How many real numbers solutions are there in the equation?<br> 0=-3x^2+x-4?

"The last time I bought this product, it cost $20.00 but it looks like it costs $29.60 today. Why such a large increase?"

Oksanka [162]

Answer:

The company decided to increase the price by $9.60, or by 48%.

Step-by-step explanation:

To measure the price difference in dollars is simple, just subtract the old price from the new one:

$29.60 - 20 = 9.60$

Divide the price change by the old price to find the price increase in percentage:

$9.60 \div 20 = 0.48 \\ 0.48 \times 100\% = 48\%$

5 0

3 years ago

A tennis ball in the shape or a sphere has a diameter of 6.7 centimeters which is closest to the volume

Lynna [10]

In your question it seems like there should be choices too! Either way, how does one find the volume of a sphere?

8 0

2 years ago

The sum of 13 divided by a number and that number divided by 13

7nadin3 [17]

Sum means add

the number is x

13 divided by a number is 13/x
the number divided by the number is x/13

so
$\frac{13}{x}+\frac{x}{13}$
also can be simplified to
$\frac{169}{13x}+\frac{x^2}{13x}$ or
$\frac{x62+169}{13x}$

translated it is $\frac{13}{x}+\frac{x}{13}$ where x is the number

5 0

2 years ago

Which term completes the product so that it is the difference of squares?<br>(-5x-3)(-5x+<br>)

lilavasa [31]

Hello,

(-5x-3)(-5x+ 3) = (-5x)² - 3²

7 0

3 years ago

Prove or disprove (from i=0 to n) sum([2i]^4) <= (4n)^4. If true use induction, else give the smallest value of n that it doe

ddd [48]

Answer:

The statement is true for every n between 0 and 77 and it is false for $n\geq 78$

Step-by-step explanation:

First, observe that, for n=0 and n=1 the statement is true:

For n=0: $\sum^{n}_{i=0} (2i)^4=0 \leq 0=(4n)^4$

For n=1: $\sum^{n}_{i=0} (2i)^4=16 \leq 256=(4n)^4$

From this point we will assume that $n\geq 2$

As we can see, $\sum^{n}_{i=0} (2i)^4=\sum^{n}_{i=0} 16i^4=16\sum^{n}_{i=0} i^4$ and $(4n)^4=256n^4$ . Then,

$\sum^{n}_{i=0} (2i)^4 \leq(4n)^4 \iff \sum^{n}_{i=0} i^4 \leq 16n^4$

Now, we will use the formula for the sum of the first 4th powers:

$\sum^{n}_{i=0} i^4=\frac{n^5}{5} +\frac{n^4}{2} +\frac{n^3}{3}-\frac{n}{30}=\frac{6n^5+15n^4+10n^3-n}{30}$

Therefore:

$\sum^{n}_{i=0} i^4 \leq 16n^4 \iff \frac{6n^5+15n^4+10n^3-n}{30} \leq 16n^4 \\\\ \iff 6n^5+10n^3-n \leq 465n^4 \iff 465n^4-6n^5-10n^3+n\geq 0$

and, because $n \geq 0$ ,

$465n^4-6n^5-10n^3+n\geq 0 \iff n(465n^3-6n^4-10n^2+1)\geq 0 \\\iff 465n^3-6n^4-10n^2+1\geq 0 \iff 465n^3-6n^4-10n^2\geq -1\\\iff n^2(465n-6n^2-10)\geq -1$