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V125BC
5 days ago
5

The team first experiments with changing the position of the curved pit. In the computer program, the vertex begins on the origi

n, and the curve is modeled by the parent quadratic equation, y = x2. Match each description with the equation that will create that pit.

Mathematics
1 answer:
Zina [9.1K]5 days ago
6 0
This explanation follows: The query lacks completeness; please refer to the complete question in the provided attachment. The curve described is represented by the quadratic equation, y = x². If the curve moves downwards by 2 units, the equation for the new curve becomes y = x² - 2. Shifting the curve 2 units to the left results in the equation y = (x + 2)². Conversely, moving the curve 2 units to the right transforms it to y = (x - 2)². Lastly, if shifted upwards by 2 units, the equation will be y = x² + 2.
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Given that Ray E B bisects ∠CEA, which statements must be true? Select three options. m∠CEA = 90° m∠CEF = m∠CEA + m∠BEF m∠CEB =
tester [8842]

Answer:

Attached is the question in consideration.

m\angle CEA =90 \ (deg)

m\angle BEF=135\ (deg)

\angle CEF forms a straight line.

\angle AEF depicts a right triangle.

The options 1,4,5,6 represent the correct answers.

Step-by-step explanation:

⇒ Given that \ ray\ AE  is ⊥FEC it constitutes a right triangle, leading to m\angle CEA =90\ (deg).

⇒ The measure for \angle BEF =135\ (deg) equals \angle BEF =\angle AEB +\angle AEF = (45+90)=135\ (deg) as \angle AEB bisects \angle AEC, implying that \angle AEB is half of \angle AEC, thus  \angle AEB = 45\ (deg).

⇒\angle CEF represents a straight line, as the angle measures across it yield 180\ (deg).

⇒ The angle measure for \angle AEF = 90\ (deg) is derived from the linear pair concept.

Since \angle CEA + \angle AEF = 180\ (deg), inserting the values of  m\angle CEA =90\ (deg) leads to \angle AEF = 90\ (deg).

The other two options are incorrect as:

  • m\angle CEF=m\angle CEA + m\angle BEF = (90+135)=225

       it surpasses 180\ (deg) while \angle CEF is a               

      straight line.

  • Also, m\angle CEB=2(m\angle CEA) is inaccurate.

     As \angle CEA = 90\ (deg) and \angle CEB=45\ (deg)

Thus, we have a total 4 valid answers.

The confirmed options are 1,4,5,6.

5 0
6 days ago
Read 2 more answers
"The difference between five-halves of a number<br> and 17 is 48."
Inessa [9006]
Let the number be x. The equation x/5 - 17 = 48 leads to (x - 85)/5 = 48. Multiplying gives x - 85 = 240, so x = 240 + 85, resulting in x = 325.
5 0
8 days ago
Describe two ways to solve the equation 2(4x-11)=10
AnnZ [9104]
2(4x-11) = 10

Method 1:
Divide both sides by 2:
\frac{2(4x-11)}{2} = \frac{10}{2}

4x-11 = 5

Then add 11 to both sides:

4x-11+11 = 11+5

4x = 16

Finally divide both sides by 4

\frac{4x}{4} = \frac{16}{4}

x = 4

Method 2:

Use the distributive property:

2(4x-11) = 10

2*4x- 2*11 = 10

8x - 22 = 10

Add 22 to both sides:

8x - 22+22 = 10+22

8x = 32

Then divide both sides by 8

\frac{8x}{8} = \frac{32}{8}

x = 4
6 0
7 days ago
Diego's club earns money for charity when members of the club perform community service after school,
babunello [8423]

Answer: Indeed, because the input value ranges from 0 to 12.

Step-by-step explanation:

To address this query, we first need to formulate a function that reflects the situation.

Every student engaging in community service earns the club $5

E(s) = 5 s

Where:

  • E= total earnings
  • n = number of students participating in community services (input value)

Indeed, since the input value n indicates how many students from the club engage in community service, and with the total number of students being 12, the input can indeed vary between 0 and 12.

6 0
20 days ago
The value of a car t years after it is purchased is given by the decreasing function V, where V(t) is measured in dollars. The r
PIT_PIT [9121]

Answer:

dV(t)/dt = kV(t)

Step-by-step explanation:

The annual change in the car's value, represented by dV(t)/dt, has a proportional relationship with V(t), the car's current value.

dV(t)/dt ∝ V(t)

dV(t)/dt = kV(t)

7 0
1 month ago
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