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Shalnov [3]
3 years ago
11

Need help Math Question Multiple Choice!

Mathematics
1 answer:
erica [24]3 years ago
8 0

Given :

A function f(x) for different range of x.

To Find :

The value of f( -3 ) .

Solution :

We have to find the value of function at x = -3.

Now, in the given figure -3 lies in range x ≤ -2 and definition of function at that range is :

f(x) = \dfrac{2}{x}

So, putting value of x = -3 in above equation, we get :

f(-3) = \dfrac{2}{-3} \\\\f(-3) = \dfrac{-2}{3}

Hence, this is the required solution.

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Which table of values will generate this graph? On a coordinate plane, points are at (negative 2, 0), (0, 1), (0, negative 4), a
zhuklara [117]

Answer:

Option D.

Step-by-step explanation:

The given points are (-2,0), (0,1), (0,-4) and (3,0).

In each ordered pair first element is x-coordinate and second is y-coordinate.

For 4 points, the table must have 2 columns and 4 rows.

So, the required table of values is

x         y

-2        0

0       -4

0        1

3        0

Therefore, the correct option is D.

5 0
4 years ago
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Approximate square root of 39 to the nearest whole number
Dominik [7]
It is about 6.24 so it rounds to 6
5 0
3 years ago
Which expression is equivalent to √48x5, if x> 0?
Lerok [7]

Answer:

C. 4x²√(3x)

Step-by-step explanation:

if x > 0

\sqrt{48x^{5}} =\sqrt{48} \times \sqrt{x^{5}}

          =\sqrt{16\times 3} \times \sqrt{x^{4}\times x}

          =\sqrt{16} \times \sqrt{3} \times \sqrt{x^{4}} \times \sqrt{x}

          =4\times \sqrt{3} \times \sqrt{\left( x^{2}\right)^{2}  } \times \sqrt{x}

          =4\times \sqrt{3} \times x^{2}\times \sqrt{x}

          =4\times x^{2} \times  \sqrt{3} \times \sqrt{x}

          =4\times x^{2}\times \sqrt{3x}

8 0
2 years ago
1. If A = {a, b, c, d}, B={c, d, e, f}, C={x, y, z} find (A-B)​
vitfil [10]

Answer

here

A={a,b,c,d}

B={c,d,e,f}

C={x,y,z}

A-B={e,f}

8 0
1 year ago
Use the Law of Sines to find the measure of angle J to the nearest degree.
rjkz [21]
\bf \textit{Law of sines}
\\ \quad \\
\cfrac{sin(\measuredangle A)}{a}=\cfrac{sin(\measuredangle B)}{b}=\cfrac{sin(\measuredangle C)}{c}\\\\
-----------------------------\\\\
\cfrac{sin(J)}{9.1}=\cfrac{sin(97^o)}{11}\implies sin(J)=\cfrac{9.1\cdot sin(97^o)}{11}
\\\\\\
\textit{now taking }sin^{-1}\textit{ to both sides}
\\\\\\
sin^{-1}\left[ sin(J) \right]=sin^{-1}\left( \cfrac{9.1\cdot sin(97^o)}{11} \right)
\\\\\\
\measuredangle J=sin^{-1}\left( \cfrac{9.1\cdot sin(97^o)}{11} \right)
8 0
4 years ago
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