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Bogdan [553]
3 years ago
12

20 pts! If Quadrilateral J K L M is congruent to quadrilateral C B D A, which pair of sides must be congruent? Segment J K and S

egment A B Segment J K and Segment C B Segment J M and Segment A D Segment J M and Segment B C
Mathematics
1 answer:
Natali5045456 [20]3 years ago
8 0

Answer:

segment I'm and segment ad

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Someone who can do this help me please.
lianna [129]

Answer:

x= 2

Step-by-step explanation:

the shapes are 6 in apart

5 0
3 years ago
What is the answer and work for these attached below
GrogVix [38]
-4(-5-b)=1/3(b+16) Multiply both sides by 3 to get rid of the fraction
-12(-5-b)=b+16  distribute the -12 to get rid of the parenthesis 
60+12b=b+16  get the b on the left side and non b values to the right side
11b=-44 solve for b
b=-44/11 simplify the fraction
b=-4

3/5(t+18)=-3(2-t)   multiply both sides by 5/3 to get rid of thefraction
t+18=-5(2-t)   distribute the -5 to get rid of the parenthises
t+18=-10+5t  get the t to the left side and non t values to the right
-4t=-28  solve for t
t=7
5 0
3 years ago
Bryan is baking cookies for Zachary's birthday party. Bryan uses 1 cups
Nikolay [14]

Answer:

1 more cup

Step-by-step explanation:

1:10

x:50

10*5=50

1*5=5

5-4=1

1 more cup

8 0
3 years ago
Help please. <br><br> Please <br><br> Please
iragen [17]

2x - 6 = 60 - x  \\ add \: x \: to \:both \: sides \\  2x + x - 6 = 60 - x + x \\ 3x - 6 = 60 \\ add \: 6 \: to \: both \: sides \\ 3x - 6 + 6 = 60 + 6 \\ 3x = 66 \\ divide \: both \: sides \: by \: 3 \\  \frac{3x}{3}  =  \frac{66}{3}  \\ x = 22

8 0
2 years ago
Solve 2x^2 + x - 4 = 0 <br> X2 +
damaskus [11]

Answer:

\large \boxed{\sf \ \ x = -\dfrac{\sqrt{33}+1}{4} \ \ or \ \ x = \dfrac{\sqrt{33}-1}{4} \ \ }

Step-by-step explanation:

Hello, please find below my work.

2x^2+x-4=0\\\\\text{*** divide by 2 both sides ***}\\\\x^2+\dfrac{1}{2}x-2=0\\\\\text{*** complete the square ***}\\\\x^2+\dfrac{1}{2}x-2=(x+\dfrac{1}{4})^2-\dfrac{1^2}{4^2}-2=0\\\\\text{*** simplify ***}\\\\(x+\dfrac{1}{4})^2-\dfrac{1+16*2}{16}=(x+\dfrac{1}{4})^2-\dfrac{33}{16}=0

\text{*** add } \dfrac{33}{16} \text{ to both sides ***}\\\\(x+\dfrac{1}{4})^2=\dfrac{33}{16}\\\\\text{**** take the root ***}\\\\x+\dfrac{1}{4}=\pm \dfrac{\sqrt{33}}{4}\\\\\text{*** subtract } \dfrac{1}{4} \text{ from both sides ***}\\\\x = -\dfrac{1}{4} -\dfrac{\sqrt{33}}{4} \ \ or \ \ x = -\dfrac{1}{4} +\dfrac{\sqrt{33}}{4}

Hope this helps.

Do not hesitate if you need further explanation.

Thank you

4 0
3 years ago
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