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Vladimir [108]
2 years ago
5

The two figures are similar. Find the ratios (red to blue) of the perimeters and of the areas. Write the ratios as fractions in

simplest form.

Mathematics
2 answers:
ivann1987 [24]2 years ago
7 0

Answer:

Ratio of perimeters = \frac{3}{4}

Ratio of areas = \frac{9}{16}

Step-by-step explanation:

Thinking process:

The ratio of the parameters is given by the following:

ratio = \frac{length of side 1}{length of side 2}

        =\frac{6}{8} \\= \frac{3}{4}

The ratio of the areas is given by:

ratio = (\frac{length of side 1}{length of side 2}) ^{2}

        = (\frac{6}{8}) ^{2}

        = \frac{36}{64}

        = \frac{9}{16}

Degger [83]2 years ago
4 0
I think the ratio of the perimeter is 6/8 which simplifies to 3/4 and I'm sorry idk the ratio of area..... hope that helped though
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juin [17]

The answer is C.

Note when Soro divides both sides by -2.5 to solve the equation. When multiplying/dividing by a negative number, one should always reverse the inequality sign, but Soro forgot to do this.

-2.5x≥-30

After this step, the answer SHOULD HAVE been:

x≤12

Let me know if you need any clarifications, thanks!

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2 years ago
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A local project being analyzed by PERT has 42​ activities, 13 of which are on the critical path. If the estimated time along the
ryzh [129]

Answer:

the probability that the project will be completed in 95 days or​ less, P(x ≤ 95) = 0.023

Step-by-step explanation:

This is a normal probability distribution question.

We'll need to standardize the 95 days to solve this.

The standardized score is the value minus the mean then divided by the standard deviation.

z = (x - xbar)/σ

x = 95 days

xbar = mean = 105 days

σ = standard deviation = √(variance) = √25 = 5

z = (95 - 105)/5 = - 2

To determine the probability that the project will be completed in 95 days or​ less, P(x ≤ 95) = P(z ≤ (-2))

We'll use data from the normal probability table for these probabilities

P(x ≤ 95) = P(z ≤ (-2)) = 0.02275 = 0.023

5 0
2 years ago
Daisy has 5 1/2 pounds of chocolate she uses for fifth of it to bake brownies how much chocolate does she use to bake brownies
oksano4ka [1.4K]
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4 0
1 year ago
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The graphs of the quadratic functions f(x) = 6 – 10x2 and g(x) = 8 – (x – 2)2 are provided below. Observe there are TWO lines si
natta225 [31]

Answer:

a) y = 7.74*x + 7.5

b)  y = 1.148*x + 6.036

Step-by-step explanation:

Given:

                                  f(x) = 6 - 10*x^2

                                  g(x) = 8 - (x-2)^2

Find:

(a) The line simultaneously tangent to both graphs having the LARGEST slope has equation

(b) The other line simultaneously tangent to both graphs has equation,

Solution:

- Find the derivatives of the two functions given:

                                f'(x) = -20*x

                                g'(x) = -2*(x-2)

- Since, the derivative of both function depends on the x coordinate. We will choose a point x_o which is common for both the functions f(x) and g(x). Point: ( x_o , g(x_o)) Hence,

                                g'(x_o) = -2*(x_o -2)

- Now compute the gradient of a line tangent to both graphs at point (x_o , g(x_o) ) on g(x) graph and point ( x , f(x) ) on function f(x):

                                m = (g(x_o) - f(x)) / (x_o - x)

                                m = (8 - (x_o-2)^2 - 6 + 10*x^2) / (x_o - x)

                                m = (8 - (x_o^2 - 4*x_o + 4) - 6 + 10*x^2)/(x_o - x)

                                m = ( 8 - x_o^2 + 4*x_o -4 -6 +10*x^2) /(x_o - x)

                                m = ( -2 - x_o^2 + 4*x_o + 10*x^2) /(x_o - x)

- Now the gradient of the line computed from a point on each graph m must be equal to the derivatives computed earlier for each function:

                                m = f'(x) = g'(x_o)

- We will develop the first expression:

                                m = f'(x)

                                ( -2 - x_o^2 + 4*x_o + 10*x^2) /(x_o - x) = -20*x

Eq 1.                          (-2 - x_o^2 + 4*x_o + 10*x^2) = -20*x*x_o + 20*x^2

And,

                              m = g'(x_o)

                              ( -2 - x_o^2 + 4*x_o + 10*x^2) /(x_o - x) = -20*x

                              -2 - x_o^2 + 4*x_o + 10*x^2 = -2(x_o - 2)(x_o - x)

Eq 2                       -2 - x_o^2 + 4*x_o+ 10*x^2 = -2(x_o^2 - x_o*(x + 2) + 2*x)

- Now subtract the two equations (Eq 1 - Eq 2):

                              -20*x*x_o + 20*x^2 + 2*x_o^2 - 2*x_o*(x + 2) + 4*x = 0

                              -22*x*x_o + 20*x^2 + 2*x_o^2 - 4*x_o + 4*x = 0

- Form factors:       20*x^2 - 20*x*x_o - 2*x*x_o + 2*x_o^2 - 4*x_o + 4*x = 0

                              20*x*(x - x_o) - 2*x_o*(x - x_o) + 4*(x - x_o) = 0

                               (x - x_o)(20*x - 2*x_o + 4) = 0  

                               x = x_o   ,     x_o = 10x + 2    

- For x_o = 10x + 2  ,

                               (g(10*x + 2) - f(x))/(10*x + 2 - x) = -20*x

                                (8 - 100*x^2 - 6 + 10*x^2)/(9*x + 2) = -20*x

                                (-90*x^2 + 2) = -180*x^2 - 40*x

                                90*x^2 + 40*x + 2 = 0  

- Solve the quadratic equation above:

                                 x = -0.0574, -0.387      

- Largest slope is at x = -0.387 where equation of line is:

                                  y - 4.502 = -20*(-0.387)*(x + 0.387)

                                  y = 7.74*x + 7.5          

- Other tangent line:

                                  y - 5.97 = 1.148*(x + 0.0574)

                                  y = 1.148*x + 6.036

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the price of a pack of gum today is 63 cents. this is 3 cents more than three times the price ten years ago. what was the price
Paul [167]
The answer would be .43 
7 0
1 year ago
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