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enyata [817]
3 years ago
10

What is the distance between (2,3) and (5,7)

Mathematics
1 answer:
valentinak56 [21]3 years ago
4 0
This is the rule for any calculation of this type. Note that you can replace X(B) by X(A) and the same for Y because the answers are squared and you'll always get the same value

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Lubov Fominskaja [6]
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I really hope this help
8 0
3 years ago
Write a TRUE statement using the words acute angle and regular octagan.<br><br> THANK YOU!!
marishachu [46]
In a regular octagon it is impossible to have an acute angle!
Hope this helps!!
3 0
3 years ago
Read 2 more answers
If f(x)= x-3 over x and g(x)= 5x-4 what is the domain of (f•g)(x)
marusya05 [52]

Answer:

I attached the work to your problem below.

I hope it helps.

8 0
3 years ago
Solve the following and explain your steps. Leave your answer in base-exponent form. (3^-2*4^-5*5^0)^-3*(4^-4/3^3)*3^3 please st
Naily [24]

Answer:

\boxed{2^{\frac{802}{27}} \cdot 3^9}

Step-by-step explanation:

<u>I will try to give as many details as possible. </u>

First of all, I just would like to say:

\text{Use } \LaTeX !

Texting in Latex is much more clear and depending on the question, just writing down without it may be confusing or ambiguous. Be together with Latex! (*^U^)人(≧V≦*)/

$(3^{-2} \cdot 4^{-5} \cdot 5^0)^{-3} \cdot (4^{-\frac{4}{3^3} })\cdot 3^3$

Note that

\boxed{a^{-b} = \dfrac{1}{a^b}, a\neq 0 }

The denominator can't be 0 because it would be undefined.

So, we can solve the expression inside both parentheses.

\left(\dfrac{1}{3^2}  \cdot \dfrac{1}{4^5}  \cdot 5^0 \right)^{-3} \cdot \left(\dfrac{1}{4^{\frac{4}{3^3} } }\right)\cdot 3^3

Also,

\boxed{a^{0} = 1, a\neq 0 }

\left(\dfrac{1}{9}  \cdot \dfrac{1}{1024}  \cdot 1 \right)^{-3} \cdot \left(\dfrac{1}{4^{\frac{4}{27} } }\right)\cdot 27

Note

\boxed{\dfrac{1}{a} \cdot \dfrac{1}{b}= \frac{1}{ab} , a, b \neq  0}

\left(\dfrac{1}{9216}   \right)^{-3} \cdot \left(\dfrac{1}{4^{\frac{4}{27} } }\right)\cdot 27

\left(\dfrac{1}{9216}   \right)^{-3} \cdot \left(\dfrac{27}{4^{\frac{4}{27} } }\right)

\left( \dfrac{1}{\left(\dfrac{1}{9216}\right)^3} \right)\cdot \left(\dfrac{27}{4^{\frac{4}{9} } }\right)

\left( \dfrac{1}{\left(\dfrac{1}{9216}\right)^3} \right)\cdot \left(\dfrac{27}{4^{\frac{4}{27} } }\right)

Note

\boxed{\dfrac{1}{\dfrac{1}{a} }  = a}

9216^3\cdot \left(\dfrac{27}{4^{\frac{4}{9} } }\right)

\left(\dfrac{ 9216^3\cdot 27}{4^{\frac{4}{27} } }\right)

Once

9216=2^{10}\cdot 3^2 \implies  9216^3=2^{30}\cdot 3^6

\boxed{(a \cdot b)^n=a^n \cdot b^n}

And

$4^{\frac{4}{27}} = 2^{\frac{8}{27} $

We have

\left(\dfrac{ 2^{30} \cdot 3^6\cdot 27}{2^{\frac{8}{27} } }\right)

Also, once

\boxed{\dfrac{c^a}{c^b}=c^{a-b}}

2^{30-\frac{8}{27}} \cdot 3^6\cdot 27

As

30-\dfrac{8}{27} = \dfrac{30 \cdot 27}{27}-\dfrac{8}{27}  =\dfrac{802}{27}

2^{30-\frac{8}{27}} \cdot 3^6\cdot 27 = 2^{\frac{802}{27}} \cdot 3^6 \cdot 3^3

2^{\frac{802}{27}} \cdot 3^9

4 0
3 years ago
The linear regression line for Tampa Tribune Database is y=8x+367. In approximately what year is the Tampa Tribune expecting to
lakkis [162]

The Tampa Tribune expecting to add 700 new pictures per year to their database in 2041

<h3>The linear equation of the graph</h3>

The equation of the line of best fit is given as:

y = 8x + 367

When the number of pictures added is 700, we have:

y = 700

Substitute 700 for y in y = 8x + 367

700 = 8x + 367

Subtract 367 from both sides of the equation

333= 8x

Rewrite the above equation as:

8x = 333

Divide both sides by 8

x = 41.625

Remove decimal (do not approximate)

x = 41

This means that:

Year = 2000 +41

Year = 2041

Hence, the Tampa Tribune expecting to add 700 new pictures per year to their database in 2041

Read more about linear regression at:

brainly.com/question/26137159

8 0
3 years ago
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