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Free_Kalibri [48]
2 years ago
12

Consider the first five steps of the derivation of the Quadratic Formula.

Mathematics
1 answer:
lara31 [8.8K]2 years ago
6 0

Answer:

Full proof below

Step-by-step explanation:

\displaystyle ax^2+bx+c=0\\\\ax^2+bx=-c\\\\x^2+\biggr(\frac{b}{a}\biggr)x=-\frac{c}{a}\\\\x^2+\biggr(\frac{b}{a}\biggr)x+\biggr(\frac{b}{2a}\biggr)^2=-\frac{c}{a}+\biggr(\frac{b}{2a}\biggr)^2\\\\x^2+\biggr(\frac{b}{a}\biggr)x+\frac{b^2}{4a^2}=-\frac{c}{a}+\frac{b^2}{4a^2}\\ \\\biggr(x+\frac{b}{2a}\biggr)^2=-\frac{4ac}{4a^2}+\frac{b^2}{4a^2}\\ \\\biggr(x+\frac{b}{2a}\biggr)^2=\frac{b^2-4ac}{4a^2}\\ \\x+\frac{b}{2a}=\pm\frac{\sqrt{b^2-4ac}}{2a}\\ \\x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

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A force of 18 lb is required to hold a spring stretched 8 in. beyond its natural length. How much work W is done in stretching i
Wittaler [7]

Answer: 18.75 lb.ft

Step-by-step explanation:

Given

Force required to stretch spring 8 in. is 18 lb

it can be written

\Rightarrow F=kx\\\Rightarrow 18=k(8)\\\\\Rightarrow k=\dfrac{18}{8}=\dfrac{9}{2}\ lb/in.

Work done in stretching from its natural length to 10 in.

\Rightarrow W=\dfrac{1}{2}kx^2\\\\\Rightarrow W=0.5\times \dfrac{9}{2}\times (10)^2\\\\\Rightarrow W=225\ lb.in.\ or\\\Rightarrow W=18.75\ lb.ft

8 0
3 years ago
How to work 48÷(4+2×10)-1​
Mashutka [201]

Answer:

The answer is 1

Step-by-step explanation:

In order to do this equation you use PEMDAS Parentheses, Exponents, Multiplication , Division, Addition and Subtraction

So,

48 / (4+2X10)-1\\  48 /(4+20)-1\\ 48/(24)-1\\48/24-1\\2-1\\1

Hoped this helped :D

5 0
3 years ago
Given the following hypotheses: H0: μ = 490 H1: μ ≠ 490 A random sample of 15 observations is selected from a normal population.
Elenna [48]

Answer:

We conclude that the population mean is equal to 490.

Step-by-step explanation:

We are given that a random sample of 15 observations is selected from a normal population. The sample mean was 495 and the sample standard deviation 9.

Let \mu = <u><em>population mean</em></u>.

So, Null Hypothesis, H_0 : \mu = 490      {means that the population mean is equal to 490}

Alternate Hypothesis, H_A : \mu\neq 490     {means that the population mean is different from 490}

The test statistics that will be used here is <u>One-sample t-test statistics</u> because we don't know about population standard deviation;

                               T.S.  =  \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }  ~ t_1_4

where, \bar X = sample mean = 495

            s = sample standard deviation = 9

             n = sample of observations = 15

So, <em>the test statistics </em>=   \frac{495-490}{\frac{9}{\sqrt{15} } }  ~ t_1_4

                                     =  2.152

The value of t-test statistics is 2.152.

Now, at a 0.01 level of significance, the t table gives a critical value of -2.977 and 2.977 at 14 degrees of freedom for the two-tailed test.

Since the value of our test statistics lies within the range of critical values of t, so <u><em>we have insufficient evidence to reject our null hypothesis</em></u> as the test statistics will not fall in the rejection region.

Therefore, we conclude that the population mean is equal to 490.

4 0
3 years ago
An employee earns an hourly wage shown in the graph below
vitfil [10]

Answer: the answer is 40

Step-by-step explanation:

5 0
4 years ago
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