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3 years ago

Dixon leaves school to go home.he walks 6 blocks north and 8 blocks west. How far is Dixon from the school

Mathematics

2 answers:

GenaCL600 [577]3 years ago

6 0

I don’t know I’m really sorry

erica [24]3 years ago

5 0

Answer:

2 blocks would be the correct answer

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3 years ago

The average profit a local store owner earns on a given day is $960 and is growing

Ostrovityanka [42]

The average profit of the local store is an illustration of an exponential function

The profit function is: $y = 960 * (0.9894)^{x}$

<h3>How to determine the function</h3>

From the question, we have:

Initial value (a) = 960
Rate per year (b) = 88%

An exponential function is represented as:

$y = ab^x$

So, the function that represents the profit per year is:

$y = 960 * (88\%)^x$

Divide x by 12 to get the monthly rate

$y = 960 * (88\%)^{x/12}$

Evaluate the exponent

$y = 960 * (0.9894)^{x}$

Hence, the profit function is: $y = 960 * (0.9894)^{x}$

Read more about exponential functions at:

brainly.com/question/11464095

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2 years ago

8x +2 factored using the GCF

wolverine [178]

The answer is 2 because I solved it and checked it.

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3 years ago

Lagrange Multiplier Million Dollar question. Anybody please Use the method of Lagrange multipliers to find the max and min value

jekas [21]

We are given the equation: g(s,t) = t^2 e^s

Which is subject to the constraint: s^2 + t^2=3

The points (x,y) that will maximize g(s,t) will be the points that will satisfy the equation ∇f(x, y, z) = λ∇g(x, y, z), therefore:

2 t e^s = λ (2 t), so e^s = λ --> 1

t^2 e^s = λ (2 s) --> 2

s^2 + t^2 = 3 -->3

To solve this problem, note that λ cannot be zero by equation 1 since e^s can never be zero. Therefore plug in equation 1 to 2:

t^2 e^s = (e^s) (2 s)

t^2 = 2 s --> 4

Plug in equation 4 to 3:

s^2 + 2s = 3

By completing the square:

s^2 + 2s + 1 = 4

(s + 1)^2 = 4

s + 1 = ±2

s = -3, 1

Calculating for t using equation 4:

when s = -3

t^2 = 2(-3)

t = sqrt(-6)

Since t is imaginary, therefore s=-3 is not a solution

when s = 1

t^2 = 2(1)

t = sqrt(2) = ±1.414

Therefore the maxima and minima points are at:

(1, -1.414) and (1, 1.414)

g(1, -1.414)=(-1.414)^2 e^(1) = 5.434

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3 years ago

Which is the correct input-output table for the function g(x)=-1/2(4x+6)

Sonbull [250]

Answer: The third choice.

Step-by-step explanation: To find the first output, start with your first input. In this case, it's -2, so your equation would be g(-2)= -1/2(4(-2)+6). Start by solving inside the parenthesis. 4x-2=-8 and -8+6=-2. -1/2x-2= 1, so your first output should be 1. The only choice that has this is the third one, so that is your answer. Hope I could help :)

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