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2 years ago

Help help help help help help help help!!!!

Mathematics

2 answers:

mina [271]2 years ago

8 0

Answer:

Step-by-step explanation:

stealth61 [152]2 years ago

8 0

Answer:

c is the answer have a nice day

Step-by-step explanation:

You might be interested in

Two trains leave a city at 3 P.M. One train goes east at a constant rate of 70 mph. The other goes west at a rate of 62 mph. How

andrey2020 [161]

Answer:

4 hours

Step-by-step explanation:

The general formula for distance, rate and time is:

d = rt, where distance is the product of rate multiplied by time

Given that one train is moving at a rate of 70 mph and one is moving at a rate of 62 mph in the opposite direction, the combined distance over time is equal to 528 miles. Using the variable 't' for time:

70t + 62t = 528

Combine like terms: 132t = 528

Divide both sides by 132: 132t/132 = 528/132 or t = 4 hours

8 0

2 years ago

Read 2 more answers

A group of college students are going to a lake house for the weekend and plan on renting small cars and large cars to make the

Alexus [3.1K]

4x+6y=56 and y=4x can be used to determine the number of small cars rented and the number of large cars rented, where x represents number of small cars and y represents number of large cars.

Step-by-step explanation:

Number of people hold by small cars = 4

Number of people hold by large cars = 6

Let,

Number of small cars = x

Number of large cars = y

The students rented 4 times as many large cars as small cars,

y = 4x Eqn 1

which altogether can hold 56 people.

4x+6y=56 Eqn 2

4x+6y=56 and y=4x can be used to determine the number of small cars rented and the number of large cars rented, where x represents number of small cars and y represents number of large cars.

Keywords: linear equations, addition

Learn more about linear equations at:

#LearnwithBrainly

6 0

3 years ago

What is the missing number in the expression 3(q+7)=27

Murljashka [212]

Answer:

Step-by-step explanation:7+2 is 9 and 9x3 is 27

3 0

3 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

Answer this pls<br>don't send links

Irina18 [472]

Answer:

a 20 times

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers