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3 years ago

What are the solutions of the equation? x² – 21 = –4x

Mathematics

2 answers:

Effectus [21]3 years ago

6 0

Here are two solutions for x2-4x-21=0

x = 7
x = -3

antiseptic1488 [7]3 years ago

5 0

<span> x = 7<span>
x = -3
i used a caluclator im not to good at this either but i hope this helps

</span></span>

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This is probably easy

Vinvika [58]

Answer:

5 hours

Step-by-step explanation:

Let's express each scenario as a function in a slope-intercept form, y = mx + b,

Where,

y = total amount of water

x = number of hours

m = rate of at which water is drained or filled per hour

b = initial quantity of water in the tank

✔️Equation for Tank A:

m = 2 gallons/hour = 2

b = 100 gallons = 100

Substitute m = 2 and b = 100 into y = mx + b:

y = 2x + 100

✔️Equation for Tank B:

m = 6 gallons/hour = -6 (this is negative 6 because water drains)

b = 140 gallons = 140

Substitute the values into y = mx + b

y = -6x + 140

✔️Set the two equation equal together to determine the number of hours (x) that must pass for the total amount of water (y) in tank A and tank B to be the same.

2x + 100 = -6x + 140

Collect like terms

2x + 6x = -100 + 140

8x = 40

Divide both sides by 8

x = 40/8

x = 5

Therefore, 5 hours must pass for the total amount of water in both tanks to be the same.

3 0

2 years ago

Estimate the integral ∫6,0 x^2dx by the midpoint estimate, n = 6

Anettt [7]

Splitting up the interval [0, 6] into 6 subintervals means we have

$[0,1]\cup[1,2]\cup[2,3]\cup\cdots\cup[5,6]$

and the respective midpoints are $\dfrac12,\dfrac32,\dfrac52,\ldots,\dfrac{11}2$ . We can write these sequentially as ${x_i}^*=\dfrac{2i+1}2$ where $0\le i\le5$ .

So the integral is approximately

$\displaystyle\int_0^6x^2\,\mathrm dx\approx\sum_{i=0}^5({x_i}^*)^2\Delta x_i=\frac{6-0}6\sum_{i=0}^5({x_i}^*)^2=\sum_{i=0}^5\left(\frac{2i+1}2\right)^2$

Recall that

$\displaystyle\sum_{i=1}^ni^2=\frac{n(n+1)(2n+1)}6$
$\displaystyle\sum_{i=1}^ni=\frac{n(n+1)}2$
$\displaystyle\sum_{i=1}^n1=n$

so our sum becomes

$\displaystyle\sum_{i=0}^5\left(\frac{2i+1}2\right)^2=\sum_{i=0}^5\left(i^2+i+\frac14\right)$
$=\displaystyle\frac{5(6)(11)}6+\frac{5(6)}2+\frac54=\frac{143}2$

8 0

2 years ago

Suppose your restaurant bill comes to $28.35 .Find your total cost if the tax is 6.25% and you leave a 20% tip on the amount bef

Anon25 [30]

Your total after tax is $30.12
A 20% tip is approximately $5.50 ($5.67)

3 0

3 years ago

A clinical trial tests a method designed to increase the probability of conceiving a girl. In the study 340 babies were born, a

Bas_tet [7]

Answer:

The 99% confidence interval estimate of the percentage of girls born is (0.8, 0.9).

b)Yes, the proportion of girls is significantly different from 0.5.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence interval $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

Z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

For this problem, we have that:

In the study 340 babies were born, and 289 of them were girls. This means that $n = 340, \pi = \frac{289}{340} = 0.85$

Use the sample data to construct a 99% confidence interval estimate of the percentage of girls born.

So $\alpha = 0.01$ , z is the value of Z that has a pvalue of $1 - \frac{0.01}{2} = 0.995$ , so $z = 2.575$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{340}} = 0.85 - 2.575\sqrt{\frac{0.85*0.15}{400}} = 0.80$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{340}} = 0.85 + 2.575\sqrt{\frac{0.85*0.15}{400}} = 0.90$

The 99% confidence interval estimate of the percentage of girls born is (0.8, 0.9).

Does the method appear to be effective?

b)Yes, the proportion of girls is significantly different from 0.5.

4 0

3 years ago

What is the sum?

RSB [31]

It is C) 2.55 because i was taught that you can subtract the negative from the positive. 5.35-2.8=2.55 and it is positive

5 0

3 years ago

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