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mario62 [17]
3 years ago
9

Hey can anyone help me solve this

Mathematics
2 answers:
VashaNatasha [74]3 years ago
7 0

Answer:

24

Step-by-step explanation:

Sati [7]3 years ago
6 0

Answer and Step-by-step explanation:

I think the answer is 48.

We should add the 12 and 4, then multiply by 3. The 16 times 3 is to get the original amount of candies. To check, we go through the process using the number 48.

48 times 1/3 is 16. 4 of them are given to the brother, which leaves us with 12 remaining candies.

<em><u>#teamtrees #PAW (Plant And Water)</u></em>

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6th grade math help me pleaseee
Mashcka [7]

Answer:

negative s will equal -16

8 0
3 years ago
Read 2 more answers
Identify the equation of the graph shown. A) y = 2x + 3 B) y = 2x - 3 C) y = -2x + 3 D) y = -2x - 3 4)
Anna35 [415]

Answer:

The answers provided aren't reasonable for this graph.

Step-by-step explanation:

The graph's y-intercept is -2, since that is the point where the line intersects the y-axis. The slope is 3x, because the distance from one point to the next is 3 up, 1 right. So the equation for the graph is y = 3x - 2.

4 0
2 years ago
Anyone know how to do #21?
Andreas93 [3]
Would it be y2 ? idk im not sure but i tried


6 0
3 years ago
Luis has $4.00 from his allowance. He visits an arcade where soda costs $0.75 and video games cost $0.25. Which of the following
nordsb [41]

Answer: c

Step-by-step explanation:

3 0
2 years ago
Calculate this reflection of the triangle:
Mice21 [21]

Answer:

a = 3, b = 0, c = 0, d = -2

Step-by-step explanation:

<em>To find the reflection Multiply the matrices</em>

∵ The dimension of the first matrix is 2 × 2

∵ The dimension of the second matrix is 2 × 3

<em>1. Multiply the first row of the 1st matrix by each column in the second matrix add the products of each column to get the first row in the 3rd matrix.</em>

2. Multiply the second row of the 1st matrix by each column in the second matrix add the products of each column to get the second row of the 3rd matrix

\left[\begin{array}{ccc}1&0\\0&-1\end{array}\right]  × \left[\begin{array}{ccc}0&3&0\\0&0&2\end{array}\right]  = \left[\begin{array}{ccc}(1*0+0*0)&(1*3+0*0)&(1*0+0*2)\\(0*0+-1*0)&(0*3+-1*0)&(0*0+-1*2)\end{array}\right]=\left[\begin{array}{ccc}0&3&0\\0&0&-2\end{array}\right]

Compare the elements in the answer with the third matrix to find the values of a, b, c, and d

∴ a = 3

∴ b = 0

∴ c = 0

∴ d = -2

7 0
3 years ago
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