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3 years ago

What does the residual of -3 mean?

Mathematics

1 answer:

Zielflug [23.3K]3 years ago

6 0

Answer:

Plz mark brainlest!

Step-by-step explanation:

Residual describes what remains after most of something is gone. It's an almost formal word for what's leftover. If you've gotten over your breakup but you still have the urge to kick your ex, then you have some residual bitterness.Residual. The vertical distance between a data point and the graph of a regression equation. The residual is positive if the data point is above the graph. Hope this helps! Ask me anything if you have any quistions!

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A motor bike was originally listed at

Radda [10]

Answer:

Answer

muddassirmkhan95

Beginner

5 answers

97 people helped

Answer:

Single Equivalent Discount % on final price = 41.58%

Explanation:

The grill was initially priced at $769.99 and was discounted to $550.54 which is 28.5% ((550.54/769.99)-1)*100

and then further discounted by 18.3% ((449.79/550.54)-1)*100 to $449.79.

but if we want to get the single discount rate for $449.79 (final price), we have to compare it with very first original price of $769.99 so the formula will be:

((449.79/769.99)-1)*100 = 41.58% which shows the actual discount % versus the original price of the grill.

Step-by-step explanation:

7 0

3 years ago

Rationalize the denominator and simplify.

cestrela7 [59]

If you're using the app, try seeing this answer through your browser: brainly.com/question/3151018

——————————

$\mathsf{\dfrac{3\sqrt{6}+5\sqrt{2}}{4\sqrt{6}-3\sqrt{2}}}$

Multiply and divide by the conjugate of the denominator, which is $\mathsf{4\sqrt{6}+3\sqrt{2}:}$

$\mathsf{=\dfrac{\big(3\sqrt{6}+5\sqrt{2}\big)\cdot \big(4\sqrt{6}+3\sqrt{2}\big)}{\big(4\sqrt{6}-3\sqrt{2}\big)\cdot \big(4\sqrt{6}+3\sqrt{2}\big)}}$

Expand those products and eliminate the brackets:

$\mathsf{=\dfrac{\big(3\sqrt{6}+5\sqrt{2}\big)\cdot 4\sqrt{6}+ \big(3\sqrt{6}+5\sqrt{2}\big)\cdot 3\sqrt{2}}{\big(4\sqrt{6}-3\sqrt{2}\big)\cdot 4\sqrt{6}+ \big(4\sqrt{6}-3\sqrt{2}\big)\cdot 3\sqrt{2}}}\\\\\\ \mathsf{=\dfrac{12\cdot \big(\sqrt{6}\big)^2+20\sqrt{2}\cdot \sqrt{6}+9\cdot \sqrt{6}\cdot \sqrt{2}+15\cdot \big(\sqrt{2}\big)^2}{16\cdot \big(\sqrt{6}\big)^2-12\cdot \sqrt{2}\cdot \sqrt{6}+ 12\cdot \sqrt{6}\cdot \sqrt{2}-9\cdot \big(\sqrt{2}\big)^2}}\\\\\\ \mathsf{=\dfrac{12\cdot 6+20\sqrt{2\cdot 6}+9\sqrt{6\cdot 2}+15\cdot 2}{16\cdot 6-12\sqrt{2\cdot 6}+ 12\sqrt{6\cdot 2}-9\cdot 2}}\\\\\\ \mathsf{=\dfrac{72+20\sqrt{12}+9\sqrt{12}+30}{96-12\sqrt{12}+12\sqrt{12}-18}}$

$\mathsf{=\dfrac{72+29\sqrt{12}+30}{96-18}}\\\\\\ \mathsf{=\dfrac{102+29\sqrt{2^2\cdot 3}}{78}}\\\\\\ \mathsf{=\dfrac{102+29\cdot \sqrt{2^2}\cdot \sqrt{3}}{78}}\\\\\\ \mathsf{=\dfrac{102+29\cdot 2\sqrt{3}}{78}}$

$\mathsf{=\dfrac{102+58\sqrt{3}}{78}}\\\\\\ \mathsf{=\dfrac{\,\diagup\!\!\!\! 2\cdot \big(51+29\sqrt{3}\big)}{\diagup\!\!\!\! 2\cdot 39}}$

$\mathsf{=\dfrac{51+29\sqrt{3}}{39}}\quad\longleftarrow\quad\textsf{this is the answer.}$

I hope this helps. =)

4 0

3 years ago

Please help, i have been doing this for a while and it is still wrong.

adelina 88 [10]

Answer:

51.7661748226 degrees; 38.2338251774 degrees

Step-by-step explanation:

First, draw a diagram to illustrate this situation. Looking at angle A, Eric's height, 1.65 m, corresponds to OPPOSITE, and Eric's shadow, 1.30 m, corresponds to ADJACENT. Using SOH CAH TOA, the equation is:

tan(A) = 1.65/1.30

A = arctan(1.65/1.30)

A = 51.7661748226 deg

Looking at angle B, 1.65 m is ADJACENT and 1.30 m is OPPOSITE. Therefore, the equation is:

tan(B) = 1.30/1.65

B = arctan(1.30/1.65)

B = 38.2338251774 deg

3 0

2 years ago