What is the greatest common factor of 3062, 8062, and 2013?

Yanka [14]

The question is too general....but it could be acute, obtuse

3 0

3 years ago

A solid oblique pyramid has an equilateral triangle as a base with an edge length of 4 cm and an area of 12 cm2. What is the vol

Pavel [41]

Solution:

Let me define pyramid first.A pyramid is a Polyhedron , whose base can be any polygon and it's faces are triangular which meet at a point called vertex.

Volume of Pyramid = $\frac{1}{2}\times{\text{Area of Base}\times{\text{Height}$

Side of base which is an equilateral triangle= 4 cm

Area of whole pyramid = 12 square cm

Area of Equilateral triangle having side 4 cm= $\frac{\sqrt3}{4}\times(Side)^2=\frac{\sqrt3}{4}\times(4)^2=4 \sqrt3$ cm²

So, Area of three triangle which are faces of pyramid = (12 - 4 √3)cm²

Area of triangle = $\frac{1}{2} \times (Base)\times(Height)$

12 - 4 √3= $\frac{1}{2} \times (4)\times(Height)$

Height = $\frac{12-4\sqrt3}{2}$

Height=(6-2√3) cm

Volume of pyramid = Area of Equilateral triangle which is it's Base × Height of Pyramid

=4√3 × (6-2√3) cm³

=(24 √3 -24)

= 24 (1.732-1)

=24 × 0.732

=17.568 cm³

8 0

3 years ago

Read 2 more answers

Which three lengths could be the lengths of the sides of a triangle?

natali 33 [55]

Answer:

D. 15 cm, 9 cm, 24 cm

Step-by-step explanation:

The three lengths that could be the lengths of the sides of a triangle, must satisfy the following condition;

---sum of any two smaller sides chosen must be equal to the third side.

Test option A : 6 cm + 11 cm = 17 cm ( 17 cm ≠ 23 cm)

Test option B: 10 cm + 15 cm = 25 cm (25 cm ≠ 24 cm)

Test option C: 6 cm + 6 cm = 12 cm (12 cm ≠ 22 cm)

Test option D: 15 cm + 9 cm = 24 cm (24 cm = 24 cm) ---correct option.

Thus, the three lengths that could be the lengths of the sides of a triangle are 15 cm, 9 cm, and 24 cm

5 0

3 years ago

How much money should be invested at 5% interest, compounded quarterly, so that 7 years later the investment will be worth $5000

vampirchik [111]

$\bf \qquad \textit{Compound Interest Earned Amount}
\\\\
A=P\left(1+\frac{r}{n}\right)^{nt}
\quad 
\begin{cases}
A=\textit{accumulated amount}\to &\$5000\\
P=\textit{original amount deposited}\\
r=rate\to 5\%\to \frac{5}{100}\to &0.05\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{quarterly, four times}
\end{array}\to &4\\
t=years\to &7
\end{cases}
\\\\\\
5000=P\left(1+\frac{0.05}{4}\right)^{4\cdot 7}\implies 5000=P(1.0125)^{28}\\\\\\ \cfrac{5000}{(1.0125)^{28}}=P$

8 0

3 years ago

Factor the expressions below using the greatest

Elena-2011 [213]

Answer:

a=4m

b=63x¹2 - 35x6 maybe you formatted/typed this wrong, but this is impossible..?

c=4

3 0

2 years ago