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4 years ago

F(x)= x^2 -3 find f(-2)

Mathematics

2 answers:

mel-nik [20]4 years ago

8 0

Answer:

Step-by-step explanation:

F(x) =2 when x^2-3

(2)^2-3=4-3=1

gregori [183]4 years ago

5 0

Use the substitution method

F(-2)=x^2-3

=(-2)^2-3

Use PEMDAS

P= parenthesis

E= exponents

M=multiplication

D=division

A=addition

S=subtraction

Do (-2)^2 first then -3

Put down (-2)(-2) two times because of the exponent

-negative number * -negative number= +positive number

4-3

Answer: 1

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___ can be found by making x=0

vfiekz [6]

Answer:

y intercept

Step-by-step explanation:

At first I didn't think there was enough information, but there is. In general terms, for any polynomial which is also a function, making x = 0 gives the point where (0,b) = the y intercept.

For example

using f(x) = x^2 + 4x + 6 when x = 0 the result left is f(0) = 6 so the y intercept is (0,6)

If you have x^2 + 6x the y intercept is (0,0)

4 0

3 years ago

Help please ?.......

algol13

Answer:

Step-by-step explanation:

It is simply the sumation of a triangle and a rectangle minus a half circle.

area triangle + area rectangle - area half circle = Total area

1/2( bh ) + bh - (π (d/2)²) / 2 = Total area

0.5(7)(6) + (10)(7) - (π (7/2)² ) / 2 = Total area cm²

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7 0

4 years ago

A study of long-distance phone calls made from General Electric Corporate Headquarters in Fairfield, Connecticut, revealed the l

Anna [14]

Answer:

a) 0.4452

b) 0.0548

c) 0.0501

d) 0.9145

e) 6.08 minutes or greater

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 4.7 minutes

Standard Deviation, σ = 0.50 minutes.

We are given that the distribution of length of the calls is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

a) P(calls last between 4.7 and 5.5 minutes)

$P(4.7 \leq x \leq 5.5) = P(\displaystyle\frac{4.7 - 4.7}{0.50} \leq z \leq \displaystyle\frac{5.5-4.7}{0.50}) = P(0 \leq z \leq 1.6)\\\\= P(z \leq 1.6) - P(z$

$P(4.7 \leq x \leq 5.5) = 44.52\%$

b) P(calls last more than 5.5 minutes)

$P(x > 5.5) = P(z > \displaystyle\frac{5.5-4.7}{0.50}) = P(z > 1.6)\\\\P( z > 1.6) = 1 - P(z \leq 1.6)$

Calculating the value from the standard normal table we have,

$1 - 0.9452 = 0.0548 = 5.48\%\\P( x > 5.5) = 5.48\%$

c) P( calls last between 5.5 and 6 minutes)

$P(4.7 \leq x \leq 5.5) = P(\displaystyle\frac{5.5 - 4.7}{0.50} \leq z \leq \displaystyle\frac{6-4.7}{0.50}) = P(1.6 \leq z \leq 2.6)\\\\= P(z \leq 2.6) - P(z$

$P(5.5 \leq x \leq 6) = 5.01\%$

d) P( calls last between 4 and 6 minutes)

$P(4 \leq x \leq 6) = P(\displaystyle\frac{4 - 4.7}{0.50} \leq z \leq \displaystyle\frac{6-4.7}{0.50}) = P(-1.4 \leq z \leq 2.6)\\\\= P(z \leq 2.6) - P(z$