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alina1380 [7]
2 years ago
8

PLEASE I NEED HELP

Mathematics
1 answer:
MArishka [77]2 years ago
4 0
D. Decreases in Quadrant ||| , i’m pretty sure
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Write an equation for the description
MissTica

Answer:

131=6w

Step-by-step explanation:

l=length\\\\w=width\\\\\Longrightarrow l=2w\\\\\Rightarrow Perimeter=l+l+w+w\\\\P=2l+2w\\\\P=2(2w)+2w\\\\P=4w+2w\\\\P=6w\\\\P=131\\\\131=6w\\\\w=\frac{131}{6}\\\\w=21.8\overline{3}\\\\l=43.\overline{6}

5 0
3 years ago
Read 2 more answers
Given quadrilateral EFGH at e(-4,7) f(8,4) G(t,-5) and h(-7,-1) using coordinate geometry prove EFGH is a rectangle
Tcecarenko [31]

The above question was not properly written.

Complete Question

Given quadrilateral EFGH with vertices at E(-4,8), F(8,4), G(5,-5) and H(-7,-1), prove using coordinate geometry that EFGH is a rectangle.

Answer:

Quadrilateral EFGH is a rectangle.l because:

EF = GH and FG = EH

Step-by-step explanation:

The formula for coordinate geometry is given as :

√(x2 - x1)² + (y2 - y1)² when we have coordinates: (x1, y1) and (x2 , y2)

For the quadrilateral EFGH with given coordinates above to be a rectangle,

EF = GH

FG = EH

Hence:

For side EF

E(-4,8), F(8,4)

= √(8 - (-4))² + (4 - 8)²

= √12² + -4²

= √144 + 16

= √160 units

For side FG

F(8,4), G(5,-5)

=√(5 - 8)² + (-5 - 4)²

= √-3² + -9²

= √9 + 81

= √90 units

For Side GH

G(5,-5) , H(-7,-1)

= √(-7 - 5)² + (-1 - (-5))²

= √-12² + 4²

= √144 + 16

= √160 units

For side EH

E(-4,8), H(-7,-1)

= √(-7 -(-4))² +(-1 - 8)²

= √-3² + -9²

= √9 + 81

= √90 units

From the above calculation, we can see that truly,

EF = GH

FG = EH

Therefore, quadrilateral EFGH is a rectangle.

6 0
2 years ago
How do you round 3.36 to the tenths place
bagirrra123 [75]

Answer: 3.4

Step-by-step explanation: Because, you're rounding to the nearst tenth and .36 has a 6 in it anything above a 5 you round up and anything under you round down.

4 0
3 years ago
Read 2 more answers
TIMED QUESTION NEED HELP FASTWhat values of b satisfy 3(2b + 3)2 = 36?
egoroff_w [7]

Answer:

\frac{-3+ 2\sqrt{3}}{2}\textrm{ and }\frac{-3- 2\sqrt{3}}{2}

Step-by-step explanation:

3(2b+3)^{2}=36\\\frac{3(2b+3)^{2}}{3}=\frac{36}{3}\\(2b+3)^{2}=12

Taking square root both sides, we get

\sqrt{(2b+3)^{2}}=\pm \sqrt{12}\\2b+3=\pm \sqrt{4\times 3}\\2b+3=\pm 2\sqrt{3}\\2b+3-3=\pm 2\sqrt{3}-3\\2b=-3\pm 2\sqrt{3}\\b=\frac{-3\pm 2\sqrt{3}}{2}\\b=\frac{-3+ 2\sqrt{3}}{2}\textrm{ or }b=\frac{-3- 2\sqrt{3}}{2}

Therefore, the values of b are:

\frac{-3+ 2\sqrt{3}}{2}\textrm{ and }\frac{-3- 2\sqrt{3}}{2}

5 0
3 years ago
Read 2 more answers
What is the probability of rolling a number greater than 4 on a fair number cube (dice)? (Simplify to lowest terms)
Reptile [31]

Answer:

1/3

Step-by-step explanation:

there are 2 numbers greater than 4 on dice and there are 6 numbers on a die

3 0
3 years ago
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