9. How many planes are shown in the figure?

Mathematics

1 answer:

Paraphin [41]2 years ago

3 0

<h3>9)</h3><h3>5 planes</h3>

<h3>10)</h3><h3>2 planes</h3>

<h3>12)</h3><h3>Yes , they belong to plane (w)</h3>

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HELP NO LINKS OR TROLLING

jekas [21]

Answer:

20.28 cm²

Step-by-step explanation:

The rectangle on the bottom is 7 cm by 2 cm.

The semicircle has a diameter of 7 cm - 1 cm - 2 cm = 4 cm

diameter = 4 cm

radius = diameter/2 = 2 cm

total area = area of semicircle + area of rectangle

total area = (1/2)πr² + LW

total area = (1/2)(3.14)(2 cm)(2 cm) + 7 cm × 2 cm

total area = 20.28 cm²

3 0

3 years ago

According to Descartes's rule of sign, how many possible positive and negative roots are there for the equation

damaskus [11]

<h2>Descartes Rule of Sign</h2><h3>Answer:</h3>

There are $3$ Real Roots for the Equation

<h3>Step-by-step explanation:</h3>

According to Descartes Rule of Sign, The maximum number of real roots a polynomial equation has is the number of times the sign of term changes from a higher power term to a lower power term.

In the given equation, $0=-4x^6-3x^5+2x^2-4x+1$ , we can already see that from $\red -4x^6$ to $\red -3x^5$ we can see that the sign didn't change so there might be no real roots. From $\red -3x^5$ to $\blue +2x^2$ , we can see that the signs change so we now know that there might be $1$ real root for equation. From $\blue +2x^2$ to $\red -4x$ , the signs changed so there might be $2$ real roots. From $\red -4x$ to $\blue +1$ , it changed again so there might be $3$ real roots. There are no other term with lesser power to compare for $\blue +1$ so we are left with $3$ real roots.