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const2013 [10]
2 years ago
10

ACTIVITY 1: LET'S PRACTICE!

Mathematics
1 answer:
V125BC [204]2 years ago
8 0

Answer:

1. YES

1. YES2. YES

1. YES2. YES3.YES

4.NO

Hope it helps

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A system of equations is graphed on the coordinate plane.
ANTONII [103]
You have 2y=3x-1 and 4y=6x-2. first step is to get the y by itself . first equation you divide everything by 2 which gives you the equation: y= 3/2x-1/2
or y= 1.5x-.5, for the second equation you divide everything by 4, which gives you y=6/4x-2/4 (1/2) or y=1.5-.5. both equations are equal to one another: 
the answer is therefore C Infinitely many solutions.

8 0
3 years ago
Solve the right triangle.<br> Round your answers to the nearest tenth.
Nadya [2.5K]

*☆*――*☆*――*☆*――*☆*――*☆*――*☆*――*☆*――*☆**☆*――*☆*――*☆*――*☆

Answer: c=38.89

b=29.79

<A=40

Explanation:

I hope this helped!

<!> Brainliest is appreciated! <!>

- Zack Slocum

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5 0
3 years ago
Read 2 more answers
Write the equation of the line that is parallel to y+2=5(x-1) and passes through the point (-1, 0)
zavuch27 [327]

Answer:

Step-by-step explanation:

slope of required line=5

point is (-1,0)

reqd. eq. of line is

y-0=5(x-(-1))

y=5(x+1)

y=5x+5

8 0
2 years ago
M is between L and N. LM = 7x-1, MN = 2x +4, and LN = 12. find the value of ‘x’ an determine if M is a bisector.
Sholpan [36]

LM + MN = LN

7x - 1 + 2x + 4 = 12

9x + 3 = 12

9x = 12 - 3

9x = 9

x = 9/9

x = 1

7 0
3 years ago
Due Soon Need Help Geometry!
AveGali [126]

 

Some basic formulas involving triangles

\ a^2 = b^2 + c^2 - 2bc \textrm{ cos } \alphaa  2 =b  2+2 + c 2

−2bc cos α

\ b^2 = a^2 + c^2 - 2ac \textrm{ cos } \betab   2=

 

m_b^2 = \frac{1}{4}( 2a^2 + 2c^2 - b^2 )m   b2 = 41(2a 2 + 2c 2-b 2)

b

Bisector formulas

\ \frac{a}{b} = \frac{m}{n}  ba =nm  

​  

 

\ l^2 = ab - mnl  2=ab-mm

A = \frac{1}{2}a\cdot b = \frac{1}{2}c\cdot hA=  

\ A = \sqrt{p(p - a)(p - b)(p - c)}A=  

p(p−a)(p−b)(p−c)

​  

 

\iits whatever  A = prA=pr with r we denote the radius of the triangle inscribed circle

\ A = \frac{abc}{4R}A=  

4R

abc

​  

 - R is the radius of the prescribed circle

\ A = \sqrt{p(p - a)(p - b)(p - c)}A=  

p(p−a)(p−b)(p−c)

​  

5 0
3 years ago
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