All categories

4 years ago

What happens to an image when it is enlarged using scale factor between -1and1

Mathematics

1 answer:

adell [148]4 years ago

6 0

I have done this question before but i can’t remember

You might be interested in

PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!

aev [14]

Answer: (B) -625

<u>Step-by-step explanation:</u>

Given the sequence {-500, -100, -20, -4, -0.8, ... }, we know that that the first term (a) is -500 and the ratio (r) is $\dfrac{-100}{-500}=\dfrac{1}{5}$

Input those values into the Sum formula:

$S_n=\dfrac{a(1-r^n)}{1-r}\\\\S_8=\dfrac{-500\bigg(1-\dfrac{1}{5}^n\bigg)}{1-\dfrac{1}{5}}\\\\.\quad=\dfrac{-500\bigg(1-\dfrac{1}{390,625}\bigg)}{\dfrac{4}{5}}\\\\.\quad=\dfrac{-500\bigg(\dfrac{390,624}{390,625}\bigg)}{\dfrac{4}{5}}\\\\.\quad=\dfrac{5(-500)(390,624)}{4(390,625)}\\\\.\quad=-624.998\\\\.\quad=-625\quad \text{(rounded to 2 decimal places)}$

4 0

4 years ago

The 6-lb particle is subjected to the action of its weight and forces F1 = 52i + 6j - 2tk6 lb, F2 = 5t 2 i - 4tj - 1k6 lb, and F

olga2289 [7]

Answer:

r=294.9m

Step-by-step explanation:

The forces on the particle are

$W=mg\hat{j}\\F_{1}=52\hat{i}+6\hat{j}-2t\hat{k}\\F_{2}=5t^{2}\hat{i}-4t\hat{j}-1\hat{k}\\F_{3}=(5-2t)\hat{i}$

Now , we sum all these forces to get the net force

$F_{T}=W+F_{1}+F_{2}+F_{3}\\F_{T}=(52+5t^{2}+5-2t)\hat{i}+((6+6-4t)\hat{j}+(-2t-1)\hat{k}\\F_{T}=(57-2t+5t^{2})\hat{i}+(12-4t)\hat{j}+(-2t-1)\hat{k}\\$

we can use the fact F=m*a and integrate the acceleration

$a(t)=\frac{1}{m}F(t)\\\\v(t)=\int a(t)dt=\frac{1}{m}\int{F_{T}}dt\\\\v(t)=\frac{1}{m}[(57t-t^{2}+\frac{5}{3}t^{3})\hat{i}+(12t-2t^{2})\hat{j}+(-t^{2}-t)\hat{k}]\\\\r(t)=\int v(t)dt=\frac{1}{m}[(\frac{57}{2}t^{2}-\frac{1}{3}t^{3}}+\frac{5}{4}t^{4})\hat{i}+(6t^{2}-\frac{2}{3}t^{3})\hat{j}+(-\frac{1}{3}t^{3}-\frac{1}{2}t^{2})]$

and we evaluate in r(2) an we take the norm to obtain the distance

$r(2)=\frac{1}{m}[\frac{394}{3}\hat{i}+\frac{56}{3}\hat{j}-\frac{14}{3}\hat{k}]\\|r(2)|=\frac{1}{m}\sqrt{[(\frac{394}{3})^{2}+(\frac{56}{3})^{2}+(\frac{14}{3})^{2}]}\\|r(2)|=\frac{132.73}{0.45}=294.9m$

I hope this is useful for you

regards

8 0

4 years ago

Myles is tracking the growth of his palm tree. it was 62 inches when he bought it, and it has grown 6 inches per year. what does

Gre4nikov [31]

The initial value represents the height of the palm tree when the tree was first bought.
The input for the initial value is 0.
The output for the initial value is 62 inches.

What is initial value?

It is given that,

The height of the palm tree at the time Myles bought it was 62 inches.

He is tracking the growth of the tree and he has observed that it has grown 6 inches per year.

So, the initial value of the tree, here, is representing the height of the palm tree when it was bought.

As can be seen clearly, there is no input for the initial value. And, the output for the initial value is given to be 62 inches.

Learn more about initial value here:

brainly.com/question/8223651

#SPJ4

6 0

2 years ago

Indicate in standard form the equation of the line through the given points.<br> P(0, -4), Q(5.1)

Elan Coil [88]

keeping in mind that standard form for a linear equation means

• all coefficients must be integers, no fractions

• only the constant on the right-hand-side

• all variables on the left-hand-side, sorted

• "x" must not have a negative coefficient

$\bf P(\stackrel{x_1}{0}~,~\stackrel{y_1}{-4})\qquad Q(\stackrel{x_2}{5}~,~\stackrel{y_2}{1}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{1}-\stackrel{y1}{(-4)}}}{\underset{run} {\underset{x_2}{5}-\underset{x_1}{0}}}\implies \cfrac{1+4}{5}\implies \cfrac{5}{5}\implies 1$

$\bf \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{(-4)}=\stackrel{m}{1}(x-\stackrel{x_1}{0})\implies y+4=x \\\\\\ y=x-4\implies -x+y=-4\implies x-y=4$