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AlladinOne [14]
2 years ago
13

Study the solutions of the three equations on the right. Then, complete the statements below. There are two real solutions if th

e radicand is There is one real solution if the radicand is There are no real solutions if the radicand is 1. y = negative 16 x squared + 32 x minus 10. x = StartFraction negative 32 plus-or-minus StartRoot 384 EndRoot Over negative 32 EndFraction. 2. y = 4 x squared + 12 x + 9. x = StartFraction negative 12 plus-or-minus StartRoot 0 EndRoot Over 8 EndFraction. 3. y = 3x squared minus 5 x + 4. x = StartFraction 5 plus-or-minus StartRoot negative 23 EndRoot Over 6 EndFraction.
Mathematics
1 answer:
SashulF [63]2 years ago
6 0

Answer:

There are two real solutions if the radicand is

✔ positive.

There is one real solution if the radicand is

✔ zero.

There are no real solutions if the radicand is

✔ negative.

Step-by-step explanation:

ON MY DOG KIDS DIS RIGHT

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HELP PLEASE. A pizza shop charges customers based on the area of the pizza they order. The cost of a pizza in dollars can be rep
san4es73 [151]

For this case we have that the cost of each pizza is given by the following function:

S (r) = 0.1 \pi * r ^ 2

Where:

r: It's the radius of the pizza

IF a pizza cost $26.32 we have:

26.32 = 0.1 \pi * r ^ 2

We cleared the radius, using \pi = 3.14:

\frac {26.32} {0.1} = \pi * r ^ 2\\263.2 = 3.14r ^ 2\\\frac {263.2} {3.14} = r ^ 2\\83.82 = r ^ 2\\r = \pm \sqrt {83.82}

We choose the positive value:

r = 9.15

So, the radius of the pizza is 9.16 \ in

Then, the diameter is: 2r = 2 * 9.15 = 18.30 \ in

Finally, the diameter of the pizza is:18.30 \ in

Answer:

Option B

4 0
2 years ago
Read 2 more answers
(Ross 5.15) If X is a normal random variable with parameters µ " 10 and σ 2 " 36, compute (a) PpX ą 5q (b) Pp4 ă X ă 16q (c) PpX
pochemuha

Answer:

(a) 0.7967

(b) 0.6826

(c) 0.3707

(d) 0.9525

(e) 0.1587

Step-by-step explanation:

The random variable <em>X</em> follows a Normal distribution with mean <em>μ</em> = 10 and  variance <em>σ</em>² = 36.

(a)

Compute the value of P (X > 5) as follows:

P(X>5)=P(\frac{x-\mu}{\sigma}>\frac{5-10}{\sqrt{36}})\\=P(Z>-0.833)\\=P(Z

Thus, the value of P (X > 5) is 0.7967.

(b)

Compute the value of P (4 < X < 16) as follows:

P(4

Thus, the value of P (4 < X < 16) is 0.6826.

(c)

Compute the value of P (X < 8) as follows:

P(X

Thus, the value of P (X < 8) is 0.3707.

(d)

Compute the value of P (X < 20) as follows:

P(X

Thus, the value of P (X < 20) is 0.9525.

(e)

Compute the value of P (X > 16) as follows:

P(X>16)=P(\frac{x-\mu}{\sigma}>\frac{16-10}{\sqrt{36}})\\=P(Z>1)\\=1-P(Z

Thus, the value of P (X > 16) is 0.1587.

**Use a <em>z</em>-table for the probabilities.

8 0
2 years ago
Which statement and explanation about the product of a 3-digit number and a 2-digit number is correct?
Murrr4er [49]
The answer is D- 999*99 is 98,901 (which is 5 digits. It cannot go higher), but 100*10 is 1,000, so it can go lower.
8 0
2 years ago
Is it possible for an odd function to have the interval 0 infinity) as its domain?
mojhsa [17]

NO!

Prove by contradiction. First, define the the following terms:

odd function: f(-x) = -f(x)] and domain: all values of x

Proof: Suppose there f(-x) = -f(x) such that x is a negative integer, then f(-x) is a positive number <em>(-(-x) = +x)</em> and -f(x) is a negative number <em>(negative of a positive number is a negative number). </em>Since a positive number cannot equal a negative number, then this is not possible.

Answer: No

3 0
2 years ago
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WILL GIVE BRAINLIEST!!!!
Kay [80]

Answer:

the answer is C

Step-by-step explanation:

4 0
2 years ago
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