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3 years ago

What is the minimum possible value for y in the equation y = x^2 + 12x + 5?

Mathematics

1 answer:

Dafna11 [192]3 years ago

4 0

Answer:

Step-by-step explanation:

hello :

y = x² + 12x + 5

y = ( x²+12x+36) -36+5

y= (x+6)²-31.......vertex form when the minimum possible value is : - 31

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Please help math middle school level

pychu [463]

Answer:

7 yd

Step-by-step explanation:

You are finding the perimeter. Your answer should include yd (yd² is for area, yd³ is for 3d images).

Note that the two lines segments (? & 9) is parallel to the other line segment (16). Find the ?:

? + 9 = 16

? + 9 (-9) = 16 (-9)

? = 16 - 9

? = 7

7 yd is your answer

3 0

3 years ago

Loni needs between 3.5 and 7.2 volts of energy, not inclusive, to make an experiment work correctly. Write a compound inequality

Evgesh-ka [11]

Answer:

The compound inequality that shows the voltage amount the necessary range is $3.5$

Step-by-step explanation:

Loni needs between 3.5 and 7.2 volts of energy, not inclusive, to make an experiment work correctly.

That means, the voltage amount should be either greater than 3.5 volts or less than 7.2 volts.

Suppose, the amount of voltage is $V$ volts to make the experiment work correctly.

Now, Voltage amount is greater than 3.5 volts means: $V>3.5$ or $3.5$

and voltage amount is less than 7.2 volts means: $V$

Thus, the compound inequality by combining the above two inequalities will be: $3.5$

5 0

3 years ago

Which expression is equivalent to the one below?

Alex17521 [72]

Answer:

D . 9 .1/13

Step-by-step explanation:

$9\div 13 = \frac{9}{13} \\\\9\cdot \frac{1}{13}\\\\=\frac{9}{1}\cdot \frac{1}{13}\\\\=\frac{9\cdot \:1}{1\cdot \:13}\\\\=\frac{9}{13}$

4 0

3 years ago

Location is known to affect the number, of a particular item, sold by an auto parts facility. Two different locations, A and B,

Mama L [17]

We have two samples, A and B, so we need to construct a 2 Samp T Int using this formula:

$\displaystyle \overline {x}_1 - \overline {x}_2 \ \pm \ t^{*} \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} }$

In order to use t*, we need to check conditions for using a t-distribution first.

Random for both samples -- NOT STATED in the problem ∴ proceed with caution!
Independence for both samples: 130 < all items sold at Location A; 180 < all items sold at Location B -- we can reasonably assume this is true
Normality: CLT is not met; n < 30 for both locations A and B ∴ proceed with caution!

Since 2/3 conditions aren't met, we can still proceed with the problem but keep in mind that the results will not be as accurate until more data is collected or more information is given in the problem.

Solve for t*:

We need the tail area first.

$\displaystyle \frac{1-.9}{2}= .05$

Next we need the degree of freedom.

The degree of freedom can be found by subtracting the degree of freedom for A and B.

The general formula is df = n - 1.

df for A: 13 - 1 = 12
df for B: 18 - 1 = 17
df for A - B: |12 - 17| = 5

Use a calculator or a t-table to find the corresponding t-score for df = 5 and tail area = .05.

t* = -2.015

Now we can use the formula at the very top to construct a confidence interval for two sample means.

$\overline {x}_A=39$
$s_A=8$
$n_A=13$
$\overline {x}_B = 55$
$s_B=2$
$n_B=18$
$t^{*}=-2.015$

Substitute the variables into the formula: $\displaystyle \overline {x}_1 - \overline {x}_2 \ \pm \ t^{*} \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} }$ .