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wlad13 [49]
4 years ago
9

Polynomial function in standard form with zeros 5,-4,1

Mathematics
1 answer:
Oksanka [162]4 years ago
5 0

Answer:

\boxed{\sf \ \ \  x^3-2x^2-19x+20 \ \ \  }

Step-by-step explanation:

hello,

by definition we can write

(x-5)(x+4)(x-1)

as 5,-4,1 are the zeroes

now we have to write it in the standard form, let's do it

(x-5)(x+4)(x-1)=(x^2+4x-5x-20)(x-1)\\=(x^2-x-20)(x-1)=x^3-x^2-20x-x^2+x+20\\=x^3-2x^2-19x+20

hope this helps

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What is he least common multiple of 5 and 6
jeyben [28]

Answer:

30.

Step-by-step explanation:

6 = 2 * 3

5 = 5

LCM = 2 * 3 * 5 = 30.

7 0
4 years ago
What i the answer to <br> 25(15-0.05)
IrinaK [193]

Answer:

373.75

Step-by-step explanation:

25\left(15-0.05\right)

Subtract:- 15- 0.05= 14.95

Multiply:-

25\times \:14.95

373.75

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3 years ago
Find two positive real numbers whose product is a maximum. the sum is 80
Eduardwww [97]
Let's say the numbers are "a" and "b"

thus \bf y=ab\qquad &#10;\begin{cases}&#10;a+b=80\\&#10;b=80-a&#10;\end{cases}\implies y=a(80-a)\implies y=80a-a^2\\\\&#10;-----------------------------\\\\&#10;\cfrac{dy}{da}=80-2a

set the derivative to 0, and check the critical points, there's only one anyway

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4 0
3 years ago
Read 2 more answers
The Skittles jar has a radius of 3.5 cm and a height of 11.5 cm. Using the two given boxes for reference what would be a good es
nevsk [136]

Answer:

1052

Step-by-step explanation:

The two given boxes for reference are presented below.

Since we know the dimensions of the first container, we can calculate its volume.

V_1= 5 \text{cm} \cdot 4 \text{cm} \cdot 4 \text{cm} = 80 \text{cm}^3

The first container held 192 skittles. Using that fact, now that we know the volume of the first box, we can calculate the volume of a single skittle.

v = \frac{80}{192} = 0.416 \approx 0.42 \text{cm}^3

The second container held 258 skittles.

The volume of the second box is:

V_2 = 12 \text{cm}\cdot 3 \text{cm} \cdot 3 \text{cm} = 108 \text{cm}^3

Using this facts, the volume of a single skittle would be

v = \frac{108}{258} = 0.418 \approx 0.42 \text{cm}^3

Therefore,  the volume of  single skittle is around to 0.42 cm³.

A skittle jar is a cylinder. Its volume is calculated by the formula

V = r^2h \pi,

where r is the radius and h is the height. From the given data, we have r = 3.15 \text{cm} and h = 11.5 \text{cm}. Hence, the volume of the jar is

V = 3.5^2 \cdot 11.5 \cdot 3.14 = 442 \text{cm}^3

Now, we can determine the number of skittles in the jar. Let n be the number of skittles in the jar. Then,

n = \frac{\text{volume of the jar}}{\text{volume of a skittle}} = \frac{V}{v} = \frac{442}{0.42} = 1052.38 \approx 1052\\

4 0
3 years ago
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