4 years ago

Which point in a triangle is equidistant from the vertices of the triangle

Mathematics

1 answer:

jekas [21]4 years ago

3 0

Circumcenter is the point on the triangle

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(PLEASE HELP) <br> Find the area of the parallelogram.

allochka39001 [22]

Answer:

Step-by-step explanation:

A = bh

B = 7

H=10

7x10 =70m

7 0

2 years ago

both brian and jermaine solve the problem. brian says the answer is $40,704. jermaine's answer is $4,604.

Studentka2010 [4]

Could you possibly be more specific?what math problem did they solve? i can help but just be a little more specific so i can answer your question

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4 years ago

3x over 2x-6 plus 9 over 6-2x

Mars2501 [29]

Answer: 3/2
Explanation:
3x/2x-6 + 9/6-2x
= 3x/2x-6 + 9/-(2x+6)
= 3x/2x-6 + -9/2x-6
= 3x/2x-6 - 9/2x-6
= 3x - 9/2x-6
= 3(x-3)/2(x-3) (cancel out x-3)
= 3/2

7 0

3 years ago

Read 2 more answers

Find the following integral. <br> Integrate x + 5/x^2 +10x +26 dx

sweet-ann [11.9K]

Given :

Polynomial , $f(x)=x+\dfrac{5}{x^2}+10x+26$ .

To Find :

Integration of f(x) .

Solution :

We know , integration of $x^n$ is :

$\int x^n \, dx = \dfrac{x^{n+1}}{n+1}+C$

( Here , is C is a constant )

Now ,

$\int f(x)\ dx=\int( x+\dfrac{5}{x^2}+10x+26)dx\\ \\ \int f(x)\ dx= \int x\ dx+\int 5x^{-2}\ dx+\int 10 x \ dx +26\ dx\\ \\ \int f(x)\ dx=\dfrac{x^2}{2}+\dfrac{5\times x^{-1}}{(-1)}+\dfrac{10x^2}{2}+26x+C\\\\ \int f(x)\ dx=\dfrac{11x^2}{2}+26x-5x^{-1}+C$

Therefore , the integration is $\dfrac{11x^2}{2}+26x-5x^{-1}+C$ .

Hence , this is the required solution .

6 0

3 years ago

72÷6374<br><img src="https://tex.z-dn.net/?f=72%20%5Cdiv%206374" id="TexFormula1" title="72 \div 6374" alt="72 \div 6374" align=

Ivan

72 divided by 6374 is equal to 0.0113

Hope it helps :)

6 0

3 years ago