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Alex777 [14]
3 years ago
11

Can someone please help me with these 2 questions?

Mathematics
1 answer:
Aleonysh [2.5K]3 years ago
5 0

Answer:

1. The correct answer is graph b

2. the correct answer is t>100

Step-by-step explanation:


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Step-by-step explanation:

8 0
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3 3/4 + 2 1/8 = (make sure you simplify the fraction)
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Read 2 more answers
A geometric sequence is defined by the general term tn = 75(5n), where n ∈N and n ≥ 1. What is the recursive formula of the sequ
andreyandreev [35.5K]
The correct answer is C) t₁ = 375, t_n=5t_{n-1}.

From the general form,
t_n=75(5)^n, we must work backward to find t₁.

The general form is derived from the explicit form, which is
t_n=t_1(r)^{n-1}.  We can see that r = 5; 5 has the exponent, so that is what is multiplied by every time. This gives us

t_n=t_1(5)^{n-1}

Using the products of exponents, we can "split up" the exponent:
t_n=t_1(5)^n(5)^{-1}

We know that 5⁻¹ = 1/5, so this gives us
t_n=t_1(\frac{1}{5})(5)^n
\\
\\=\frac{t_1}{5}(5)^n

Comparing this to our general form, we see that
\frac{t_1}{5}=75

Multiplying by 5 on both sides, we get that
t₁ = 75*5 = 375

The recursive formula for a geometric sequence is given by
t_n=t_{n-1}(r), while we must state what t₁ is; this gives us

t_1=375; t_n=t_{n-1}(5)

3 0
3 years ago
Evaluate the integral e^xy w region d xy=1, xy=4, x/y=1, x/y=2
LUCKY_DIMON [66]
Make a change of coordinates:

u(x,y)=xy
v(x,y)=\dfrac xy

The Jacobian for this transformation is

\mathbf J=\begin{bmatrix}\dfrac{\partial u}{\partial x}&\dfrac{\partial v}{\partial x}\\\\\dfrac{\partial u}{\partial y}&\dfrac{\partial v}{\partial y}\end{bmatrix}=\begin{bmatrix}y&x\\\\\dfrac1y&-\dfrac x{y^2}\end{bmatrix}

and has a determinant of

\det\mathbf J=-\dfrac{2x}y

Note that we need to use the Jacobian in the other direction; that is, we've computed

\mathbf J=\dfrac{\partial(u,v)}{\partial(x,y)}

but we need the Jacobian determinant for the reverse transformation (from (x,y) to (u,v). To do this, notice that

\dfrac{\partial(x,y)}{\partial(u,v)}=\dfrac1{\dfrac{\partial(u,v)}{\partial(x,y)}}=\dfrac1{\mathbf J}

we need to take the reciprocal of the Jacobian above.

The integral then changes to

\displaystyle\iint_{\mathcal W_{(x,y)}}e^{xy}\,\mathrm dx\,\mathrm dy=\iint_{\mathcal W_{(u,v)}}\dfrac{e^u}{|\det\mathbf J|}\,\mathrm du\,\mathrm dv
=\displaystyle\frac12\int_{v=}^{v=}\int_{u=}^{u=}\frac{e^u}v\,\mathrm du\,\mathrm dv=\frac{(e^4-e)\ln2}2
8 0
3 years ago
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