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2 years ago

PLSSS HELP I don’t get how to answer this question I’m having trouble with this question

Mathematics

2 answers:

Oksi-84 [34.3K]2 years ago

4 0

Since 5 of the sample deer are infected out of 50 you can use the fraction 5/50 to represent the population. Now the whole population would be 5250/525 since 5/50 would be 10% of that given population. 10% of 5250 is 525.

jek_recluse [69]2 years ago

3 0

Answer:

525 I think correct me if im wrong

Step-by-step explanation:

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What is the answer to x

Alex_Xolod [135]

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3 years ago

At how many points does the graph of the function below intersect the x-axis?

Ne4ueva [31]

Answer:

A.0

Step-by-step explanation:

y=4x²- 9x + 9

The function will intersect the x-axis when y = 0.

4x² - 9x + 9 = 0

a=4, b= -9, c=9

D=b² - 4ac = (-9)²- 4*4*9=81 - 144= - 63

D= -63, D<0, so the equation

4x² - 9x + 9 = 0 does not have real roots. That means that function y=4x²- 9x + 9 does not intersect x- axis. So, answer is A. 0.

5 0

3 years ago

Read 2 more answers

Let a = 60. What is the value of a ÷ 5? equation

Lerok [7]

Answer:

Step-by-step explanation:

Just substitute 60 for a and divide it by 5

Ta Da!!

8 0

3 years ago

A function is shown below: f(x) = x3 + 2x2 - x - 2 Part A: What are the factors of f(x)? Show your work. (3 points) Part B: What

Lena [83]

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2 years ago

This extreme value problem has a solution with both a maximum value and a minimum value. Use Lagrange multipliers to find the ex

azamat

Answer:

Maximum at points (8,0),(-8,0).Minimum at points (0,8), (0,-8).

Step-by-step explanation:

There are multiple ways of using lagrange multipliers. Most of them are equivalent.

Consider the function $F(x,y) = x^2-y^2-\lambda(x^2+y^2-64)$ . We want the following $\frac{\partial F}{\partial x} = \frac{\partial F}{\partial y} = \frac{\partial F}{\partial \lambda} = 0$ .

Then, we have

$\frac{\partial F}{\partial x} = 2x-2x\lambda= 2x(1-\lambda)=0$

$\frac{\partial F}{\partial y} = -2y-2y\lambda = -2y(1+\lambda)=0$

$\frac{\partial F}{\partial \lambda} = x^2+y^2-64=0$

From the first two equations, we can see that if $\lambda =1$ then necessarily y=0. IN that case, from the third equation (which is the restriction) gives us that $x=\pm 8$ .

On the other hand, if $\lambda=-1$ then necessarily x=0. Again, using the restriction this gives us that $y=\pm 8$ .

if we evaluate the original function in this points, we have that $f(0,\pm 8) = -64, f(\pm 8,0)=64$ . Then, we have Maximum at points (8,0),(-8,0) and Minimum at points (0,8), (0,-8).

3 0

3 years ago