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butalik [34]
4 years ago
8

The figure shows three quadrilaterals on a coordinate grid:

Mathematics
1 answer:
xxTIMURxx [149]4 years ago
7 0

Answer:

M and O are similar but not congruent.

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By road, without any stopovers, it would take Saad 14 hours and 27 minutes to reach Islamabad from Karachi. He leaves Karachi at
Furkat [3]

Answer:

8:48 p.m.

Step-by-step explanation:

First order half an hour plus 19 minutes used to refuel you will get 49 minutes then add it to the 14 hours 27 minutes you get dinner was 16 minutes then added to the time when he leaves Karachi which is 5:34 am which you get 8:48 p.m.

4 0
3 years ago
The following cube has a volume of 8000 cm3. Find the length of one side of the cube.
Artemon [7]
<span>Vcube = Ledge^3
8000cm^3 = Ledge^3

use the cube root of each side and you have the measure of an edge in cm.

8000^(1/3) = 20
  every edge is 20 cm

20^2 = 400
The area of each face is 400 cm^2. </span>
6 0
3 years ago
Please anyone answer me
ollegr [7]

Let's divide the shaded region into two areas:

area 1: x = 0 ---> x = 2

ares 2: x = 2 ---> x = 4

In area 1, we need to find the area under g(x) = x and in area 2, we need to find the area between g(x) = x and f(x) = (x - 2)^2. Now let's set up the integrals needed to find the areas.

Area 1:

A\frac{}{1}  = ∫g(x)dx =  ∫xdx =  \frac{1}{2}  {x}^{2}  | \frac{2}{0}  = 2

Area 2:

A\frac{}{2}  = ∫(g(x) - f(x))dx

= ∫(x -  {(x - 2)}^{2} )dx

=  ∫( - {x}^{2}  + 5x - 4)dx

= ( - \frac{1}{3}{x}^{3} +   \frac{5}{2} {x}^{2}  - 4x)   | \frac{4}{2}

= 2.67 - ( - 0.67) = 3.34

Therefore, the area of the shaded portion of the graph is

A = A1 + A2 = 5.34

3 0
3 years ago
what measuring tool did ancient mathematics have and how do they compare to the measuring tools we have today?​
german

Answer:

Search it up my guy itsthere

Step-by-step explanation:

3 0
3 years ago
How to divide a given ratio into a given amount
jonny [76]
JfpeiJKFHGIFWjfowednklsv WEOIHGVFOEWDBISFKCJPFGVUOEIDSVB PIFGHFIAGHRKFNBLKGVBJ;ALKFBGO'klVFBFJSKDGPV;KFJDZNBAJGIJGJNVL;ZGJEORITUJLDNVZ;.JDFG[OAEJRGNA'JGBj'gaklrfmgo'ajgb'pzang'ajrebgmkza'pgjozn;klfdbgkl'fznkp'fgnko'lgkfv.

4 0
3 years ago
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