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Bad White [126]
3 years ago
9

PLEASE HELP QUICK!!!! WILL GIVE BRAINIEST!!!!!

Mathematics
1 answer:
Kipish [7]3 years ago
8 0

Answer:

From top to bottom:

Slope intercept Equation: y = \frac{3}{4} x+2   y=3x-2

Slope: 4    2

Y-int: (0,-2)  (0,10)

Step-by-step explanation:

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How to solve this substitution​
Lena [83]
-5x+4y=11
substitute for x which is 5
(-5)(5)+4y=11

simplify :
4y-25=11

add 25 on both sides
4y-25=11
+25 +25
the 25 cancels out

so it would be 4y=36
and you divide both sides by 4

4/4 =36/4

which gives you (5,9)
x= 5 , y =9
6 0
3 years ago
Help! locus points!
Liula [17]

Answer:

  1. Circle X with radius 2 cm.
  2. Either of two lines parallel to AB.

Step-by-step explanation:

1. The definition of a circle is all the points in a plane that are at some radius r from a given point (the circle center). That is what you have, with a radius of 2 cm.

__

2. Parallel lines are the same distance apart everywhere. Each line will have two parallel lines at some given distance from it, one on each side. Here, the separation distance from AB is 1 cm, so your locus of points is the two lines parallel AB that are 1 cm from it on either side.

5 0
4 years ago
Help me plzz I need help with this!!
Anni [7]

Answer:

expanded it would be 0.05 x 50 + 0.05 x 40

Step-by-step explanation:

5 0
3 years ago
3/5 ÷ 4/15 = 9/4. Is this true?
Rasek [7]

Answer:

yes

Step-by-step explanation:

it is 9/4 indeed.........

4 0
4 years ago
Read 2 more answers
4. Use the process outlined in the lesson to approximate the number 2√3. Use the approximation √3 ≈ 1.732 050 8.
jeka57 [31]

Answer:

sequence of five intervals

(1) 2  < 2^{\sqrt{3} }   < 2^{2}

(2) 2^{1.7}  < 2^{\sqrt{3} }    < 2^{1.8}

(3) 2^{1.73}  < 2^{\sqrt{3} }    < 2^{1.74}

(4) 2^{1.732}  < 2^{\sqrt{3} }    < 2^{1.733}

(5) 2^{1.7320}  < 2^{\sqrt{3} }    < 2^{1.7321}

Step-by-step explanation:

as per question given data      

√3 ≈ 1.732 050 8 

to find out      

sequence of five intervals

solution      

as we have given that √3 value that is here

√3 ≈ 1.732 050 8            ........................1

so  

when we find 2^{\sqrt{3} }           ................2

put here √3 value in equation number  2  

we get  2^{\sqrt{3} }   that is  3.322    

so    

sequence of five intervals

(1) 2  < 2^{\sqrt{3} }   < 2^{2}

(2) 2^{1.7}  < 2^{\sqrt{3} }    < 2^{1.8}

(3) 2^{1.73}  < 2^{\sqrt{3} }    < 2^{1.74}

(4) 2^{1.732}  < 2^{\sqrt{3} }    < 2^{1.733}

(5) 2^{1.7320}  < 2^{\sqrt{3} }    < 2^{1.7321}

8 0
3 years ago
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