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Anna11 [10]
3 years ago
11

Forget how to do this please someone help

Mathematics
1 answer:
tangare [24]3 years ago
6 0

Answer:

9.830

Step-by-step explanation:

I'm not sure if it's right but hope this helps you

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Helps mes plzs!!! Its really hard
Likurg_2 [28]

Answer:

0.6

Step-by-step explanation:

-0.2=s+(-0.8)

-0.2=s-0.8

-0.2+0.8=s

s=0.6

5 0
3 years ago
Draw a rough sketch of a quadrilateral KLMN. State, (a) two pairs of opposite sides, (b) two pairs of opposite angles, (c) two p
kogti [31]

Answer:

Please find attached the drawing of quadrilateral KLMN created with MS Whiteboard using the Ink to Shape command

(a) Two pairs of opposite sides are \overline {KN}, \overline {LM} and \overline {KL}, \overline {NM}

(b) Two pairs of opposite angles are ∠LKN, ∠LMN, and ∠KLM and ∠KNM

(c) Two pairs of adjacent sides are \overline {KN}, \overline {NM} and \overline {KL}, \overline {LM}

(d) Two pairs of adjacent angles are ∠LKN, ∠KLM and ∠LMN, ∠KNM

Step-by-step explanation:

7 0
3 years ago
X^2+y^2=169, 3x+2y=39 please answer using substitution and each step
Oksanka [162]

Answer:

x=13,y=0

Step-by-step explanation:

We are given a system of equations

{x}^{2}  +  {y}^{2}  = 169 -  - eqn \: 1 \\ 3x + 2y = 39 -  - eqn \: 2

For equation 1,square all terms to reduce it

√x²+√y²=169

x+y=13

Make x the subject as required by the question to use substitution method

x=13-y

Plug x=13-y into eqn 2

3(13-y)+2y=39

39-3y+2y=39

39-y=39

y=39-39=0

Plug y=0 into equation 1

x+0=13

x=0

7 0
2 years ago
Layla went shopping for camping equipment. After looking at several different stores, she bought a tent for $163.63 and a sleepi
Ghella [55]

Answer:

$297.09

Step-by-step explanation:

To answer this question, just add:

163.63 + 133.46 = 297.09

So, Layla spent $297.09 in all

Hope this helps :)

5 0
3 years ago
find the perimeter and diagonal of a square whose area is equal to the area of a rectangle whose length is 9m and breadth is 4m
aalyn [17]
Area\ of\ a\ rectangle:A_{\fbox{ }}=9m\times4m=36m^2\\\\a-side\ of\ a\ square\\\\Area\ of\ a\ square:A_{\fbox{}}=a^2\\\\A_{\fbox{ }}=A_{\fbox{}}\Rightarrow a^2=36m^2\Rightarrow a=\sqrt{36m^2}\Rightarrow a=6m\\\\Perimeter\ of\ the\ square:P=4a\Rightarrow P=4\times6m=\boxed{24m}\\\\Diagonal\ of\ the\ square:d=a\sqrt2\Rightarrow d=\boxed{6\sqrt2\ m}
4 0
3 years ago
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