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Angelina_Jolie [31]
3 years ago
6

1. Find the distance between each pair of points. Round your answer to

Mathematics
1 answer:
zlopas [31]3 years ago
7 0

Step-by-step explanation:

\sqrt{(3 - ( - 4))^{2}  + ( - 7 - 6)^{2} }  \\  =  \sqrt{(3 + 4)^{2}  +  {( - 13)}^{2} }  \\  =   \sqrt{ 49 + 169}  \\   = \sqrt{218} \\  = 14.7648

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Question attached below
tamaranim1 [39]

Answer:

B . f(x+h)=\frac{x+h}{1+x+h}

Step-by-step explanation:

Given function is:

f(x)=\frac{x}{1+x}

Sall h is used to denote minor change in function. The value of the new function f(x+h) will be calculated by putting x+h in place of x in the function.

So putting x+h in place of x

f(x+h)= \frac{x+h}{1+(x+h)}\\=\frac{x+h}{1+x+h}

So, option B is the correct answer ..

6 0
4 years ago
Which of the following equations describes the graph?
Strike441 [17]

y-3x^2+4 is the answer of your question

8 0
2 years ago
Which solid figure has six faces (including the base(s)) and six vertices?
Mamont248 [21]

hexagon has 6 sides and 6 bases


7 0
3 years ago
Read 2 more answers
for a triangle to be obtuse only one of the angles must br shown to be obtuse. (in fact only one of the angles can possibly be o
Alchen [17]

Answer:

The explanation is given below.

Step-by-step explanation:

Given for a triangle to be obtuse only one of the angles must be shown to be obtuse but for a triangle to be acute all three angles must be shown to be acute.

According to Pythagoras theorem for a right angled triangle which has one angle of 90° rest of two are acute.

a^{2}+b^{2}=c^{2}

Hence, by Pythagorean inequality

a^{2}+b^{2}>c^{2}

which gives the triangle is acute.

Hence, one acute angle with the Pythagorean inequalities theorem shows that the triangle is acute.



5 0
3 years ago
Help!! Which of the following best represents Z1 • Z2 select all that apply
AURORKA [14]

z_1=\sqrt{3}\left(cos\:\frac{\pi }{4}+i\:sin\:\frac{\pi }{4}\right)

z_2=\sqrt{6}\left(cos\:\frac{3\pi \:}{4}+i\:sin\:\frac{3\pi \:}{4}\right)

\cos \left(\frac{\pi }{4}\right)=\frac{\sqrt{2}}{2}

\sin \left(\frac{\pi }{4}\right)=\frac{\sqrt{2}}{2}

\cos \left(\frac{3\pi }{4}\right)=-\frac{\sqrt{2}}{2}

\sin \left(\frac{3\pi }{4}\right)=\frac{\sqrt{2}}{2}\\

Z_1*Z_2=\sqrt{3}\left(cos\:\frac{\pi }{4}+i\:sin\:\frac{\pi }{4}\right)\cdot \sqrt{6}\left(cos\:\frac{3\pi \:}{4}+i\:sin\:\frac{3\pi \:}{4}\right)

=\sqrt{3}\left(\frac{\sqrt{2}}{2}+\:i\frac{\sqrt{2}}{2}\right)\cdot \sqrt{6}\left(\frac{-\sqrt{2}}{2}+\:i\frac{\sqrt{2}}{2}\right)

On simplifying, we get

Z_1* Z_2 =-3\sqrt{2}

<h2>Therefore, correct option is  1st option.</h2>
3 0
3 years ago
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