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3 years ago

1.5 meters equal how many cm

2 answers:

6 0

1.5 meters equals 150 cm because 1 meter equals 100 cm so there is 1.5 meters which is half more than 1 meter there would be 150cm....hope it helped :)

hram777 [196]3 years ago

3 0

1 meter = 100 cm. So 1 1/2 meters = 150 cm.

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What is the solution to this system of equations? 4x + 5y = 7

Nutka1998 [239]

Answer:

(-2, 3)

Step-by-step explanation:

4x + 5y = 7

3x - 2y = -12

Let's solve this by elimination. We want to eliminate one variable at a time. This means we need to multiply the equations to create a common multiple to cancel out a variable.

Let's work with y.

5y and -2y: For these values to cancel out, we need to multiply each term to create a common multiple.

2(4x + 5y = 7)

5(3x - 2y = -12)

Multiply.

8x + 10y = 14

15x - 10y = -60

Eliminate.

23x = -46

Divide both sides by 23.

x = -2

Now that we know x, let's plug it back into one of equations to find y.

4x + 5y = 7

4(-2) + 5y = 7

Multiply.

-8 + 5y = 7

Add.

5y = 15

Divide.

y = 3

Now we know x and y; let's plug both back into the equation we have not checked yet.

3x - 2y = -12

3(-2) - 2(3) = -12

Multiply.

-6 - 6 = -12

Subtract.

-12 = -12

Your solution is correct.

(-2, 3)

Hope this helps!

8 0

2 years ago

Solve for y in the given equation 2y - 24 = 14y Answer are 8y -8y -2 2

ArbitrLikvidat [17]

Answer:

y = -2

Step-by-step explanation:

First, we subtract 2y on both sides:

-24 = 12y

Then, we divide 12 on both sides:

y = -2

And we're done ^^ hope this helps!

6 0

3 years ago

Elsa makes a poster that is 1.5 m tall. Greg makes a poster that is 96 cm tall. Who makes a taller poster?

Mademuasel [1]

Answer:

dont mind me just grinding for points

Step-by-step explanation:

6 0

2 years ago

Use series to verify that <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

PLS HELP ASAPPP I WILL GIVE BRAINLIEST

FrozenT [24]

Answer:

Yes, it is.

Step-by-step explanation:

6 0

3 years ago

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