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ser-zykov [4K]
3 years ago
7

What’s the vertex y=-1/2(x+3)^2-1

Mathematics
2 answers:
Gemiola [76]3 years ago
8 0
Use the vertex form,
y
=
a
(
x
−
h
)
2
+
k
y
=
a
(
x
-
h
)
2
+
k
, to determine the values of
a
a
,
h
h
, and
k
k
.
a
=
−
1
2
a
=
-
1
2
h
=
−
3
h
=
-
3
k
=
−
1
k
=
-
1
Find the vertex
(
h
,
k
)
(
h
,
k
)
.
(
−
3
,
−
1
)
Lena [83]3 years ago
7 0

Answer:

vertex = (- 3, - 1 )

Step-by-step explanation:

The equation of a parabola in vertex form is

y = a(x - h)² + k

where (h, k) are the coordinates of the vertex and a is a multiplier

y = - \frac{1}{2} (x + 3)² - 1 ← is in vertex form

with vertex = (- 3, - 1 )

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Answer:

2103 bags

Step-by-step explanation:

8412/4=2103

6 0
2 years ago
“encontrar la integral indefinida y verificar el resultado mediante derivación”
Oliga [24]

I=\displaystyle\int\frac x{(1-x^2)^3}\,\mathrm dx

Haz la sustitución:

y=1-x^2\implies\mathrm dy=-2x\,\mathrm dx

\implies I=\displaystyle-\frac12\int\frac{\mathrm dy}{y^3}=\frac1{4y^2}+C=\frac1{4(1-x^2)^2}+C

Para confirmar el resultado:

\dfrac{\mathrm dI}{\mathrm dx}=\dfrac14\left(-\dfrac{2(-2x)}{(1-x^2)^3}\right)=\dfrac x{(1-x^2)^3}

I=\displaystyle\int\frac{x^2}{(1+x^3)^2}\,\mathrm dx

Sustituye:

y=1+x^3\implies\mathrm dy=3x^2\,\mathrm dx

\implies I=\displaystyle\frac13\int\frac{\mathrm dy}{y^2}=-\frac1{3y}+C=-\frac1{3(1+x^3)}+C

(Te dejaré confirmar por ti mismo.)

I=\displaystyle\int\frac x{\sqrt{1-x^2}}\,\mathrm dx

Sustituye:

y=1-x^2\implies\mathrm dy=-2x\,\mathrm dx

\implies I=\displaystyle-\frac12\int\frac{\mathrm dy}{\sqrt y}=-\frac12(2\sqrt y)+C=-\sqrt{1-x^2}+C

I=\displaystyle\int\left(1+\frac1t\right)^3\frac{\mathrm dt}{t^2}

Sustituye:

u=1+\dfrac1t\implies\mathrm du=-\dfrac{\mathrm dt}{t^2}

\implies I=-\displaystyle\int u^3\,\mathrm du=-\frac{u^4}4+C=-\frac{\left(1+\frac1t\right)^4}4+C

Podemos hacer que esto se vea un poco mejor:

\left(1+\dfrac1t\right)^4=\left(\dfrac{t+1}t\right)^4=\dfrac{(t+1)^4}{t^4}

\implies I=-\dfrac{(t+1)^4}{4t^4}+C

4 0
3 years ago
Verify the following identity and show steps<br><br> (Cos2 θ)/(1+sin2 θ)= (cot θ-1)/(cot θ+1)
Paul [167]

Answer:

Verified below

Step-by-step explanation:

We want to show that (Cos2θ)/(1 + sin2θ) = (cot θ - 1)/(cot θ + 1)

In trigonometric identities;

Cot θ = cos θ/sin θ

Thus;

(cot θ - 1)/(cot θ + 1) gives;

((cos θ/sin θ) - 1)/((cos θ/sin θ) + 1)

Simplifying numerator and denominator gives;

((cos θ - sin θ)/sin θ)/((cos θ + sin θ)/sin θ)

This reduces to;

>> (cos θ - sin θ)/(cos θ + sin θ)

Multiply top and bottom by ((cos θ + sin θ) to get;

>> (cos² θ - sin²θ)/(cos²θ + sin²θ + 2sinθcosθ)

In trigonometric identities, we know that;

cos 2θ = (cos² θ - sin²θ)

cos²θ + sin²θ = 1

sin 2θ = 2sinθcosθ

Thus;

(cos² θ - sin²θ)/(cos²θ + sin²θ + 2sinθcosθ) gives us:

>> cos 2θ/(1 + sin 2θ)

This is equal to the left hand side.

Thus, it is verified.

5 0
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tamaranim1 [39]
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