2^3×4^7+63 = 131135
2^3×4^7-63 = 131009
Answer:
Step-by-step explanation:
1
. Start by making a "let statement."
Let
x
represent the weight of the package.
2
. Create an algebraic expression.
5
7
x
=
40.5
i
pounds
3
. Isolate for
x
by dividing both sides by
5
7
.
5
7
x
÷
5
7
=
40.5
i
pounds
÷
5
7
4
. Recall that in order to divide a fraction by another fraction, take the reciprocal of the divisor and change the division sign to a multiplication sign.
5
7
x
⋅
7
5
=
40.5
i
pounds
⋅
7
5
5
. Solve for
x
.
5
⋅
7
7
⋅
5
x
=
40.5
⋅
7
5
i
pounds
x
=
∣
∣
∣
∣
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
a
a
56.7
i
pounds
a
a
∣
∣
−−−−−−−−−−−−−−−
Answer: no solution
Step-by-step explanation:
2x2=4x-7
4x=4x−7
0=−7
Since 0=−7 is false, the answer is no solution.
D. NI = AC is the needed information to prove ΔINF = ΔCAT by the ASA Postulate.
For the triangles to be congruent by ASA, the measurements of two angles and one side must be proven congruent. Since two sets of congruent angles are already given, one side must also be congruent.
When proving congruency using ASA, the congruent parts of the triangles must be in this order: Angle, Side, Angle. So, you have to the side that is between the two sets of congruent angles.
Answer:
<h2>
32</h2>
Step-by-step explanation:
