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4 years ago

Which is a solution to (x - 2)(x + 10) = 13? O x = 3 Ox=8 x = 10 x = 11

Mathematics

1 answer:

Dmitriy789 [7]4 years ago

4 0

The answer gonna be 3
For the equation

(X-2)(x+10)=13

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Write in slope intercept form given two points (1,2) and (-2,5)

Anvisha [2.4K]

I’m pretty sure it’s -1
How you solve this is y2-y1/x2-x1
So 5-2/-3-1 which would be -1

4 0

3 years ago

HELP ME PLEASEEEEEEEEEEEEEEEE

Veronika [31]

Answer:

D im not sure but i hope its right

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

I need help finding y

denpristay [2]

Answer:

y = 9.8995

Step-by-step explanation:

This question requires the use of basic trigonometry.

In this problem, we know one of the sides and two of the angles.

Since one of the angles is 45 degrees and the sum of all internal angles of a triangle is 180 degrees, the other angle can only be 45 degrees, making this an isosceles triangle - that is, the length of the other leg, x, is also 7.

You could use the soh cah toa mnemonic to figure it out, but here just using the a^2 + b^2 = c^2 should be sufficient since we know both legs of the triangle.

7^2 + 7^2 = 98, and the square root of 98 gives you about 9.8995, which sounds about right for this situation.

Hope this helped!

4 0

3 years ago

How many triangles can be constructed with sides measuring 5 m, 16 m, and 5 m?

lianna [129]

There are two measures 5 cm then the triangle will be isosceles or 16 is far beyond 5 + 5 = 10 so we cannot build any triangle so te answer is <span>B none
</span>

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

3 years ago