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Stels [109]
3 years ago
13

Determine if the following point is a solution to the system of equations:

Mathematics
1 answer:
Hunter-Best [27]3 years ago
3 0

Answer:

Yes it is

Step-by-step explanation:

3 + 4 =7

4 = 2(3) - 2

4 = 6 - 2

4 = 4

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a living room is 18ft by 14ft wide on a scale drawing for the house the living room is 3.5 in long how wide is the living room i
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18/3.5=5.1 so you do 14/5.1=2.7

so your answer is 2.7 in
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Given the trinomial 2x*2 + 4x − 2, what is the value of the discriminant? a. 0 b. 10 c. 32 d. 36
Komok [63]
2x^{2}+4x-2 \\  \\ \Delta=b^{2}-4ac \\  \\ \Delta=4^{2}-4*2*(-2)=16+16=32

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7 0
3 years ago
Read 2 more answers
What should happen to both sides of the equation to isolate the variable?
LUCKY_DIMON [66]

Answer:

If two expressions are equal to each other, and you add the same value to both sides of the equation, the equation will remain equal. When you solve an equation, you find the value of the variable that makes the equation true. In order to solve the equation, you isolate the variable

Step-by-step explanation:

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3 years ago
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mote1985 [20]

Step-by-step explanation:

ay guey como estas yo no entiendo este tipo de matematica para nerds

6 0
3 years ago
A cylinder is inscribed in a right circular cone of height 4.5 and radius (at the base) equal to 5.5 . What are the dimensions o
cluponka [151]

Answer:

r = 3.667

h = 1.5

Step-by-step explanation:

Given:-

- The base radius of the right circular cone, R = 5.5

- The height of the right circular cone, H = 4.5

Solution:-

- We will first define two variables that identifies the volume of a cylinder as follows:

                                r: The radius of the cylinder

                                h: The height of cylinder

- Now we will write out the volume of the cylinder ( V ) as follows:

                                V = \pi*r^2h

- We see that the volume of the cylinder ( V ) is a function of two variables ( don't know yet ) - ( r,h ). This is called a multi-variable function. However, some multi-variable functions can be reduced to explicit function of single variable.

- To convert a multi-variable function into a single variable function we need a relationship between the two variables ( r and h ).

- Inscribing, a cylinder in the right circular cone. We will denote 5 points.

              Point A: The top vertex of the cone

              Point B: The right end of the circular base ( projected triangle )

              Point C: The center of both cylinder and base of cone.

              Point D: The top-right intersection point of cone and cylinder

              Point E: Denote the height of the cylinder on the axis of symmetry of both cylinder and cone.  

- Now, we will look at a large triangle ( ABC ) and smaller triangle ( ADE ). We see that these two triangles are "similar". Therefore, we can apply the properties of similar triangles as follows:

                              \frac{AC}{AE} = \frac{BC}{DE}  \\\\\frac{H}{H-h} = \frac{R}{r}

- Now we can choose either variable variable to be expressed in terms of the other one. We will express the height of cylinder ( h ) in term of radius of cylinder ( r ) as follows:

                             H- h = r\frac{H}{R} \\\\h = \frac{H}{R}*(R-r)

- We will use the above derived relationship and substitute into the formula given above:

                            V = \pi r^2 [ \frac{H}{R}*(R - r )]\\\\V = \frac{\pi H}{R}.r^2.(R-r)

- Now our function of volume ( V ) is a single variable function. To maximize the volume of the cylinder we need to determine the critical points of the function as follows:

                            \frac{dV}{dr} =  \frac{\pi H}{R}*(2rR-2r^2 - r^2 )\\\\\frac{dV}{dr} =  \frac{\pi H}{R}*(2rR-3r^2 ) = 0\\\\(2rR-3r^2 ) = 0\\\\2R -3r = 0\\\\r = \frac{2}{3}*R

- We found the limiting value of the function. The cylinder volume maximizes when the radius ( r ) is two-thirds of the radius of the right circular cone.

- We can use the relationship between the ( r ) and ( h ) to determine the limiting value of height of cylinder as follows:

                          h = \frac{H}{R} * ( R - \frac{2}{3}R)\\\\h = \frac{H}{3}

- The dimension of the inscribed cylinder with maximum volume are as follows:

                         r = \frac{2}{3}*5.5 = 3.667\\\\h = \frac{4.5}{3} = 1.5

Note: When we solved for the critical value of radius ( r ). We actually had two values: r = 0 , r = 2R/3. Where, r = 0 minimizes the volume and r = 2R/3 maximizes. Since the function is straightforward, we will not test for the nature of critical point ( second derivative test ).

7 0
4 years ago
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