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klemol [59]
2 years ago
10

Please find the value of x in the figure

Mathematics
1 answer:
notka56 [123]2 years ago
8 0

Answer:

Step-by-step explanation:

E is the midpoint of Ac & D is the midpoint of AB

DE = (1/2)BC        {Midpoint theorem}

x +1 = \dfrac{1}{2}(4x - 6)\\\\\\
Multiply \ the \ whole \ equation \  by 2\\\\
2*(x+1) =2*\dfrac{1}{2}*(4x - 6)

2*x + 2*1 = 4x - 6

2x + 2 = 4x - 6

Add 6 to both sides

2x + 2 + 6 = 4x

2x + 8 = 4x

Subtract 2x from both sides

8 = 4x - 2x

2x = 8

Divide both sides by 2

x = 8/2

x = 4

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Answer:

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3 years ago
I neep help asap deadline 4:00pm
Sunny_sXe [5.5K]

Answer:

x = -2 to x = 1

Step-by-step explanation:

A function is constant when the y value does not change (i.e. the graph is a horizontal line). This is true between x = -2 and x = 1.

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3 years ago
Read 2 more answers
Please answer this math question.
Molodets [167]

Answer:

21 minutes

Step-by-step explanation:

To cut a 5-ft log into 5 equal peices, you need to make 4 cuts (each cut cuts one piece from a whole log, the last cut divides the remaining log into two pieces).

If it takes 5\dfrac{1}{4} minutes to make one cut, then it takes

5\dfrac{1}{4}\cdot 4=\dfrac{5\cdot 4+1}{4}\cdot 4=\dfrac{21}{4}\cdot 4=21

minutes to cut the log into 5 equal pieces.

7 0
4 years ago
The table shows the viscosity of an oil as a function of temperature. Identify a quadratic model for the viscosity, given the te
777dan777 [17]

Answer:

The viscosity at 140°C is predicted to be 7.2

Step-by-step explanation:

The function that model the relationship between viscosity and temperature = Quadratic model

The general form of a quadratic equation is y = a·x² + b·x + c

Therefore, we have;

When y = 10.8, x = 110, which gives;

10.8 = a·110² + b·110 + c = 12100·a + 110·b + c

10.8 = 12100·a + 110·b + c

When y =8.2, x = 130

8.2 = a· 130² + b· 130 + c = 16900·a + 130·b + c

8.2 = 16900·a + 130·b + c

When y = 160, x = 5.8

5.8 = a·160² + b·160 + c = 25600·a + 160·b + c

5.8 = 25600·a + 160·b + c

The three equations above can be listed as follows;

10.8 = 12100·a + 110·b + c

8.2 = 16900·a + 130·b + c

5.8 = 25600·a + 160·b + c

Solving using matrices gives;

\begin{bmatrix}12100 & 110 & 1\\ 16900 & 130 & 1\\ 25600 & 160 & 1\end{bmatrix} \begin{bmatrix}a\\ b\\ c\end{bmatrix} = \begin{bmatrix}10.8\\ 8.2\\ 5.8\end{bmatrix}

\begin{bmatrix}a\\ b\\ c\end{bmatrix} = -\dfrac{1}{3000}\begin{bmatrix}3 & -5 & 2\\ -867 & 1350 & -480\\ 62400 & -88000 & 28600\end{vmatrix} \begin{bmatrix}10.8\\ 8.2\\ 5.8\end{bmatrix}

From which we have;

a = 0.001, b = -0.37, c = 39.4

Substituting gives;

y = 0.001·x² - 0.37·x + 39.4

When x = 140

y = 0.962·140² - 0.37·140 + 39.4= 7.2

The viscosity at 140°C = 7.2.

8 0
3 years ago
HELP !!!!!!!!!! What percent and degrees of the pie have been eaten? What’s your reasoning
Dmitry_Shevchenko [17]

Answer:

1/6 of the pie has been eaten and it should be 17.5°

8 0
3 years ago
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