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
∣
∣
−−−−−−−−−−−−−−−
523.6 m3 i hope this helps mark me as brainliest
Answer:
6
Step-by-step explanation:
57=15+7x --> 57-15=7x --> 42=7x --> 42/7=x --> 6
7.5x18.74= 140.55 is the correct answer
Answer:
2.62
Step-by-step explanation:

First, write the square root as exponent.

Move the denominator to the numerator and negate the exponent.

Use log product property.

Use log exponent property.

Substitute values.
