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Igoryamba
2 years ago
5

Can someone help me with this please

Mathematics
1 answer:
d1i1m1o1n [39]2 years ago
4 0

9514 1404 393

Answer:

  1. attendance (y), time (x)

  2. 1,2: increasing; 3: constant; 4: decreasing

  3. attendance increases, remains constant, then decreases

  4. increasing; pre-heating the oven

Step-by-step explanation:

It is always difficult to read the mind of the author of questions like these. Usually, specific wording is expected. If your curriculum material has examples you can copy from, you should do that.

__

1. The variables and what they stand for are listed on the axes of the graph. The y-variable stands for attendance. The x-variable stands for time. It isn't clear whether the desired answer is "y and x" or "attendance and time".

__

2. The graph slopes upward in intervals 1 and 2. It can be described as "increasing."

The graph remains level in interval 3. It can be described as "constant."

The graph slopes downward in interval 4. It can be described as "decreasing."

__

3. The attendance is adequately described in question 2. Attendance increases, remains constant, then decreases. It isn't clear how general or specific the answer is supposed to be.

__

4. The graph of the function is "increasing." We don't know if the indicated temperature is the temperature of the bread, the oven, or the room containing the oven. If we assume the indicated temperature is that of the oven, we expect an increase during the oven pre-heat interval.

If the indicated temperature is that of the bread, then the increase would be following insertion of the bread into the oven.

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if trapezium of area 140 cm horsepower sides 10 cm apart and one of them decides is 16 cm longfind the area find the length of t
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Given :

  • Area of the trapezium is 140 cm².
  • Its height is 10 cm.
  • One of the parallel side is 16 cm.

To Find :

  • The other parallel side.

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  • Let the other parallel side be x

We know that,

{ \qquad \: \pmb{ \dfrac{1}{2} \times (sum \: of \: the \: parallel \: sides) \times height} = \pmb{Area_{(trapezium)}}}

Now, Substituting the given values in the formula :

\qquad { \dashrightarrow \: { \sf{ \dfrac{1}{2} \times (16 + x)\: \times 10 \: = {140}}}}

\qquad { \dashrightarrow \: { \sf{ 5 \times (16 + x)\  \: = {140}}}}

\qquad { \dashrightarrow \: { \sf{ 80 + 5x = {140}}}}

\qquad { \dashrightarrow \: { \sf{  5x = {140 - 80}}}}

\qquad { \dashrightarrow \: { \sf{  5x = {60}}}}

\qquad { \dashrightarrow \: { \sf{  x = { \dfrac{60}{5} }}}}

\qquad { \dashrightarrow \: { \pmb{  x = { 12 }}}}

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