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2 years ago

Describe a situation in which you could use scaling to solve a real-world problem.

Mathematics

1 answer:

balu736 [363]2 years ago

6 0

Answer:

You want to know how much lager a full sized candy bar is to see if you are getting a good price.

Step-by-step explanation:

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Suppose a triangle has sides a, b, and c, and that a2 + b2 < c. Let o be the

siniylev [52]

Step-by-step explanation:

if a^2 +b^2 <c^2,

then abc is not a right triangle since for a right triangle a^2+ b^2 = c^2

8 0

3 years ago

(a) Tom says that 1/5 + 7/20 = 8/25

mezya [45]

numerator = top part of the fraction

denominator = bottom part of the fraction

* means multiply

you can only add the numerator when the denominator is the same

cant add the denominator

denominator has to be the same

1/5 = 4/20

4/20 + 7/20 = 11/20

1 7/8 = 15/8

3/4 ÷ 15/8 =

3/4 * 8/15 =

24/60 = divide both top and bottom by 12 =

2/5

7 0

3 years ago

Read 2 more answers

Which answer best describes the complex zeros of the polynomial function?

marysya [2.9K]

Answer-

D. The function has one real zero and two nonreal zeros. The graph of the function intersects the x-axis at exactly one location.

Solution-

The given polynomial is,

$f(x)=x^3-x^2+6x-6$

The zeros of the polynomials are,

$\Rightarrow f(x)=0$

$\Rightarrow x^3-x^2+6x-6=0$

$\Rightarrow x^2(x-1)+6(x-1)=0$

$\Rightarrow (x^2+6)(x-1)=0$

$\Rightarrow (x^2+6)=0,(x-1)=0$

$\Rightarrow x=\sqrt{-6},\ x=1$

$\Rightarrow x=1,\ \pm \sqrt6i$

Therefore, this function has only one real zero i.e 1 and two nonreal zeros i.e ±√6i . The graph of the function intersects the x-axis at exactly one location i.e at x = 1

3 0

3 years ago

<img src="https://tex.z-dn.net/?f=%20%5Cint%20%5Cfrac%7Bdx%7D%7Bxln%5E%7Bp%7Dx%20%7D" id="TexFormula1" title=" \int \frac{dx}{xl

Eva8 [605]

Substitute $u=\ln x$ , so that $\mathrm du=\dfrac{\mathrm dx}x$ . The integral is then equivalent to

$\displaystyle\int\frac{\mathrm dx}{x\ln^px}=\int\frac{\mathrm du}{u^p}=\begin{cases}\dfrac{u^{p+1}}{p+1}+C&\text{for }p\neq1\\\\\ln|u|+C&\text{for }p=1\end{cases}$

Then transforming back to $x$ gives

$\displaystyle\int\frac{\mathrm dx}{x\ln^px}=\begin{cases}\dfrac{\ln^{p+1}x}{p+1}+C&\text{for }p\neq1\\\\\ln|\ln x|+C&\text{for }p=1\end{cases}$

4 0

3 years ago

you send 12 photos to a company that makes personalized wall calendars the company enlarges the photos and inserts one for each

Assoli18 [71]

Answer:

15 inches

Step-by-step explanation:

We have that,

Length of the photo = 4 inches and Width of the photo = 6 inches

Also, the width of the image on the calendar = 10 inches

Let, the length on the image of the calendar = x inches.

Since, the ratio of the photo and its image will remain same, we get,

$\frac{4}{6}=\frac{10}{x}$

i.e. $x=\frac{60}{4}$

i.e. x = 15

Hence, the length of the image of the photo on the calendar is 15 inches.

3 0

3 years ago