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vlabodo [156]
3 years ago
7

Shawna and her best friend found some

Mathematics
2 answers:
Tomtit [17]3 years ago
8 0
I think $10 cause basically 2 people find money and they spilt it in half so they each get $5 so it would mostly likely be $10 if I’m wrong just let me know :)
Nutka1998 [239]3 years ago
4 0
I think it is $10 also. Because they both got $5 and 5+5=10.
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Select all the possible input-output pairs for the function
mrs_skeptik [129]

Answer:

A.(-1,-1)

E.(4,64)

Step-by-step explanation:

We are given that

Function

y=x^3

We have to find all possible input-output pairs for the function.

Substitute x=-1

y=(-1)^3=-1

Then, the pair is (-1,-1)

Now, substitute x=-2

y=(-2)^3=-8

Then, pair is (-2,-8)

Now, substitute x=3

y=3^3=27

Now, substitute x=4

y=4^3=64

Now, substitute x=1

Then, y=1

Hence, Option A and E are true.

A.(-1,-1)

E.(4,64)

7 0
2 years ago
Read 2 more answers
Find the weighted average of the numbers -1 and 1, with weight 2/3 on the first number and 1/3 on the second number.
jasenka [17]

The  weighted average of the numbers -1 and 1, with weight 2/3 on the first number and 1/3 on the second number is 0.33

What is weighted average?

Weighted average is the sum of the two numbers multiplied by their respective weights, in other words, the weighted average can be determined using the below formula :

Weighted average=(first number*its weight)+(second number*its weight)_

weighted average=(1*2/3)+(-1*1/3)

weighted average=0.33

Find out more about weighted average on:brainly.com/question/28042295

#SPJ1

6 0
1 year ago
4. The slope of the tangent for the function y = 1x is 1/(2(x). Find the equation of the tangent line at the point x = 1. Illust
Marianna [84]

Answer:

x-2y+1=0    

Step-by-step explanation:

We are given the following information in the question:

y = \sqrt{x}

Differentiating y with respect to x:

\displaystyle\frac{dy}{dx} = \frac{1}{2\sqrtx}

At x = 1

\displaystyle\frac{dy}{dx}\Bigr|_{\substack{x=1} }= \frac{1}{2\sqrt{1}}=\frac{1}{2}

y(1) = \sqrt{1} = 1

Equation of tangent:

(y-y_0) = \displaystyle\frac{dy}{dx}(x-x_0)

Putting the values:

(y-y(1)) = \displaystyle\frac{dy}{dx}\Bigr|_{\substack{x=1} }(x-1)\\\\y - 1= \frac{1}{2}(x-1)\\2y -2 = x - 1\\x-2y+1=0  

The above equations are the required equation of the tangent.

3 0
3 years ago
4.
tatyana61 [14]
1.6 chapters per hour lmk what it says
3 0
2 years ago
3 ways to get forty nine
IceJOKER [234]
7 x 7
2 + 47
56 - 7

Hope you needed this!
3 0
3 years ago
Read 2 more answers
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