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Softa [21]
3 years ago
7

Find the area 3yd/10yd

Mathematics
2 answers:
tamaranim1 [39]3 years ago
6 0

Answer: 15yd^2

Step-by-step explanation : 1/2•10•3 =15yd^2

Soloha48 [4]3 years ago
3 0
The area of a rectangle 3 yd by 10 yd is
   (3 yd) × (10 yd) = 30 yd²
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We can't do this one because we don't know terrance's and jessie's unit rate

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There are 2 rows of 8 chairs use the distributive property to solve
vitfil [10]
2(8) 2×8 =16 that is how u do it
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15 POINTS PLEASE HELP!!! WILL MARK BRAINLIEST
dangina [55]

Given: ∠ DEF

To construct: ∠TSZ ≅ ∠DEF

Construction: Consider the attachment

Step-01: Draw a line XY and choose a point S on it as a vertex of the required angle. Further marks point T such that DE = ST

Step-02: Take an arc AB from point E in ∠DEF of any length and draw at point S which cuts at point P on XY line.

Step-03: Take another arc of length AB from point B in ∠DEF and draw from point P which cuts to the previous arc at Q.

Step-04: Now, join the point SQ and extend up to Z such that EF = SZ

Hence, ∠ TSZ will be the required congruent constructed angle to∠DEF

6 0
3 years ago
Whats the answer to 3+2h=7
Brut [27]

Greetings! Hope this helps!

Answer

3 + 2h = 7

-3           -3

2h = 4

/2    /2

h = 2

Have a good day!

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A brainliest would help tons! :D

5 0
3 years ago
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Greg is trying to solve a puzzle where he has to figure out two numbers, x and y. Three less than two-third of x is greater than
Fittoniya [83]
Inequation 1: 

\frac{2}{3}x-3 \geq y

to plot the pairs (x, y) for which the inequation holds, draw the line y=\frac{2}{3}x-3

then pick a point in either side of the line. If that point is a solution of the inequation, than color that region of the line, if that point is not a solution, then color the other part of the line.

we do the same for the second inequation. Then the solution, is the region of the x-y axes colored in both cases.

inequation 2: 

y+ \frac{2}{3}x\ \textless \ 4

y\ \textless \ - \frac{2}{3} x+ 4



draw the lines 

i)  y=\frac{2}{3}x-3          use points (0, -3),  (3, -1)

ii)y=- \frac{2}{3} x+ 4       use points ( 0, 4),   (3, 2)


let's use the point P(3, 3) to see what region of the lines need to be coloured:

\frac{2}{3}x-3 \geq y  ; 
\frac{2}{3}(3)-3 \geq 3
2-3 \geq 3, not true so we color the region not containing this point


y+ \frac{2}{3}x\ \textless \ 4
(3)+ \frac{2}{3}(3)\ \textless \ 4
3+ 5\ \textless \ 4 not true, so we color the region not containing the point (3, 3)

The graph representing the system of inequalities is the region colored both red and blue, with the blue line not dashed, and the red line dashed.



4 0
3 years ago
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