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34kurt
3 years ago
11

Solve the system using substitution 4x+3y=23 x-5=0

Mathematics
1 answer:
iragen [17]3 years ago
3 0

Answer:

y=1, x=5

Step-by-step explanation:

Since x-5=0 is given already, give the x a value and switch the 5 to the other side x-5=0 ----> x=5

Now we have a value for x, so plug x into the original equation,

4(5)+3y=23,

20+3y=23

Subtract both sides by 20

20-20+3y=23-20

3y=3

dvide both sides by 3

y=1

use the value for x we got earlier

y=1, x=5

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C=2m+d solve for m plz
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C (-d) = 2m + d (-d)
C - d = 2m
(C - d)/2 = 2m/2
m = (C - d)/2

m = (C - d)/2 is your answer

hope this helps
 
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4 years ago
The two cones are congruent
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Answer:

3.1

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4 years ago
Can some one help me with 1,3,and 5
FrozenT [24]

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8 0
3 years ago
Factor 26r3s + 52r5 – 39r2s4.
algol [13]

Answer:

13r²(2rs + 4r³ - 3s⁴)

Step-by-step explanation:

In equation 26r³s + 52r⁵ - 39r²s⁴;

The GCF of 26, 52, and 39 = 13

The GCF of r³, r⁵ and r² = r²

The GCF of s, (no "s"), and s⁴ = no "s" (Since one of the number doesn't have "s")

Now we can factor out 13r² from all three expressions;

26r³s + 52r⁵ - 39r²s⁴

=> <em>13r²(2rs) + 13r²(4r³) - 13r²(3s⁴)</em>

To factor it all together;

<u>13r²(2rs + 4r³ - 3s⁴)</u>

Hope this helps!

7 0
3 years ago
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aalyn [17]

AnsS

Step-by-step explanation:

Is The C

4 0
3 years ago
Read 2 more answers
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