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vesna_86 [32]
3 years ago
12

HELP PLEASE!!!!!! I will give anyone who knows the answer a brainliest!

Mathematics
2 answers:
vredina [299]3 years ago
6 0
\bf \textit{middle point of 2 points }\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
%  (a,b)
H&({{ 10}}\quad ,&{{ 1}})\quad 
%  (c,d)
K&({{ 4}}\quad ,&{{ 7}})
\end{array}\qquad
%   coordinates of midpoint 
\left(\cfrac{{{ x_2}} + {{ x_1}}}{2}\quad ,\quad \cfrac{{{ y_2}} + {{ y_1}}}{2} \right)
Alla [95]3 years ago
3 0
The answer is B ( 7, 4)

The formula is x1 (10 ) + x2 (4 ) / 2

So the x coordinate is 7.

Now, y1 (1 ) + y2 (7 ) / 2

The y coordinate is 4

Therefore your answer is (7,4)

Please Mark Brainliest if helped!

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Answer:

25.12 inches

Step-by-step explanation:

To find the circumference or the perimeter of a circle it is the diameter times pi. (pi = 3.14) Or you can do radius times 2 times pi.

Answer is 4 times 2 times 3.14 which is 25.12

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3 years ago
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x = 21

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Hope this helps!

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0.61m - 1.51m = 9 Solve for x
worty [1.4K]
X=(-10 ) Hope this helps!!! :)
4 0
3 years ago
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What type of function best models the data in the graph?
a_sh-v [17]

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Step-by-step explanation:

3 0
3 years ago
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State whether f is a function
Alenkinab [10]
<h2>Answer:</h2>

y=\pm (\frac{x}{6})^{\frac{1}{4}} \ is \ not \ a \ function

<h2>Step by step solution:</h2>

A function f from a set A to a set B is a relation that assigns to each element x in the set A exactly one element y in the set B. The set A is the domain (also called the set of inputs) of the function and the set B contains the range (also called the set of outputs). On the other hand, a function has an inverse function if and only if passes the Horizontal Line Test for Inverse Functions. This test tells us that a function f has an inverse function if and only if there is no any horizontal line that intersects the graph of f at more than one point. So the function is called one-to-one. The graph of f is shown below. As you can see, this function does not pass the Horizontal Line Test, therefore the inverse is not a function. However, let's find f^-{1}(x):

f(x)=6x^4 \\ \\ Substitute \ f(x) \ by \ y \\ \\ y=6x^4 \\ \\ Interchange \ x \ and \ y: \\ \\ x=6y^4 \\ \\ Solve \ for \ y: \\ \\ y^4=\frac{x}{6} \\ \\ Solving \\ \\y=\pm \sqrt[4]{\frac{x}{6}} \\ \\ \boxed{y=\pm \left(\frac{x}{6}\right)^{\frac{1}{4}}}

and this is not a function because there are elements in the set of inputs that match with two elements in the set of outputs.

5 0
3 years ago
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