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2 years ago

Combine any like terms in the expression. If there are no like terms, rewrite the expression.

Mathematics

1 answer:

Alona [7]2 years ago

6 0

Answer:

2x+x

Step-by-step explanation:

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Which inequality sign makes the statement true ?

masha68 [24]

Your answer is the last one!

6 0

2 years ago

What is the area of a triangle with vertices (3,0) (9,0) (7,6)

Nezavi [6.7K]

Answer:

The area of the triangle is of 21 units of area.

Step-by-step explanation:

The area of a triangle with three vertices $(x_1,y_1),(x_2,y_2),(x_3,y_3)$ is given by the determinant of the following matrix:

$A = \pm 0.5 \left|\begin{array}{ccc}x_1&y_1&1\\x_2&y_2&1\\1_3&y_3&1\end{array}\right|$

In this question:

Vertices (3,0) (9,0) (7,6). So

$A = \pm 0.5 \left|\begin{array}{ccc}3&0&1\\9&0&1\\7&7&1\end{array}\right|$

$A = \pm 0.5(3*0*1+0*1*7+1*9*7-0*7*1-0*9*1-3*1*7)$

$A = \pm 0.5*(63-21)$

$A = \pm 0.5*42 = 21$

The area of the triangle is of 21 units of area.

5 0

2 years ago

Help I will make you a brainlist!!

enyata [817]

You answer would be Oy= 7x-1. I hope this helps!

4 0

2 years ago

2/3 times 3 times 2/7

fomenos

Answer:

4/7

Step-by-step explanation:

2/7 x 3= 6/7. 6/7× 2/3=4/7

7 0

2 years ago

Read 2 more answers

Given that −4i is a zero, factor the following polynomial function completely. Use the Conjugate Roots Theorem, if applicable. f

GalinKa [24]

Answer:

$\large \boxed{\sf \bf \ \ f(x)=(x-4i)(x+4i)(x+3)(x-5) \ \ }$

Step-by-step explanation:

Hello, the Conjugate Roots Theorem states that if a complex number is a zero of real polynomial its conjugate is a zero too. It means that (x-4i)(x+4i) are factors of f(x).

$\text{Meaning that } (x-4i)(x+4i) =x^2-(4i)^2=x^2+16 \text{ is a factor of f(x).}$

The coefficient of the leading term is 1 and the constant term is -240 = 16 * (-15), so we a re looking for a real number such that.

$f(x)=x^4-2x^3+x^2-32x-240\\\\ =(x^2+16)(x^2+ax-15)\\\\ =x^4+ax^3-15x^2+16x^2+16ax-240$

We identify the coefficients for the like terms, it comes

a = -2 and 16a = -32 (which is equivalent). So, we can write in $\mathbb{R}$ .

$\\f(x)=(x^2+16)(x^2-2x-15)$

The sum of the zeroes is 2=5-3 and their product is -15=-3*5, so we can factorise by (x-5)(x+3), which gives.

$f(x)=(x^2+16)(x^2-2x-15)\\\\=(x^2+16)(x^2+3x-5x-15)\\\\=(x^2+16)(x(x+3)-5(x+3))\\\\=\boxed{(x^2+16)(x+3)(x-5)}$

And we can write in $\mathbb{C}$

$f(x)=\boxed{(x-4i)(x+4i)(x+3)(x-5)}$

Hope this helps.

Do not hesitate if you need further explanation.

Thank you

7 0

2 years ago