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trapecia [35]
2 years ago
10

Find the surface area of the prism 3m 4m 8m 5m

Mathematics
2 answers:
juin [17]2 years ago
6 0

Answer:

Step-by-step explanation:

Two triangles.

one triangle = 1/2 * small leg * middle sized leg

one triangle = 1/2 * 3 * 4

One triangle = 6

Two triangles = 6 * 2 =                                            12

Front

Area = L * W

L = 8

W = 5

Area = 8*5                                                              40

Left Side

L = 8

W = 3

Area = 8*3

Area = 24

Area=                                                                     24

Right Side

L = 8

W = 4

Area =                                                         <u>          32</u>

Total Area = 32 + 24 + 40 + 12                          108

There is another way to do this.

Since the height is the same for all three faces, you can do it like this

Area = 8*(3 + 4 + 5) = 8(j12) = 96

Area of the two tiangles     = 12

Total area =                          108

Rashid [163]2 years ago
4 0

Answer:

3 x 4 = 12, 12 divided by 2 ) x 1/2) = 6, 6 x 2 = 12(area of two triangles)

8 x 4 = 32

8 x 3 = 24

8 x 5 = 40

Let's add all the given areas:

12 + 32 + 24 + 40 = 108 m^2 is your answer.

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3) 8p + 7q = 43<br>2 -7=-q<br>how can I solve this with substitution?​
Bess [88]

Answer:

p=1, q=5. (1, 5).

Step-by-step explanation:

8p+7q=43

2-7=-q

------------------

-q=-5

q=5

--------

8p+7(5)=43

8p+35=43

8p=43-35

8p=8

p=8/8=1

6 0
3 years ago
I don't know to find the answer to 8. Can someone explain to me?
lapo4ka [179]
Perhaps the easiest way to find the midpoint between two given points is to average their coordinates: add them up and divide by 2.

A) The midpoint C' of AB is
.. (A +B)/2 = ((0, 0) +(m, n))/2 = ((0 +m)/2, (0 +n)/2) = (m/2, n/2) = C'
The midpoint B' is
.. (A +C)/2 = ((0, 0) +(p, 0))/2 = (p/2, 0) = B'
The midpoint A' is
.. (B +C)/2 = ((m, n) +(p, 0))/2 = ((m+p)/2, n/2) = A'

B) The slope of the line between (x1, y1) and (x2, y2) is given by
.. slope = (y2 -y1)/(x2 -x1)
Using the values for A and A', we have
.. slope = (n/2 -0)/((m+p)/2 -0) = n/(m+p)

C) We know the line goes through A = (0, 0), so we can write the point-slope form of the equation for AA' as
.. y -0 = (n/(m+p))*(x -0)
.. y = n*x/(m+p)

D) To show the point lies on the line, we can substitute its coordinates for x and y and see if we get something that looks true.
.. (x, y) = ((m+p)/3, n/3)
Putting these into our equation, we have
.. n/3 = n*((m+p)/3)/(m+p)
The expression on the right has factors of (m+p) that cancel*, so we end up with
.. n/3 = n/3 . . . . . . . true for any n

_____
* The only constraint is that (m+p) ≠ 0. Since m and p are both in the first quadrant, their sum must be non-zero and this constraint is satisfied.

The purpose of the exercise is to show that all three medians of a triangle intersect in a single point.
7 0
3 years ago
A roadside vegetable stand sells pumpkins for $5 each and tomatoes for $3 each. The
stealth61 [152]

Answer:

Step-by-step explanation: I need help

3 0
2 years ago
the half-life of chromium-51 is 38 days. If the sample contained 510 grams. How much would remain after 1 year?​
madam [21]

Answer:

About 0.6548 grams will be remaining.  

Step-by-step explanation:

We can write an exponential function to model the situation. The standard exponential function is:

f(t)=a(r)^t

The original sample contained 510 grams. So, a = 510.

Each half-life, the amount decreases by half. So, r = 1/2.

For t, since one half-life occurs every 38 days, we can substitute t/38 for t, where t is the time in days.

Therefore, our function is:

\displaystyle f(t)=510\Big(\frac{1}{2}\Big)^{t/38}

One year has 365 days.

Therefore, the amount remaining after one year will be:

\displaystyle f(365)=510\Big(\frac{1}{2}\Big)^{365/38}\approx0.6548

About 0.6548 grams will be remaining.  

Alternatively, we can use the standard exponential growth/decay function modeled by:

f(t)=Ce^{kt}

The starting sample is 510. So, C = 510.

After one half-life (38 days), the remaining amount will be 255. Therefore:

255=510e^{38k}

Solving for k:

\displaystyle \frac{1}{2}=e^{38k}\Rightarrow k=\frac{1}{38}\ln\Big(\frac{1}{2}\Big)

Thus, our function is:

f(t)=510e^{t\ln(.5)/38}

Then after one year or 365 days, the amount remaining will be about:

f(365)=510e^{365\ln(.5)/38}\approx 0.6548

5 0
2 years ago
Which ordered pairs is a solution to -5x + 3y &gt; 12?
GaryK [48]

9514 1404 393

Answer:

  (-5, 5), (2, 8), (-6, 0)

Step-by-step explanation:

It is convenient to graph the solution, then plot the points to see which fall in the solution area.

The point (3, 9) falls on the boundary line, which is <em>not</em> part of the solution set.

The points that are solutions are ...

  B(-5, 5)

  E(2, 8)

  F(-6, 0)

5 0
3 years ago
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