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2 years ago

Is (x+5) a factor of f(x)=x^3 - 4x^2 + 3x + 7? use either the remainder theorem or the factor theorem to explain your reasoning.

Mathematics

1 answer:

Rufina [12.5K]2 years ago

3 0

Answer:

Step-by-step explanation:

<u>Use synthetic division</u>

-5 | 1 -4 3 7

____<u>-5 45 -240</u>

1 -9 48 | -233

Since the remainder is not 0, then x+5 is not a factor of the polynomial

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Stephen and five of his friends went out to eat. They decided to split the bil

kirill [66]

Stephen and his five friends = 6 people at the restaurant

Let's call X the amount of the bill.

so if X/6 = 13.21

X = 13.21 * 6

X = 79.26

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3 years ago

A triangle with vertices at A(0,0), B(0,4), and C(6,0) is dilated to yield a triangle with vertices at A(0,0), (B(0,10), and C (

dezoksy [38]

The required scale factor for the dialed triangle is given by SF= 6.28

A triangle with vertices at A(0,0), B(0,4), and C(6,0) is dilated to yield a triangle with vertices at A(0,0), (B(0,10), and C (C(15,0).

<h3>What is scale factor?</h3>

The scale factor is defined as the ratio of modified change in length to the original length.

Since, the height and base of the triangle is y coordinate of B and x coordinate of c is given by 4, 6
Area of triangle =1/2(base x height)
= 1/2 (4 x 6)
= 4.8
Similarly, for dilated triangle
Area of dilated triangle = 30

Now, scale factor(SF) = area of dilated triangle \ area of triangle
SF = 30/4.8
SF = 6.25

Thus, the required scale factor is SF = 6.25

Learn more about line Scale factor:

brainly.com/question/22312172

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4 0

2 years ago

I'm stuck, please help me

Sav [38]

60/48
x100
60/48=1.25
1.25x100
125%

3 0

3 years ago

Read 2 more answers

NEED HELP ASAP, WILL GIVE BRAINLIEST

Svetlanka [38]

Domain: [-5, 5)
range:(-4,4]
hope this helps

5 0

2 years ago

4. Using the geometric sum formulas, evaluate each of the following sums and express your answer in Cartesian form.

nikitadnepr [17]

Answer:

$\sum_{n=0}^9cos(\frac{\pi n}{2})=1$

$\sum_{k=0}^{N-1}e^{\frac{i2\pi kk}{2}}=0$

$\sum_{n=0}^\infty (\frac{1}{2})^n cos(\frac{\pi n}{2})=\frac{1}{2}$

Step-by-step explanation:

$\sum_{n=0}^9cos(\frac{\pi n}{2})=\frac{1}{2}(\sum_{n=0}^9 (e^{\frac{i\pi n}{2}}+ e^{\frac{i\pi n}{2}}))$

$=\frac{1}{2}(\frac{1-e^{\frac{10i\pi}{2}}}{1-e^{\frac{i\pi}{2}}}+\frac{1-e^{-\frac{10i\pi}{2}}}{1-e^{-\frac{i\pi}{2}}})$

$=\frac{1}{2}(\frac{1+1}{1-i}+\frac{1+1}{1+i})=1$

2nd

$\sum_{k=0}^{N-1}e^{\frac{i2\pi kk}{2}}=\frac{1-e^{\frac{i2\pi N}{N}}}{1-e^{\frac{i2\pi}{N}}}$

$=\frac{1-1}{1-e^{\frac{i2\pi}{N}}}=0$

3th

$\sum_{n=0}^\infty (\frac{1}{2})^n cos(\frac{\pi n}{2})==\frac{1}{2}(\sum_{n=0}^\infty ((\frac{e^{\frac{i\pi n}{2}}}{2})^n+ (\frac{e^{-\frac{i\pi n}{2}}}{2})^n))$

$=\frac{1}{2}(\frac{1-0}{1-i}+\frac{1-0}{1+i})=\frac{1}{2}$

What we use?

We use that

$e^{i\pi n}=cos(\pi n)+i sin(\pi n)$

and

$\sum_{n=0}^k r^k=\frac{1-r^{k+1}}{1-r}$

6 0

3 years ago