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creativ13 [48]
3 years ago
7

HELP !!! Evaluate f(-2) for the function f(x) = 3x2 + 2x - 5

Mathematics
1 answer:
Phoenix [80]3 years ago
4 0

Answer:

3

Step-by-step explanation:

So we have the function:

f(x)=3x^2+2x-5

To evaluate the value of f(-2), substitute -2 for x. Thus:

f(-2)=3(-2)^2+2(-2)-5

Square first:

f(-2)=3(4)+2(-2)-5

Multiply:

f(-2)=12-4-5

Now, subtract:

f(-2)=8-5=3

So:

f(-2)=3

And we're done!

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HELP ILL AWARD BRAINLIEST
inna [77]

Answer:

Think about it

Step-by-step explanation:

If you made $20 in 2 hrs you made $10 per hour

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The equation is y = 10x + 0 <-- this is also answer to number 4

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5 0
4 years ago
Read 2 more answers
Find the slant height of a square pyramid with base edges of 12 cm and an altitude of 8 cm?
liberstina [14]
To find the slant height we must take apart the pyramid first. Let us cut it in half. There we can easily see that the slant height is really just the hypotenuse of the triangle formed by half the base and the altitude.

Half the base length would be 6 cm.

Using the Pythagorean therom:

a² + b² = c²
6² + 8² = c²
36 + 64 = c²
100 = c²
c = 10

The slant height should be 10 cm. Hope this helps!
8 0
4 years ago
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Find two unit vectos that are orthogonal to both [0,1,2] and [1,-2,3]
alekssr [168]

Answer:

Let the vectors be

a = [0, 1, 2] and

b = [1, -2, 3]

( 1 ) The cross product of a and b (a x b) is the vector that is perpendicular (orthogonal) to a and b.

Let the cross product be another vector c.

To find the cross product (c) of a and b, we have

\left[\begin{array}{ccc}i&j&k\\0&1&2\\1&-2&3\end{array}\right]

c = i(3 + 4) - j(0 - 2) + k(0 - 1)

c = 7i + 2j - k

c = [7, 2, -1]

( 2 ) Convert the orthogonal vector (c) to a unit vector using the formula:

c / | c |

Where | c | = √ (7)² + (2)² + (-1)²  = 3√6

Therefore, the unit vector is

\frac{[7,2,-1]}{3\sqrt{6} }

or

[ \frac{7}{3\sqrt{6} } , \frac{2}{3\sqrt{6} } , \frac{-1}{3\sqrt{6} } ]

The other unit vector which is also orthogonal to a and b is calculated by multiplying the first unit vector by -1. The result is as follows:

[ \frac{-7}{3\sqrt{6} } , \frac{-2}{3\sqrt{6} } , \frac{1}{3\sqrt{6} } ]

In conclusion, the two unit vectors are;

[ \frac{7}{3\sqrt{6} } , \frac{2}{3\sqrt{6} } , \frac{-1}{3\sqrt{6} } ]

and

[ \frac{-7}{3\sqrt{6} } , \frac{-2}{3\sqrt{6} } , \frac{1}{3\sqrt{6} } ]

<em>Hope this helps!</em>

7 0
3 years ago
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