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QuestionEvaluate 3x2+4y0⋅x−13 x^{2}+4 y^{0} \cdot x-1 for x=4x=4 and y=5y=5.

Studdy Solution

STEP 1

Assumptions1. The given expression is 3x+4y0⋅x−13x^{}+4y^{0} \cdot x-1 . The value of xx is43. The value of yy is5

STEP 2

We need to substitute the values of xx and yy into the expression.x2+4y0⋅x−1=(4)2+4(5)0⋅4−1x^{2}+4y^{0} \cdot x-1 =(4)^{2}+4(5)^{0} \cdot4-1

STEP 3

Now, we need to simplify the expression. First, we evaluate the exponent.
3()2+(5)0⋅−1=3(16)+(1)⋅−13()^{2}+(5)^{0} \cdot-1 =3(16)+(1) \cdot-1

STEP 4

Next, we perform the multiplication operations.
3(16)+4(1)⋅4−1=48+4⋅4−13(16)+4(1) \cdot4-1 =48+4 \cdot4-1

STEP 5

Continue with the multiplication operation.
48+4⋅4−1=48+16−148+4 \cdot4-1 =48+16-1

STEP 6

Finally, we perform the addition and subtraction operations.
48+16−1=64−148+16-1 =64-1

STEP 7

Continue with the subtraction operation.
64−1=6364-1 =63So, when x=4x=4 and y=5y=5, the expression 3x2+4y0⋅x−13x^{2}+4y^{0} \cdot x-1 evaluates to63.

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