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2 years ago

Solve the equation by factoring. z2 − 6z − 27 = 0

1 answer:

7 0

The answer is x= 9 and -3

You must fine 2 number that multiply to equal -27 and add to equal -6

(x-9) (x+3)
Then equal the parts of the binomial to zero

x - 9 = 0
x + 3 = 0

You must get x by its self.
When you do so, you'll have the answer as -3 and 9

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What is the perimeter for 3f+3f+8+8+2f-1

Montano1993 [528]

Answer:

To find the Perimeter just add all the numbers

=3f+3f+8+8+2f-1

=8f+15

5 0

3 years ago

Read 2 more answers

Which function has a domain of x 25 and a range of y s3?

lions [1.4K]

Answer:

Step-by-step explanation:

The function has a domain x ≥ 5.

This is because the function remains real for (x - 5) ≥ 0 as negative within the square root is imaginary.

Hence, (x - 5) ≥ 0

⇒ x ≥ 5

Now, for all x values that are greater than equal to 5 the value of will be negative.

So,

⇒

⇒ y ≤ 3

Therefore, the range of the function is y ≤ 3. (Answer)

Note: Thanks rani

3 0

2 years ago

Suppose the sediment density (g/cm3 ) of a randomly selected specimen from a certain region is known to have a mean of 2.80 and

Irina18 [472]

Answer:

0.918 is the probability that the sample average sediment density is at most 3.00

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 2.80

Standard Deviation, σ = 0.85

Sample size,n = 35

We are given that the distribution of sediment density is a bell shaped distribution that is a normal distribution.

Formula:

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

Standard error due to sampling:

$=\dfrac{\sigma}{\sqrt{n}} = \dfrac{0.85}{\sqrt{35}} = 0.1437$

P(sample average sediment density is at most 3.00)

$P( x \leq 3.00) = P( z \leq \displaystyle\frac{3.00 - 2.80}{0.1437}) = P(z \leq 1.3917)$

Calculation the value from standard normal z table, we have,

$P(x \leq 3.00) = 0.918$

0.918 is the probability that the sample average sediment density is at most 3.00

4 0

2 years ago

The sum of the diagonals of a rhombus is 5√2.

Alex

Greetings!
Let ABCD be a rhombus
AC + BD = 5√2 cm

Area of ABCD = Ar△ABD + Ar △BCD
$= \frac{1}{2} \times BD \times AO + \frac{1}{2} \times BD \times OC$
$= \frac{1}{2} BD (AO + OC)$
$\frac{1}{2} BD \times AC$

$So, \frac{ BD \times AC}{2} = 4 \: cm {}^{2}$
$BD \times AC = 8$
$Now, AC + \: BD = 5 \sqrt{2}$

Squaring both sides, we get
$AC {}^{2} + BD {}^{2} + 2 AC.BD =50$
$AC {}^{2} + BD {}^{2} + 2 \times 8 = 50$
$AC {}^{2} + BD {}^{2} = 50 - 16$
$AC {}^{2} + BD {}^{2} = 34$

In △AOB, we have
$OA {}^{2} + OB {}^{2} =AB {}^{2}$
$( \frac{ AC }{2} ) {}^{2} + ( \frac{ BD}{2} ) {}^{2} = AB {}^{2}$
$\frac{ AC {}^{2} } {4} + \frac{ BD {}^{2} }{4} = AB {}^{2}$
$AC {}^{2} + BD {}^{2} = 4 AB {}^{2}$
$34 = 4 AB {}^{2}$

Square rooting both sides
$\sqrt{34} = 2 AB$
$Perimeter = 4 \: AB \\ = 2 \times 2 \: AB \\ = 2 \: \times \sqrt{34 } \\ = 2 \sqrt{34 \: } units$ .

Hope it helps!

7 0

3 years ago

What is the volume of the triangular right prism?

Minchanka [31]

Answer:D

Step-by-step explanation:

5 0

1 year ago