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3 years ago

Which expression is another way to write the expression shown here

Mathematics

1 answer:

nordsb [41]3 years ago

7 0

Answer:

5(a+b+c+d)

Step-by-step explanation:

this is correct another way of writing this expression

because 5 is common to all and we can put it out

hope it helps

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The drawing of a building has a scale of 1 inch to 30 feet. what is the actual heght, in feet of the building

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A triangular rug has sides that are 5 ft and 4 ft long with an included angle of 79°. what is the area of this rug? enter your a

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Simplifying rational expressions x^2-16/x^2+6x+8 = x-4/x+2 x^2-x-6/x^2-3x-10 = x-3/x-5

enot [183]

$\dfrac{x^2-16}{x^2+6x+8}$

Decompose the numerator and denominator into multipliers

To simplify the numerator we use the formula of difference of squares

$x^2-y^2=(x-y)(x+y)$

$x^2-16=(x-4)(x+4)$

To decompose the denominator into multipliers solve the square equation

$x^2+6x+8=0\\D=6^2-4*8=4=2^2\\x_1=\dfrac{-6+2}{2} =-2\\x_2=\dfrac{-6-2}{2} =-4$

Formula for factoring a square equation

$(x-x_1)(x-x_2)$

Substituting the found roots of the equation into the formula

$(x-(-2))(x-(-4))=(x+2)(x+4)$

After simplifying the numerator and denominator we get a fraction

$\dfrac{(x-4)(x+4)}{(x+2)(x+4)}=\dfrac{x-4}{x+2}$ , so

$\dfrac{x^2-16}{x^2+6x+8}=\dfrac{(x-4)(x+4)}{(x+2)(x+4)}=\dfrac{x-4}{x+2}$

$\dfrac{x^2-x-6}{x^2-3x-10}$

Decompose the numerator and denominator into multipliers

To decompose the numerator into multipliers solve the square equation

$x^2-x-6=0\\D=(-1)^2-4*(-6)=25=5^2\\x_1=\dfrac{1+5}{2} =3\\x_2=\dfrac{1-5}{2} =-2$

Formula for factoring a square equation

$(x-x_1)(x-x_2)$

Substituting the found roots of the equation into the formula

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

To decompose the denominator into multipliers solve the square equation

$x^2-3x-10=0\\D=(-3)^2-4*(-10)=49=7^2\\x_1=\dfrac{3+7}{2} =5\\x_2=\dfrac{3-7}{2} =-2$

Formula for factoring a square equation

$(x-x_1)(x-x_2)$

Substituting the found roots of the equation into the formula

$(x-5)(x-(-(-2))=(x-5)(x+2)$

After simplifying the numerator and denominator we get a fraction

$\dfrac{(x-3)(x+2)}{(x-5)(x+2)}=\dfrac{x-3}{x-5}$ , so

$\dfrac{x^2-x-6}{x^2-3x-10}=\dfrac{(x-3)(x+2)}{(x-5)(x+2)}=\dfrac{x-3}{x-5}$

Hello from Russia:^)

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3 years ago

Can someone please help me !

Schach [20]

$2)~3x\to\text{Linear Monomial, because it's of 1st degree.}\to B\\\\
3)~2x^4-x+5\to\text{Quartic Trinomial, because it's of 4th degree.}\to C\\\\
4)~(2x^2-5x+6)+(5x^2-3x+4)=(2x^2+5x^2)-(5x+3x)+(6+4)\\\\
=7x^2-8x+10\to A$

8 0

3 years ago