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2 years ago

Find the roots of the following equations. a) 4x+9=34 b) (x-4)(x+2)=0 c) 2x^2-6x+4=0

1 answer:

7 0

A. 4x + 9 = 34
 - 9 - 9
 4x = 25
 4 4
 x = 6.25

B. (x - 4)(x + 2) = 0
 x - 4 = 0 U x + 2 = 0
 + 4 + 4 - 2 - 2
 x = 4 x = -2

C. 2x² - 6x + 4 = 0
 2(x²) - 2(3x) + 2(2) = 0
2(x² - 3x + 2) = 0
 2 2
 x² - 3x + 2 = 0
 x = -(-3) +/- √((-3)² - 4(1)(2))
 2(1)
 x = 3 +/- √(9 - 8)
 2
 x = 3 +/- √(1)
 2
 x = 3 +/- 1
 2
 x = 3 + 1 U x = 3 - 1
 2 2
 x = 4 x = 2
 2 2
 x = 2 x = 1

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Compare -5/6 _ -3/7 choose from 1) > 2)

vfiekz [6]

Answer

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Step-by-step explanation

First you need to find the GCF of the donominators which are 6 and 7. The GCF is 42.

6*x=42. x=7 5*7=35. -35/42 is the first fraction.

7*y=42 y=6 3*6=18. -18/42 is the second fraction.

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3 years ago

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Find a polynomial with integer coefficients that satisfies the given conditions. R has degree 4 and zeros 3 − 3i and 2, with 2 a

dolphi86 [110]

Answer:

The required polynomial is $P(x)=x^4-10x^3+46x^2-96x+72$ .

Step-by-step explanation:

If a polynomial has degree n and $c_1,c_2,...,c_n$ are zeroes of the polynomial, then the polynomial is defined as

$P(x)=a(x-c_1)(x-c_2)...(x-x_n)$

It is given that the polynomial R has degree 4 and zeros 3 − 3i and 2. The multiplicity of zero 2 is 2.

According to complex conjugate theorem, if a+ib is zero of a polynomial, then its conjugate a-ib is also a zero of that polynomial.

Since 3-3i is zero, therefore 3+3i is also a zero.

Total zeroes of the polynomial are 4, i.e., 3-3i, 3_3i, 2,2. Let a=1, So, the required polynomial is

$R(x)=(x-3+3i)(x-3-3i)(x-2)(x-2)$

$R(x)=((x-3)+3i)((x-3)-3i)(x-2)^2$

$R(x)=(x-3)^2-(3i)^2((x-3)-3i)(x-2)^2$ $[a^2-b^2=(a-b)(a+b)]$

$R(x)=(x^2-6x+9-9(i)^2((x-3)-3i)(x-2)^2$

$R(x)=(x^2-6x+18)(x^2-4x+4)$ $[i^2=-1]$

$R(x)=(x^2-6x+18)(x^2-4x+4)$

$R(x)=x^4-10x^3+46x^2-96x+72$

Therefore the required polynomial is $P(x)=x^4-10x^3+46x^2-96x+72$ .

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2 years ago

The radius of a circle is 14 meters. What is the circle circumference

seropon [69]

Step-by-step explanation:

HERE,

redius of a circle (r)=14 meters.

we know that,

$\boxed{\red{\sf{circle~ circumference=2\times \pi \times r }}}$

According to the question, 

circle circumference=

$\tt{2\times \pi \times 14 }$

$\tt{2 \times \dfrac{22}{7} \times 14 }$

$\tt{2 \times \dfrac{22}{\cancel{7}} \times \cancel{14}}$

$\tt{2\times 22 \times 2 }$

$\tt{ 88~meters }$

so, the circumference of that circle is 88 meters.

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2 years ago

I need help If you know this answer plz give it ta meh

Oksana_A [137]

Answer:

4.9 units

Step-by-step explanation:

We are dealing with a Right angle triangle, and the c is actually the hypotenuse. We are given the measures of a and b.

a being 4, and b being 3.

We need to use the Pythagorean Theorem to solve this.

$a^2 + b^2 = c^2$

we can rewrite this as;

$4^2 + 3^2 = c^2\\$

Calculation time

15 + 9 = c

c = 24

Now we find the square root of each equation to determine what $c^2$ is.

$\sqrt{15}+\sqrt{9}=\sqrt{24}$

4 + 3 = 4.89897948557

Let's round that to the nearest tenth.

4.9 units

Hope this helped :)

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