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2 years ago

What is the given point on the line of this equation y +2 = 5/3(x-3)

Mathematics

1 answer:

8_murik_8 [283]2 years ago

4 0

Answer:

Step-by-step explanation:

Solve x then y after..

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Find the slope of the lines graphed below

Nadusha1986 [10]

Answer:

1. 2

2. -3/5

3. Undefined

Step-by-step explanation:

Using the rise over run method you get 4/2, -3/5, and then for the straight vertical line, its undefined.

8 0

2 years ago

Which expression is equivalent to cot t sec t?

Solnce55 [7]

Answer: $csc(t)$

Step-by-step explanation:

Alright, lets get started.

The given expression is given as :

$cot( t) \times sec (t)$

We know quotient identity as :

$cot(t)=\frac{cos(t)}{sin(t)}$

Similarly, we know reciprocal identity as :

$sec(t)=\frac{1}{cos(t)}$

lets plug the value of cot and sec in given expression

$\frac{cos(t)}{sin(t)} \times \frac{1}{cos(t)}$

cos will be cancelled, remaining will be

$\frac{1}{sin(t)}$

Using reciprocal identity again, that will equal to :

$csc(t)$ ................... Answer (A)

8 0

2 years ago

Use series to verify that<br><br> <img src="https://tex.z-dn.net/?f=y%3De%5E%7Bx%7D" id="TexFormula1" title="y=e^{x}" alt="y=e^{

SVETLANKA909090 [29]

$y = e^x\\\\\displaystyle y = \sum_{k=1}^{\infty}\frac{x^k}{k!}\\\\\displaystyle y= 1+x+\frac{x^2}{2!} + \frac{x^3}{3!}+\ldots\\\\\displaystyle y' = \frac{d}{dx}\left( 1+x+\frac{x^2}{2!} + \frac{x^3}{3!}+\frac{x^4}{4!}+\ldots\right)\\\\$

$\displaystyle y' = \frac{d}{dx}\left(1\right)+\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(\frac{x^2}{2!}\right) + \frac{d}{dx}\left(\frac{x^3}{3!}\right) + \frac{d}{dx}\left(\frac{x^4}{4!}\right)+\ldots\\\\\displaystyle y' = 0+1+\frac{2x^1}{2*1} + \frac{3x^2}{3*2!} + \frac{4x^3}{4*3!}+\ldots\\\\\displaystyle y' = 1 + x + \frac{x^2}{2!}+ \frac{x^3}{3!}+\ldots\\\\\displaystyle y' = \sum_{k=1}^{\infty}\frac{x^k}{k!}\\\\\displaystyle y' = e^{x}\\\\$

This shows that $y' = y$ is true when $y = e^x$

-----------------------

Note 1: A more general solution is $y = Ce^x$ for some constant C.
Note 2: It might be tempting to say the general solution is $y = e^x+C$ , but that is not the case because $y = e^x+C \to y' = e^x+0 = e^x$ and we can see that $y' = y$ would only be true for C = 0, so that is why $y = e^x+C$ does not work.

6 0

2 years ago

12x⁴y²divided by 3x³y to the 5 th power....Please help me out....

ki77a [65]

To divide the expression we proceed as follows:
12x⁴y²÷3x³y⁵
=12x⁴y²/3x³y⁵
=(12÷3)×(x⁴÷x³)×(y²÷y⁵)
when you divide number with the same base, you subtract the numerators:
=4×(x⁴⁻³)×(y²⁻⁵)
simplifying this we get
=4xy⁻³
Answer: 4xy⁻³

5 0

3 years ago

A football player kicks a 0.94 kg football with a force of 2.4 N.

bearhunter [10]

.94 times 2.6 would be 2.4

8 0

2 years ago

Read 2 more answers