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3 years ago

Simplify the expression please

Mathematics

1 answer:

Reika [66]3 years ago

3 0

$\bf -2(3+2i)+2(4-5i)\implies 
\\\\\\
-6+8-10i-4i\implies \boxed{2-14i}$

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I need help ASAP -2x^2+x 9-25x^2+8x X^5+x^3-x^2-1

nika2105 [10]

Answer:

For $-2x^2+x$ , Leading coefficient is -2, Degree is 2, Constant term is 0

For $9-25x^2+8x$ , Leading coefficient is -25, Degree is 2, Constant term is 9

For $x^5+x^3-x^2-1$ , Leading coefficient is 1, Degree is 5, Constant term is -1

Step-by-step explanation:

Given:

Polynomials:

$-2x^2+x\\9-25x^2+8x\\x^5+x^3-x^2-1$

To find: leading coefficient, constant term and degree

Solution:

Leading coefficient is the coefficient of the variable with the highest power.

Degree is the highest power of the variable.

The term of degree 0 is the constant term of a polynomial.

For $-2x^2+x$ ,

Leading coefficient is -2

Degree is 2

Constant term is 0

For $9-25x^2+8x$

Leading coefficient is -25

Degree is 2

Constant term is 9

For $x^5+x^3-x^2-1$

Leading coefficient is 1

Degree is 5

Constant term is -1

4 0

3 years ago

Please help me ......:(

algol [13]

Answer: B) F(x) = √x and G(x) = 3x + 2

Step-by-step explanation:

The composite function G(F(x)) is when you replace every x-value in the G(x) function with the F(x) function.

$A)\ G(3x+2) = \sqrt{3x+2}\\\\B)\ G(\sqrt{x})=3(\sqrt{x})+2\quad =3\sqrt{x}+2\\\\C)\ G(\sqrt{x}+2) = 3\quad \text{there are no x-values in the G(x) function to replace}\\\\D) G(3\sqrt{x})=2\quad \text{there are no x-values in the G(x) function to replace}$

The only one that matches G(F(x)) = 3√x + 2 is OPTION B

5 0

3 years ago

10. Write the ratio as a fraction in simplest form. 12 : 42

Pavel [41]

Answer:

2/7

Step-by-step explanation:

12 : 42 is the same as 12/42

simplify the fraction by dividing the numerator and denominator by 6

you get 2/7

8 0

3 years ago

Read 2 more answers

2(h - 8) - h = h - 16

tresset_1 [31]

$2(h-8)-h=h-16$

$2h-16-h=h-16$

$h=h$

h has infinitely many solutions.

5 0

2 years ago

Three circles with centers A, B, and C are externally tangent to each other as shown in the figure. Lines EG and DG are tangent

fiasKO [112]

My solution to the problem is as follows:

EC = 15 ... draw CF = 6 (radius) ...use Pythagorean theorem to find EF.

EF^2 + CF^2 = EC^2
EF^2 = 15^2 - 6^2 = 189 .... EF = sq root 189

triangle GDE is similar to CFE ... thus proportional
GD / ED = CF / EF
GD / 18 = 6 / (sq root 189)
GD = 108 / (sq root 189)

I hope my answer has come to your help. God bless and have a nice day ahead!

5 0

3 years ago