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sattari [20]
3 years ago
12

Solving Exponential and Logarithmic Equation In exercise,solve for x or t.See example 5 and 6.

Mathematics
1 answer:
lianna [129]3 years ago
8 0

Answer:

The solution isx = 0.3863

Step-by-step explanation:

The first step to solve this equation is placing everything with the exponential to one side of the equality, and everything without the exponential to the other side. So

e^{x+1} = 4

We already have this, so the first part is ok.

Now, the ln is the interse operation to the exponential, so we apply the ln to both sides of the equality.

\ln{e^{x+1}} = \ln{4}

x + 1 = 1.3863

x = 1.3863 - 1

x = 0.3863

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The sides of a square field are 12 meters. A sprinkler in the center of the field sprays a circular area with a diameter that co
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3 years ago
The price of Philip's favorite candy has been reduced by $0.15 per piece. Philip buys 14 pieces and spends $3.22.
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6 0
3 years ago
Read 2 more answers
Let
Maksim231197 [3]

Answer:

3.3

Step-by-step explanation:

We determine the plane formed by u_1 and u_2. The normal to the plane is given by their cross product:

u_1\times u_2

\begin{pmatrix} - 2\\ - 4\\ 1 \end{pmatrix}\times\begin{pmatrix} - 4\\ 1 \\ - 4\end{pmatrix}=\begin{pmatrix} 15 \\ - 12\\ - 18\end{pmatrix}

The equation of the plane is then given by

15x-12y-18z=0

The distance between a vector \begin{pmatrix} x_1 \\ y_1 \\ z_1 \end{pmatrix} and a plane Ax+By+Cz +D =0 is given by

d=\dfrac{|Ax_1+By_1+Cz_1 +D|}{\sqrt{A^2+B^2+C^2}}

Comparing,

A = 15,\ B = - 12,\ C = -18,\ D = 0,\ x_1=3,\ y_1 =-8, \ z_1 =3

Substituting,

d=\dfrac{|(15\times3)+(-12\times-8)+(-18\times3)+0|}{\sqrt{15^2+(-12)^2+(-18)^2}}

d=\dfrac{|45+96-54|}{\sqrt{225+144+324}}=\dfrac{87}{\sqrt{693}}

d =\dfrac{87}{26.32\ldots}\approx 3.3

4 0
3 years ago
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