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Karolina [17]
2 years ago
6

Find the surface area of the figure.

Mathematics
1 answer:
JulijaS [17]2 years ago
6 0

Answer:

144

Step-by-step explanation:

Area of 1 triangle: 20 in

Area of 4 triangles: 80 in

Area of the square: 64 in

Add the 4 triangles and the square.

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Can you please help me with this
Y_Kistochka [10]

Answer:

1157 / 500

Step-by-step explanation:

8 0
3 years ago
Use factoring to solve <br> 3x^2-11x-6=0
Zielflug [23.3K]

Answer:

x = -0.62 and x = 4.29

These values are approximations !

Step-by-step explanation:

You can not use Factoring to solve this equation. Perhaps you did not use the exact notation as in the original question?

Anyway, to solve it, you can still use the ABC formula.

3x²-11x-6=0

a = 3

b =-11

c = -6

x = [11 + √{121 -4(3 *-6)} ]/ 2*3

x = [11 + √{121 -4(-24)} ]/ 6

x = [11 + √{121 + 96} ]/ 6

x = [11 + √{217} ]/ 6

x = 4.29 This is an approximation !

x = [11 - √{121 -4(3 *-6)} ]/ 2*3

x = [11 - √{121 -4(-24)} ]/ 6

x = [11 - √{121 + 96} ]/ 6

x = [11 - √{217} ]/ 6

x = -0.62 This is an approximation !

extra: see the graph.

4 0
3 years ago
On the way to the mall Miguel rides his skateboard to get to the bus stop. He then waits a few minutes for the bus to come, then
professor190 [17]
The answer would be c .
4 0
3 years ago
Read 2 more answers
The table below gives data from a linear function. find a formula for the function.
baherus [9]
Given the following table that gives data from a linear function:

\begin {tabular}&#10;{|c|c|c|c|}&#10;Temperature, $y = f(x)$ (^\circ C)&0&5&20 \\ [1ex]&#10;Temperature, $x$ (^\circ F)&32&41&68 \\ &#10;\end {tabular}

The formular for the function can be obtained by choosing two points from the table and using the formular for the equation of a straight line.

Recall that the equation of a straight line is given by
\frac{y-y_1}{x-x_1} = \frac{y_2-y_1}{x_2-x_1}

Using the points (32, 0) and (41, 5), we have:
\frac{y-0}{x-32} = \frac{5-0}{41-32}= \frac{5}{9}  \\  \\ \Rightarrow y=\frac{5}{9}(x-32)
4 0
3 years ago
A manufacturer of chocolate chips would like to know whether its bag filling machine works correctly at the 430 gram setting. It
Zanzabum

Answer: We reject H₀  we find that the machine is underflling the bottles

P ( 421 ± 4,3376 )

Step-by-step explanation:

We assume a normal distribution

Population mean 430 grs

Unknown standard deviation

We have a one tail condition "underfilling"

And our test is:

Null hypothesis      H₀         X  =  μ₀

Alternative hypothesis    Hₐ     X  < μ₀

We must use t student distribution and find the interval

X ± t*(s/√n)

In that expession   X is the sample mean  421 grs, "s"  is sample standard deviation, n is sample size, then

421 ±  t * ( 15 / √21 )          (1)

We go to t table and look for t value for  α = 0,1 and df = 21 - 1   df = 20

we get t (remember it is a one tail test)  t = 1,325, plugging this value in equation (1) we get the interval

421 ± 1,325 * ( 15/√21 )    ⇒  421 ±  1,325 * ( 3,2737)

421 ± 4,3376  

421 + 4,3376  = 425,34

421 - 4,3376  = 416,66

As we can see the mean value of the population 430 grs is not inside the interval  [ 416,66 ;  425,34 ] then we can assure the machine is underfilling the bags, and not meeting the setting spec

6 0
4 years ago
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