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3 years ago

UJUU J

Mathematics

1 answer:

Stels [109]3 years ago

3 0

Answer:

Plese read the complete procedure below:

Step-by-step explanation:

The polynomial is p(a) = (a^4 - 6a^3 + 3a^2 + 26a – 24)

1 -6 3 26 -24 | 1

1 -5 -2 24

1 -5 -2 24 0

The remainder is zero, then (a-1) is a factor of the polynomial

1 -6 3 26 -24 | 2

2 -8 10 72

1 -4 5 36 48

When p(a) is divided by (a-2) the remainder 28/p(a)

1 -6 3 26 -24 | - 4

-4 40 172 -792

1 -10 43 198 -816

When p(a) is divided by (a-2) the remainder -816/p(a)

c) I attached an image of the long division below:

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3 years ago

A rectangular field is 75 yards wide and 105 yards long

nika2105 [10]

Answer:

width = 72 yards

length = 108 yards

Step-by-step explanation:

Given:

Width = 75 yards
Length = 105 yards

Area of the field with the given values:

$\begin{aligned}\textsf{Area of a rectangle}&=\sf width \times length\\& = \sf 75 \times 105\\& = \sf 7875\:\: yd^2\end{aligned}$

To maintain the same perimeter, but change the area, either:

decrease the width and increase the length by the same amount, or
increase the width and decrease the length by the same amount.

In geometry, length pertains to the longest side of the rectangle while width is the shorter side. Therefore, we should choose:

decrease the width and increase the length by the same amount.

Define the variables:

Let x = the amount by which to decrease/increase the width and length.

Therefore:

$\implies \sf width \times length < 7875\:\:yd^2$

$\implies (75-x)(105+x) < 7875$

Solve the inequality:

$\begin{aligned}(75-x)(105+x) & < 7875\\7875-30x-x^2 & < 7875\\-x^2-30x & < 0\\-x(x+30) & < 0\\x(x+30) & > 0\\\implies x & > 0 \:\: \textsf{ or }\:\:x < - 30\end{aligned}$

Therefore, as distance is positive only and the maximum width is 75 yd (since we are subtracting from the original width):

$\begin{cases}\textsf{width} = 75 - x\\\textsf{length} = 105 + x\end{cases}$

$\textsf{where } 0 < x < 75$

Therefore, to find the width and length of another rectangular field that has the same perimeter but a smaller area than the first field, simply substitute a value of x from the restricted interval into the found expressions for width and length:

Example 1: