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exis [7]
2 years ago
5

Multiply 2 y (one-fourth y + three-fourths). 9 over 4 y + 11 over 4 one-half y + 6 over 4 one-half y squared + 6 over 4 y 9 over

4 y squared + 11 over 4 y
Mathematics
1 answer:
Illusion [34]2 years ago
6 0

The result of multiplying 2y (one-fourth y + three-fourths) is one-half y squared + 6 over 4 y

<h3>Multiplication of numbers</h3>

  • 2 y (one-fourth y + three-fourths).

2y(1/4y + 3/4)

= (2y × 1/4y) + (2y × 3/4)

= 2/4y² + 6/4y

= 1/2y² + 3/2y

Given options:

  • 9 over 4 y + 11 over 4

= 9/4y + 11/4

  • one-half y + 6 over 4

= 1/2y + 6/4

  • one-half y squared + 6 over 4 y

= 1/2y² + 6/4y

  • 9 over 4 y squared + 11 over 4 y

9/4y² + 11/4y

Therefore, the result of multiplying 2y (one-fourth y + three-fourths) is one-half y squared + 6 over 4 y

Learn more on multiplication:

brainly.com/question/10873737

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3 years ago
the height h(t) of a trianle is increasing at 2.5 cm/min, while it's area A(t) is also increasing at 4.7 cm2/min. at what rate i
nekit [7.7K]

Answer:

The base of the triangle decreases at a rate of 2.262 centimeters per minute.

Step-by-step explanation:

From Geometry we understand that area of triangle is determined by the following expression:

A = \frac{1}{2}\cdot b\cdot h (Eq. 1)

Where:

A - Area of the triangle, measured in square centimeters.

b - Base of the triangle, measured in centimeters.

h - Height of the triangle, measured in centimeters.

By Differential Calculus we deduce an expression for the rate of change of the area in time:

\frac{dA}{dt} = \frac{1}{2}\cdot \frac{db}{dt}\cdot h + \frac{1}{2}\cdot b \cdot \frac{dh}{dt} (Eq. 2)

Where:

\frac{dA}{dt} - Rate of change of area in time, measured in square centimeters per minute.

\frac{db}{dt} - Rate of change of base in time, measured in centimeters per minute.

\frac{dh}{dt} - Rate of change of height in time, measured in centimeters per minute.

Now we clear the rate of change of base in time within (Eq, 2):

\frac{1}{2}\cdot\frac{db}{dt}\cdot h =  \frac{dA}{dt}-\frac{1}{2}\cdot b\cdot \frac{dh}{dt}

\frac{db}{dt} = \frac{2}{h}\cdot \frac{dA}{dt} -\frac{b}{h}\cdot \frac{dh}{dt} (Eq. 3)

The base of the triangle can be found clearing respective variable within (Eq. 1):

b = \frac{2\cdot A}{h}

If we know that A = 130\,cm^{2}, h = 15\,cm, \frac{dh}{dt} = 2.5\,\frac{cm}{min} and \frac{dA}{dt} = 4.7\,\frac{cm^{2}}{min}, the rate of change of the base of the triangle in time is:

b = \frac{2\cdot (130\,cm^{2})}{15\,cm}

b = 17.333\,cm

\frac{db}{dt} = \left(\frac{2}{15\,cm}\right)\cdot \left(4.7\,\frac{cm^{2}}{min} \right) -\left(\frac{17.333\,cm}{15\,cm} \right)\cdot \left(2.5\,\frac{cm}{min} \right)

\frac{db}{dt} = -2.262\,\frac{cm}{min}

The base of the triangle decreases at a rate of 2.262 centimeters per minute.

6 0
2 years ago
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