3/5=x/15 help 0<br>please

Can a triangle be formed with the side lengths of 8 cm, 4 cm, and 12 cm

julia-pushkina [17]

Yes it can but not 4cm only 8cm or 12cm

5 0

2 years ago

Read 2 more answers

The vertex of a parabola is (−5,2) , and its focus is (−5,6) .

Lubov Fominskaja [6]

Answer:

Step-by-step explanation:

If you plot the focus and the vertex, you see that the focus is on the same vertical line, just 4 units up. Since the vertex is below the focus and a parabola ALWAYS wraps around the focus, the parabola is a positive y equals x-squared type. Depending upon what you call standard form will dictate how your answer "looks", although they are both the same parabola. There are 2 forms:

$(x-h)^2=4p(y-k)$ and

$ax^2+bx+c=y$

We will work on the first one, then rewrite it into the second one.

Our h value is -5, our k value is 2 (from the vertex (h,k)), and p is defined as the distance between the focus and the vertex. Our p is 4. Filling all that in:

$(x+5)^2=16(y-2)$

That's one form. If we expand the left side of that form we have

$x^2+10x+25=16(y-2)$ Now divide both sides by 16 to get

$\frac{1}{16}x^2+\frac{5}{8}x+\frac{25}{16}=y-2$

Now add 2 to both sides in the form 32/16 to get

$\frac{1}{16}x^2+\frac{5}{8}x+\frac{57}{16}=y$

They are both the same parabola; pick whichever one fits your needs.

4 0

2 years ago

A coral reef is losing 7% of its area annually. Create a table of percentage calculations for 10 years. If the coral reef starts

Genrish500 [490]

7 0

2 years ago

Evaluate the integral x^2(x^3+9)^1/2

CaHeK987 [17]

$\bf \displaystyle \int x^2(x^3+9)^{\frac{1}{2}}\cdot dx\\\\
-------------------------------\\\\
u=x^3+9\implies \cfrac{du}{dx}=3x^2\implies \cfrac{du}{3x^2}=dx\\\\
-------------------------------\\\\$

$\bf \displaystyle \int \underline{x^2}(u)^{\frac{1}{2}}\cdot \cfrac{du}{3\underline{x^2}}\implies \int \cfrac{u^{\frac{1}{2}}}{3}\cdot du\implies \cfrac{1}{3}\int u^{\frac{1}{2}}\cdot du
\\\\\\
\cfrac{1}{3}\cdot \cfrac{u^{\frac{3}{2}}}{\frac{3}{2}}\implies \cfrac{1}{3}\cdot \cfrac{2\sqrt{u^3}}{3}\implies \cfrac{2\sqrt{u^3}}{9}\implies \cfrac{2\sqrt{(x^3+9)^3}}{9}+C$

5 0

3 years ago

Consider the sequence defined recursively by an+1 = ( an − 1 if an ≥ 10 2an if an < 10 )

densk [106]

Answer:

a) a₁ = 14, a₂ = 13, a₃ = 12, a₄ = 11

b) an+4 = 16

c) Does not exist

Step-by-step explanation:

The last digit in student number is given as 7.

a) a₀ = 7

Since an < 10 we use 2an

Therefore a₁ = a₀₊₁ = 2 × a₀ = 2 × 7 = 14

a₂ = a₁₊₁ = a₁ - 1 = 14 -1 = 13

a₃ = a₂₊₁ = a₂ - 1 = 13 -1 = 12

a₄ = a₃₊₁ = a₃ - 1 = 12 -1 = 11

b) an = 1, we have an+1 = 2an

Therefore an+2 = an+1+1 = 2 × 2 = 4

an+3 = an+2+1 = 2 × 4 = 8

an+4 = an+3+1 = 2 × 8 = 16

Therefore, an+4 = 16

c) If a₀ = 3, therefore a₁ = a₀₊₁ = 2×3 = 6

a₂ = a₁₊₁ = 2×6 = 12

a₃ = a₂₊₁ = a₂₋₁ = 12- 1 =11

a₄ = a₃₊₁ = a₃₋₁ = 11- 1 =10

a₅ = a₄₊₁ = a₄₋₁ = 10- 1 = 9

a₆ = 2×9 = 18

We can therefore see that limn→[infinity] does not exist.

3 0

3 years ago

3/5=x/15 help 0please