A triangle har a perimetr of 239cm. If the 2nd side is thrice the length of the 1st side and the 3rd side is 6 less than the 2nd side. Find the length of all sides.
1st side =

Yulia
2020-11-26
Answered

You can still ask an expert for help

Jayden-James Duffy

Answered 2020-11-27
Author has **91** answers

Let x be the length of the 1st side so that 3x is the length of the 2nd side and $3x-63x-6$ is the length of the 3rd side. The perimeter is 239 cm so we can write:

$x+3x+(3x-6)=239$

Solve for x:

$7x-6=239$

$x=35$

So, the 1st side measures 35 cm, the 2nd side measures$3\left(35\right)=105cm,$ and the 3rd side measures $3\left(35\right)-6=99cm.$

Solve for x:

So, the 1st side measures 35 cm, the 2nd side measures

asked 2022-06-13

Proving that for each prime number $p$, the number $\sqrt{p}$ is irrational.

asked 2021-08-15

Make the first 6 terms of the following sequence

${d}_{1}=1,\text{}{d}_{2}=1,\text{}\text{and}\text{}{d}_{n}={\left({d}_{n-1}\right)}^{2}+{d}_{n-2}\text{}\text{for}\text{}n\ge 3$

asked 2022-04-22

Solve

$2-\frac{2x-1}{x}\le \frac{2}{2x+1}$

asked 2022-04-05

Drawing two perpendicular tangent line from the origin to $y={x}^{2}-2x+a$ .

We drew two perpendicular tangent line from the origin (the point (0,0) ) to the curve$y={x}^{2}-2x+a$ , what is the value of a?

1)$\frac{5}{4}$

2)$\frac{-5}{4}$

3)$\frac{3}{4}$

4)$\frac{-3}{4}$

We drew two perpendicular tangent line from the origin (the point (0,0) ) to the curve

1)

2)

3)

4)

asked 2022-04-04

What is $4\sqrt{3}$ minus $4\sqrt{3}$?

asked 2022-07-28

Find the domain of the function $g(u)=\sqrt{u}+\sqrt{4-u}$

asked 2022-05-27

System of equations:

$\dot{x}=x-y-x({x}^{2}+{y}^{2})+\frac{xy}{\sqrt{{x}^{2}+{y}^{2}}}\phantom{\rule{0ex}{0ex}}\dot{y}=x+y-y({x}^{2}+{y}^{2})-\frac{{x}^{2}}{\sqrt{{x}^{2}+{y}^{2}}}$

How can I get this in polar coordinates? I know that ${r}^{2}={x}^{2}+{y}^{2}$, but how can I find $\dot{r}$ or $\dot{\theta}$?

$\dot{x}=x-y-x({x}^{2}+{y}^{2})+\frac{xy}{\sqrt{{x}^{2}+{y}^{2}}}\phantom{\rule{0ex}{0ex}}\dot{y}=x+y-y({x}^{2}+{y}^{2})-\frac{{x}^{2}}{\sqrt{{x}^{2}+{y}^{2}}}$

How can I get this in polar coordinates? I know that ${r}^{2}={x}^{2}+{y}^{2}$, but how can I find $\dot{r}$ or $\dot{\theta}$?