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xxTIMURxx [149]
2 years ago
7

Can a matrix with dimensions of 2 X 4 be added to another matrix with dimensions of 2 X 4?

Mathematics
1 answer:
sineoko [7]2 years ago
3 0

Answer:

of course

Step-by-step explanation:

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The market bought grapes for $0.87 a pound and sold them for $1.09 a pound. What is the percent of increase rounded to the neare
Nimfa-mama [501]

Answer:22.0% so yoyouyoyour answer would be C


Step-by-step explanation:


8 0
2 years ago
Read 2 more answers
-4x-2y=-12 4x+8y=-24
Alla [95]

Answer:

x=6\\ \\y=-6

Step-by-step explanation:

Given the system of two equations:

\left\{\begin{array}{l}-4x-2y=-12\\ \\4x+8y=-24\end{array}\right.

Add these two equations (left side add to left side and right side add to right side):

-4x-2y+4x+8y=-12+(-24)\\ \\-2y+8y=-12-24\\ \\6y=-36\\ \\y=-6

Substitute y=-6 into the first equation:

-4x-2\cdot (-6)=-12\\ \\-4x+12=-12\\ \\-4x=-12-12\\ \\-4x=-24\\ \\x=6

8 0
2 years ago
Simplify the expression:<br> -1 + -1 + -3h + -8h
ValentinkaMS [17]

Answer:

-2 + 11h

Step-by-step explanation:

To simplify this equation, you can first combine like terms:

( -1 + -1 ) + ( -3h + -8h )

-2 + -11h

3 0
2 years ago
ABCD- parallelogram, If the perimeter of Triangle CPQ is 15cm, Find the perimeter of triangle BAQ. Find the perimeter of triangl
melamori03 [73]

Answer:

The answer is below

Step-by-step explanation:

A parallelogram is a quadrilateral (has 4 sides and 4 angle) with two pair of parallel and opposite sides. Opposite sides of a parallelogram are parallel and equal.

Given parallelogram ABCD:

AB = CD = 18 cm; BC = AD = 8 cm

∠P = ∠P, ∠PDA = ∠PCQ (corresponding angles are equal).

Hence ΔPCQ and ΔPDA are similar by angle-angle similarity theorem. For similar triangles, the ratio of their corresponding sides equal. Therefore:

\frac{CD}{PC}= \frac{AD}{CQ}\\\\\frac{18}{6}=\frac{8}{x}  \\\\x=\frac{6*8}{18}=\frac{8}{3}\ cm

Perimeter of CPQ = CP + CQ + PQ

15 = 6 + 8/3 + PQ

PQ = 15 - (6 + 8/3)

PQ = 6.33

∠CQP = ∠AQB (vertical angles), ∠QCP = ∠QBA (alternate angles are equal).

Hence ΔCPQ and ΔABQ are similar by angle-angle similarity theorem

\frac{AQ}{QP}=\frac{AB}{CP}  \\\\\frac{AQ}{6.33} =\frac{18}{6} \\\\AQ=\frac{18}{6}*6.33\\\\AQ = 19

\frac{BQ}{CQ}=\frac{AB}{CP}  \\\\\frac{BQ}{8/3} =\frac{18}{6} \\\\BQ=\frac{18}{6}*\frac{8}{3} \\\\BQ =8

Perimeter of BAQ = AB + BQ + AQ = 18 + 8 + 19 = 45cm

PA = AQ + PQ = 19 + 6.33 = 25.33

PD = CD + DP = 18 + 6 = 24

Perimeter of PDA = PA + PD + AD = 24 + 25.33 + 8 = 57.33 cm

7 0
2 years ago
Um someone pls help me
Sedaia [141]

Answer:

s : m : l

2u : 3u : 7u

small bag=15kg

medium bag=22.5kg

large bag=52.5kg

5 0
2 years ago
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