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Len [333]
4 years ago
7

A student bought 3 boxes of pencils and 2 boxes of pens for $6. He then bought 2 boxes of pencils and 4 boxes of pens for $8. Fi

nd the cost of each box of pencils and each box of pens
Mathematics
2 answers:
Annette [7]4 years ago
8 0
Let c and n represent the numbers of pencil and pen boxes, respectively
.. 3c +2n = 6.00
.. 2c +4n = 8.00

You can halve the second equation and subtract it from the first to get
.. (3c +2n) -(c +2n) = 6.00 -4.00
.. 2c = 2.00
.. c = 1.00

Then
.. 1.00 +2n = 4.00 . . . . . half the original second equation with c filled n
.. 2n = 3.00
.. n = 1.50

A box of pencils costs $1.00; a box of pens costs $1.50.
Vedmedyk [2.9K]4 years ago
7 0

Answer:

112 x 4

Step-by-step explanation:

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2x + 7 = 4 + x solve equation using tables
Alika [10]

Answer:

x=-3

Step-by-step explanation:

2x+7=4+x

2x-x+7=4

x+7=4

x=4-7

x=-3

7 0
4 years ago
HELP PLEASE (BRAINIEST)
tangare [24]

Answer: Passage 1: The sailboat bobbed on the sea as the <u>(cool) </u>afternoon tide lapped at the quaint little ship.

Step-by-step explanation: An idea or feeling that a word invokes in addition to its literal or primary meaning.

8 0
2 years ago
Parallelogram RSTU is rotated 180° clockwise
julia-pushkina [17]

Answer:

Two pairs of parallel sides

Step-by-step explanation:

The given transformation performed on parallelogram RSTU = 180° clockwise rotation

Given that a rotation is a form of rigid transformation, the shape and size of the preimage RSTU will be equal to the the shape and size of the image R'S'T'U'

Therefore, RSTU ≅ R'S'T'U' and R'S'T'U' is also a parallelogram with two pairs of parallel sides.

4 0
3 years ago
Question :-
DochEvi [55]

\qquad\qquad\huge\underline{{\sf Answer}}

Let's solve ~

\qquad \sf  \dashrightarrow \:  \bigg( \dfrac{1}{ \sqrt{7 +  \sqrt{5} } }  +  \sqrt{7 +  \sqrt{5} } \bigg) {}^{2}

\qquad \sf  \dashrightarrow \:  \bigg( \dfrac{1}{ \sqrt{7 +  \sqrt{5} } } \bigg ) {}^{2}  +  \bigg( \sqrt{7 +  \sqrt{5} } \bigg) {}^{2}  + 2 \cdot \dfrac{1}{ \sqrt{7 +  \sqrt{5} } }  \cdot \sqrt{7 +  \sqrt{5} }

\qquad \sf  \dashrightarrow \: \dfrac{1}{7 +  \sqrt{5} }  + 7 +  \sqrt{5}  + 2

\qquad \sf  \dashrightarrow \: \dfrac{49 + 7 \sqrt{5} + 7 \sqrt{5} + 5 + 14 + 2 \sqrt{5}   }{7 +  \sqrt{5} }

\qquad \sf  \dashrightarrow \: \dfrac{68+ 16 \sqrt{5}    }{7 +  \sqrt{5} }

5 0
2 years ago
How do i solve this?
Natalija [7]

Because you know the area and that the shape is a square... if the side length is "s"... to find "s" you would say...

s²=1.69

Take the root of both sides and...

s=1.3 km

8 0
3 years ago
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