average rate: ${\displaystyle \frac{f(4)-f(0)}{4-0}}={\displaystyle \frac{16-0}{4}}=4$.

$f^{\mathrm{\prime }}(x)=2x$, so $f^{\mathrm{\prime }}(c)=4⟹2c=4⟹c=2$.

$2\in (0,4)$ ✓.

$(x+2)(x-5)=x^2-5x+2x-10=x^2-3x-10$.

Set $N=1200$ and divide by $400$:

$$1200=400e^{0.5t}$$

$$e^{0.5t}=3$$

Take natural logs of both sides:

$$0.5t=\ln 3$$

$$t=2\ln 3$$

So $t=2\ln 3\approx 2.20$ hours.

Substituting $x=2$ gives the indeterminate form $$\frac{2^2-5(2)+6}{2^2-4}=\frac{0}{0}\mathrm{.}$$ Factor both numerator and denominator and cancel the shared $(x-2)$: $$\frac{x^2-5x+6}{x^2-4}=\frac{(x-2)(x-3)}{(x-2)(x+2)}=\frac{x-3}{x+2}\mathrm{.}$$ Now substitute $x=2$ into the simplified form: $$\underset{x\to 2}{\lim }\frac{x-3}{x+2}=\frac{2-3}{2+2}=-\frac{1}{4}\mathrm{.}$$

$a(t)=x^{\mathrm{\prime }\mathrm{\prime }}(t)=6t-12$. At $t=1$: $a=-6$ m/s².

${\int }_0^3{x}^{2}dx={\displaystyle \frac{27}{3}}=9$.

$\int x{e}^{x}dx=x{e}^x-\int e^{x}dx=(x-1)e^x+C$. Evaluating from $0$ to $1$: $(0)\cdot e-(-1)\cdot 1=0-(-1)=1$.

The fifth roots of unity sum to $0$. Subtracting the root $1$ leaves a sum of $-1$, and taking real parts gives $\cos \frac{2\pi }{5}+\cos \frac{4\pi }{5}+\cos \frac{6\pi }{5}+\cos \frac{8\pi }{5}=-1$.

The integral of a non-negative continuous function over $[a,b]$ is non-negative (representing the area under the curve).