Five times a number decreased by nine is equal to twice the number increased by 23. which equation could be used to solve the problem?
rachna has nickels, dimes, and quarters in her piggy bank. she has $4.65 altogether. she has three more quarters than dimes and twice as many nickels as quarters. how many quarters does rachna have?
three times a number, x, increased by four is equal to five times the number, x. which equation can be used to solve for x?
what is the value of x in the equation 0.2(x + 1) + 0.5x = –0.3(x – 4)?

Part 1)

Part 2)

Part 3)

Part 4)

Step-by-step explanation:

Part 1)

Let

x-----> the number

The expression" Five times a number decreased by nine is equal to twice the number increased by ", is equal to the algebraic expression

Part 2) we have that

Let

x------> the number of nickels

y-------> the number of dimes

z-------> the number of quarters

Multiply by both sides

------> equation A

-----> isolate the variable y

------> equation B

------> equation C

substitute equation B and equation C in equation A

Part 3) we know that

Let

x------> the number

The equation that represent the expression is equal to

Part 4) we have

Multiply by both sides

< xcy ≅ < xby

step-by-step explanation:

in δcax and δbax,

so, δcax ≅ δbax (side-angle-side or sas)

since δcax ≅ δbax, we can conclude < acx ≅ < abx

in δacy and δaby,

so, δacy ≅ δaby (side-angle-side or sas)

since δacy ≅ δaby, we can conclude < acy ≅ < aby

now,

< acy ≅ < aby

=> < acx + < xcy ≅ < abx + < xby

=> < acx + < xcy ≅ < acx + < xby   (since < acx ≅ < abx already proved above)

subtracting < acx from both sides, we get

< acx + < xcy -< acx ≅ < acx + < xby -< acx

cancelling out < acx and -< acx from both sides, we get

< xcy ≅ < xby (proved)

∡ eia and ∡ kij are congruent

step-by-step explanation: