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3 years ago

Solve the equation for x, where x is a real number (5 points): -13x^2 + 17x + 39 = 44

1 answer:

7 0

We first subtract 44 from both sides so that the equation becomes -13x^2 + 17x - 5 = 0.

To find the solutions to this equation, we can apply the quadratic formula. This quadratic formula solves equations of the form ax^2 + bx + c = 0
                     x = [ -b ± √(b^2 - 4ac) ] / (2a)
                     x = [ -17 ± √(17^2 - 4(-13)(-5)) ] / ( 2(-13) )
                     x = [ -17 ± √(289 - (260) ) ] / ( -26 )
                     x = [ -17 ± √(29) ] / ( -26)
                     x = [ -17 ± sqrt(29) ] / ( -26 )
                     x = 17/26 ± -sqrt(29)/26
The answers are 17/26 + sqrt(29)/26 and 17/26 - sqrt(29)/26.

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between 1995 and 2006, the population of samburg, usa(in thousands) can be modeled by f(x) = 0.35x(squared) - 2.1x+15.8 where x=

Juli2301 [7.4K]

$\bf f(x)=0.35x^2-2.1x+15.8\qquad 
\begin{cases}
x=\textit{year since 1995}\\
f(x)=\textit{population amount}
\end{cases}$

the equation is a quadratic one, and it has a positive coefficient on the leading term, meaning, is opening upwards, so it has a "burrow" for the vertex.

the minimum or lowest point for a quadratic opening upwards is, well, the vertex point :), the "x" value is the year, the "y" or f(x) value is the population, we're asked for the year, or the x-coordinate of the vertex

well $\bf \begin{array}{llll}
f(x)=&0.35x^2&-2.1x&+15.8\\
&\quad \uparrow &\quad \uparrow&\uparrow \\
&\quad a&\quad b &c
\end{array}
\\\\

\\\\
\qquad \textit{vertex of a parabola}\\ \quad \\
\qquad 

\left(\boxed{-\cfrac{{{ b}}}{2{{ a}}}}\quad ,\quad {{ c}}-\cfrac{{{ b}}^2}{4{{ a}}}\right)$

5 0

4 years ago

Choose the correct trig ratio you would use to solve for the missing piece of the right triangle:

Maslowich

Answer:

cos

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Jackson currently has $35 in his bank account, and he is saving $15 a week to eventually buy a new cell phone. He can represent

Ket [755]

15x7 = 105

105+35= 140

$140

8 0

3 years ago

A pitcher can hold 5. 2 liters of water. How many 0. 4 liter glasses of water can be poured from the pitcher?

gogolik [260]

The number of 0.4 liter glasses of water that will be poured in the pitcher is 13

From the information given, we are being told that a pitcher can hold a maximum amount of 5.2 liters of water.

If we are to use a 0.4 liter of glass to pour water in the pitcher, then the number of times we will use the glass to pour water in the pitcher can be estimated as:

Number of glass that can fill the pitcher = 5.2 liters/ 0.4 liters

= 13 glass

Learn more about solving word problems in mathematics here:

brainly.com/question/2121480?referrer=searchResults

7 0

2 years ago

A) Evaluate the limit using the appropriate properties of limits. (If an answer does not exist, enter DNE.)

Gelneren [198K]

For purely rational functions, the general strategy is to compare the degrees of the numerator and denominator.

$\displaystyle \lim_{x\to\infty} \frac{2x^2-5}{7x^2+x-3} = \boxed{\frac27}$

because both numerator and denominator have the same degree (2), so their end behaviors are similar enough that the ratio of their coefficients determine the limit at infinity.

More precisely, we can divide through the expression uniformly by x ²,

$\displaystyle \lim_{x\to\infty} \frac{2x^2-5}{7x^2+x-3} = \lim_{x\to\infty} \frac{2-\dfrac5{x^2}}{7+\dfrac1x-\dfrac3{x^2}}$

Then each remaining rational term converges to 0 as x gets arbitrarily large, leaving 2 in the numerator and 7 in the denominator.

B) By the same reasoning,

$\displaystyle \lim_{x\to\infty} \frac{5x-3}{2x+1} = \boxed{\frac52}$

C) This time, the degree of the denominator exceeds the degree of the numerator, so it grows faster than x - 1. Dividing a number by a larger number makes for a smaller number. This means the limit will be 0:

$\displaystyle \lim_{x\to-\infty} \frac{x-1}{x^2+8} = \boxed{0}$

More precisely,

$\displaystyle \lim_{x\to-\infty} \frac{x-1}{x^2+8} = \lim_{x\to-\infty}\frac{\dfrac1x-\dfrac1{x^2}}{1+\dfrac8{x^2}} = \dfrac01 = 0$

D) Looks like this limit should read

$\displaystyle \lim_{t\to\infty}\frac{\sqrt{t}+t^2}{3t-t^2}$

which is just another case of (A) and (B); the limit would be

$\displaystyle \lim_{t\to\infty}\frac{\sqrt{t}+t^2}{3t-t^2} = -1$

That is,

$\displaystyle \lim_{t\to\infty}\frac{\sqrt{t}+t^2}{3t-t^2} = \lim_{t\to\infty}\frac{\dfrac1{t^{3/2}}+1}{\dfrac3t-1} = \dfrac1{-1} = -1$

However, in case you meant something else, such as

$\displaystyle \lim_{t\to\infty}\frac{\sqrt{t+t^2}}{3t-t^2}$

then the limit would be different:

$\displaystyle \lim_{t\to\infty}\frac{\sqrt{t^2}\sqrt{\dfrac1t+1}}{3t-t^2} = \lim_{t\to\infty}\frac{t\sqrt{\dfrac1t+1}}{3t-t^2} = \lim_{t\to\infty}\frac{\sqrt{\dfrac1t+1}}{3-t} = 0$

since the degree of the denominator is larger.

One important detail glossed over here is that

$\sqrt{t^2} = |t|$

for all real t. But since t is approaching *positive* infinity, we have t > 0, for which |t | = t.

E) Similar to (D) - bear in mind this has the same ambiguity I mentioned above, but in this case the limit's value is unaffected -

$\displaystyle \lim_{x\to\infty} \frac{x^4}{\sqrt{x^8+9}} = \lim_{x\to\infty}\frac{x^4}{\sqrt{x^8}\sqrt{1+\dfrac9{x^8}}} = \lim_{x\to\infty}\frac{x^4}{x^4\sqrt{1+\dfrac9{x^8}}} = \lim_{x\to\infty}\frac1{\sqrt{1+\dfrac9{x^8}}} = \boxed{1}$

Again,

$\sqrt{x^8} = |x^4|$

but x ⁴ is non-negative for real x.

F) Also somewhat ambiguous:

$\displaystyle \lim_{x\to\infty}\frac{\sqrt{x+5x^2}}{3x-1} = \lim_{x\to\infty}\frac{\sqrt{x^2}\sqrt{\dfrac1x+5}}{3x-1} = \lim_{x\to\infty}\frac{x\sqrt{\dfrac1x+5}}{3x-1} = \lim_{x\to\infty}\frac{\sqrt{\dfrac1x+5}}{3-\dfrac1x} = \dfrac{\sqrt5}3$

$\displaystyle \lim_{x\to\infty}\frac{\sqrt{x}+5x^2}{3x-1} = \lim_{x\to\infty}x \cdot \lim_{x\to\infty}\frac{\dfrac1{\sqrt x}+5x}{3x-1} = \lim_{x\to\infty}x \cdot \lim_{x\to\infty}\frac{\dfrac1{x^{3/2}}+5}{3-\dfrac1x} = \frac53\lim_{x\to\infty}x = \infty$

G) For a regular polynomial (unless you left out a denominator), the leading term determines the end behavior. In other words, for large x, x ⁴ is much larger than x ², so effectively

$\displaystyle \lim_{x\to\infty}(x^4-2x) = \lim_{x\to\infty}x^4 = \boxed{\infty}$

6 0

3 years ago