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4 years ago

Find the SLOPE between (1,1) and (4,6) ?

Mathematics

2 answers:

erastova [34]4 years ago

5 0

The slope is 5/3. hope this helps

olga55 [171]4 years ago

5 0

Plug it into the equation (y₂ - y₁) / (x₂ - x₁)

The little ₁ stands for point 1, and the little ₂ stands for point 2. Let's say (1, 1) is point one and the other point is point 2.

(6 - 1) / (4 - 1) = 5 / 3.

5/3 is the slope.

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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

3 years ago

How should the decimal point in 34.05 be moved to determine the product 34.05×106 ?

r-ruslan [8.4K]

Answer: Six places to the right

Step-by-step explanation:

For this exercise it is important to remember that. by definition, the exponent of a number indicates the number of times you must use the same factor to multiply.

Given:

$b^n$

"b" is the base and "n" is the exponent.

In this case, you have the following multiplication provided in the exercise:

$34.05*10^6$

Notice that the base 10 has an exponent 6. This indicates the following:

$10^6=10*10*10*10*10*10=1,000,000$

By definition, moving the decimal point 6 places to the right (because there are six zeros), is the same as multiplying the decimal number 34.05 by 1,000,000.

Therefore, based on the explained, when you move the decimal point six places to the right, you get the following product.

$34.05*10^6=34,050,000$

3 0

3 years ago

BC is parallel to DE<br> What is AC?

DaniilM [7]

ΔABC ~ ΔADE
AC/AE = AB/AD
AC/(AC+15) = 8/18
18AC = 8(AC+15)
18AC = 8AC + 120
18AC - 8AC = 120
10AC = 120
AC = 120/10
AC = 12

3 0

4 years ago

Read 2 more answers

44. Express each of these system specifications using predicates, quantifiers, and logical connectives. a) Every user has access

DENIUS [597]

Answer:

a. ∀x (User(x) → (∃y (Mailbox(y) ∧ Access(x, y))))

b. FileSystemLocked → ∀x Access(x, SystemMailbox)

c. ∀x ∀y ((Firewall(x) ∧ Diagnostic(x)) → (ProxyServer(y) → Diagnostic(y))

d. ∀x (ThroughputNormal ∧(ProxyServer(x)∧ ¬Diagnostic(x))) → (∃y Router(y)∧Functioning(y))

Step-by-step explanation:

Let the domain be users and mailboxes. Let User(x) be “x is a user”, let Mailbox(y) be “y is a mailbox”, and let Access(x, y) be “x has access to y”.

∀x (User(x) → (∃y (Mailbox(y) ∧ Access(x, y))))

(b)

Let the domain be people in the group. Let Access(x, y) be “x has access to y”. Let FileSystemLocked be the proposition “the file system is locked.” Let System Mailbox be the constant that is the system mailbox.

FileSystemLocked → ∀x Access(x, SystemMailbox)

(c)

Let the domain be all applications. Let Firewall(x) be “x is the firewall”, and let ProxyServer(x) be “x is the proxy server.” Let Diagnostic(x) be “x is in a diagnostic state”.

∀x ∀y ((Firewall(x) ∧ Diagnostic(x)) → (ProxyServer(y) → Diagnostic(y))

(d)

Let the domain be all applications and routers. Let Router(x) be “x is a router”, and let ProxyServer(x) be “x is the proxy server.” Let Diagnostic(x) be “x is in a diagnostic state”. Let ThroughputNormal be “the throughput is between 100kbps and 500 kbps”. Let Functioning(y) be “y is functioning normally”.

∀x (ThroughputNormal ∧(ProxyServer(x)∧ ¬Diagnostic(x))) → (∃y Router(y)∧Functioning(y))

4 0

4 years ago

Callie got 37 pieces of Halloween candy. Her brother got 42 pieces. She wants to estimate to find about how many pieces of candy

Vikki [24]

Answer:

Callie needs to hide both her and her brother's candy from him so that none is eaten/stolen. Then count each stash separately. Put the candies in groups of 10 to quickly establish that over 70 pieces were collected. Callie should steal a stack from her brother's pile while he's not looking. Just calmly shift one stack of ten candies to her side. She should announce she got 47 pieces and her brother only 32.

Step-by-step explanation:

3 0

3 years ago