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3 years ago

8

Evaluate the expression 2⋅(3^6⋅3^−5) a) 2/3 b) 3 c) 6 d) 18

2 answers:

frosja888 [35]3 years ago

6 0

D be it’s all the great number for it

PSYCHO15rus [73]3 years ago

5 0

2(729 • 1/243) then 2(3) is 6 so C is the correct answer

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F(x)= -2(x-5)^2+8

geniusboy [140]

I don’t know all of it but....
Axis of symmetry is x=5
Vertex changed to (5,8)
Parabola opens down

4 0

3 years ago

Please help me !!!!!!

cestrela7 [59]

Given:

The set of pair and graphs.

To find:

The domain and range.

Solution:

We know that,

Domain is the set of x-values or input values.

Range is the set of y-values or output values.

(a)

The given set of ordered pairs is

{(-3,3),(5,5),(-3,2),(5,3)}

Here, the x-coordinates are -3, 5, -3, 5.

A set contains distinct values.

Therefore, the domain is {-3,5}.

(b)

The graph is given.

From the given graph the set of ordered pairs is

{(-2,1),(-1,0.5),(-1,3),(0,0),(0,2),(1,0.5),(1,3)(2,1)}

Here, the y-values are 1, 0.5, 3, 0, 2, 0.5, 3, 1.

Therefore, the range is {0, 0.5, 1, 2, 3}.

(c)

The graph is given.

From the given graph the set of ordered pairs is

{(-2,3),(-1,3),(0,1),(2,4)}

Here, the x-values are -2, -1,0, 2.

Therefore, the domain is {-2,-1,0,2}.

4 0

3 years ago

Consider the figure.<br> What is JL?

lisov135 [29]

The answer for the above mentioned problem is JL = 12.5

Step by step explanation:

Given:

JM = 8

KM = 6

To Find:

JL = ?

Formula to be used:

$KM^2$ = JM x ML

In order to find " JL" we must first find "ML",

$KM^2$ = JM x ML

$6^2$ = 8 x ML

36 = 8 x ML

36/8 = ML

ML = 4.5

Now JL = 4.5+8

= 12.5

Thus the value of JL = 12.5

3 0

3 years ago

Sort the ratios listed at the right into bins so that equivalent ratios are grouped together:

cestrela7 [59]

Step $1$

<u>Find the irreducible fraction in each ratio</u>

<u>case 1)</u> $\frac{40}{64}$

Divide by $8$ boths numerator and denominator

$\frac{40}{64}=\frac{5}{8}$

<u>case 2)</u> $\frac{5}{75}$

Divide by $5$ boths numerator and denominator

$\frac{5}{75}=\frac{1}{15}$

<u>case 3)</u> $\frac{14}{35}$

Divide by $7$ boths numerator and denominator

$\frac{14}{35}=\frac{2}{5}$

<u>case 4)</u> $\frac{24}{64}$

Divide by $8$ boths numerator and denominator

$\frac{24}{64}=\frac{3}{8}$

<u>case 5)</u> $\frac{6}{15}$

Divide by $3$ boths numerator and denominator

$\frac{6}{15}=\frac{2}{5}$

<u>case 6)</u> $\frac{65}{104}$

Divide by $13$ boths numerator and denominator

$\frac{65}{104}=\frac{5}{8}$

<u>case 7)</u> $\frac{66}{176}$

Divide by $22$ boths numerator and denominator

$\frac{66}{176}=\frac{3}{8}$

<u>case 8)</u> $\frac{12}{30}$

Divide by $6$ boths numerator and denominator

$\frac{12}{30}=\frac{2}{5}$

<u>case 9)</u> $\frac{15}{225}$

Divide by $15$ boths numerator and denominator

$\frac{15}{225}=\frac{1}{15}$

<u>case 10)</u> $\frac{6}{16}$

Divide by $2$ boths numerator and denominator

$\frac{6}{16}=\frac{3}{8}$

<u>case 11)</u> $\frac{15}{24}$

Divide by $3$ boths numerator and denominator

$\frac{15}{24}=\frac{5}{8}$

<u>case 12)</u> $\frac{48}{128}$

Divide by $16$ boths numerator and denominator

$\frac{48}{128}=\frac{3}{8}$

Step $2$

<u>Sort the ratios into bins</u>

1<u>) First Bin</u>

<u> $Ratio=\frac{5}{8}$ </u>

$\frac{40}{64}$

$\frac{65}{104}$

$\frac{15}{24}$

<u>2) Second Bin </u>

<u> $Ratio=\frac{1}{15}$ </u>

$\frac{5}{75}$

$\frac{15}{225}$

<u>3) Third Bin</u>

$Ratio=\frac{2}{5}$

$\frac{14}{35}$

$\frac{6}{15}$

$\frac{12}{30}$

4<u>) Fourth Bin</u>

<u> $Ratio=\frac{3}{8}$ </u>

$\frac{24}{64}$

$\frac{66}{176}$

$\frac{6}{16}$

$\frac{48}{128}$

3 0

3 years ago

Read 2 more answers

Use the surface integral in Stokes' Theorem to calculate the circulation of the field Bold Upper F equals x squared Bold i plus

Tom [10]

Stokes' theorem says the integral of the curl of $\vec F$ over a surface $S$ with boundary $C$ is equal to the integral of $\vec F$ along the boundary. In other words, the flux of the curl of the vector field is equal to the circulation of the field, such that

$\displaystyle\iint_S\nabla\times\vec F\cdot\mathrm d\vec S=\int_C\vec F\cdot\mathrm d\vec r$

We have

$\vec F(x,y,z)=x^2\,\vec\imath+4x\,\vec\jmath+z^2\,\vec k$

$\implies\nabla\times\vec F(x,y,z)=4\,\vec k$

Parameterize the ellipse $S$ by

$\vec s(u,v)=\dfrac{u\cos v}{\sqrt5}\,\vec\imath+\dfrac{u\sqrt5\sin v}4\,\vec\jmath$

with $0\le u\le1$ and $0\le v\le2\pi$ .

Take the normal vector to $S$ to be

$\dfrac{\partial\vec s}{\partial\vec u}\times\dfrac{\partial\vec s}{\partial\vec v}=\dfrac u4\,\vec k$

Then the flux of the curl is

$\displaystyle\iint_S4\,\vec k\cdot\dfrac u4\,\vec k\,\mathrm dA=\int_0^{2\pi}\int_0^1u\,\mathrm du\,\mathrm dv=\boxed{\pi}$

4 0

3 years ago