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2 years ago

7

Can anyone help please?

1 answer:

katen-ka-za [31]2 years ago

5 0

Answer:

A'(-3, 12)
B'(9, 6)
C'(-6, -6)

Step-by-step explanation:

Dilation about the origin multiplies each coordinate value by the dilation factor.

For dilation factor k, the new coordinates are ...

(x, y) ⇒ (kx, ky)

Your dilation factor is 3, so the transformation is ...

(x, y) ⇒ (3x, 3y)

A(-1, 4) ⇒ A'(-3, 12)

B(3, 2) ⇒ B'(9, 6)

C(-2, -2) ⇒ C'(-6, -6)

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Use algebraic rules of equations to predict the solution to the system of equations. Include all of your work for full credit.

bezimeni [28]

Use substitution
x+(2x-1)=-4
3x-1=-4
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3 years ago

Combine like terms. x^2y-3xy^2-6x^2y

balu736 [363]

The result after combining like terms is: $-5x^2y-3xy^2$

Step-by-step explanation:

Given expression is:

$x^2y-3xy^2-6x^2y$

The like terms are those terms with whom the variable or product of variables is same.

As in the given expression, first and third, both terms have x^2y so both terms are like terms

Combining like terms, means the indicated operation is performed on them

So,

Combining like terms:

$x^2y-6x^2y-3xy^2\\= -5x^2y-3xy^2$

Hence,

The result after combining like terms is: $-5x^2y-3xy^2$

Keywords: Expressions, polynomials

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3 0

3 years ago

The 5th term in a geometric sequence is 160. The 7th term is 40. What are possible values of the 6th term of the sequence?

omeli [17]

Answer:

C. The 6th term is positive/negative 80

Step-by-step explanation:

Given

Geometric Progression

$T_5 = 160$

$T_7 = 40$

Required

$T_6$

To get the 6th term of the progression, first we need to solve for the first term and the common ratio of the progression;

To solve the common ratio;

Divide the 7th term by the 5th term; This gives

$\frac{T_7}{T_5} = \frac{40}{160}$

Divide the numerator and the denominator of the fraction by 40

$\frac{T_7}{T_5} = \frac{1}{4}$ ----- equation 1

Recall that the formula of a GP is

$T_n = a r^{n-1}$

Where n is the nth term

So,

$T_7 = a r^{6}$

$T_5 = a r^{4}$

Substitute the above expression in equation 1

$\frac{T_7}{T_5} = \frac{1}{4}$ becomes

$\frac{ar^6}{ar^4} = \frac{1}{4}$

$r^2 = \frac{1}{4}$

Square root both sides

$r = \sqrt{\frac{1}{4}}$

r = ± $\frac{1}{2}$

Next, is to solve for the first term;

Using $T_5 = a r^{4}$

By substituting 160 for T5 and ± $\frac{1}{2}$ for r;

We get

$160 = a \frac{1}{2}^{4}$

$160 = a \frac{1}{16}$

Multiply through by 16

$16 * 160 = a \frac{1}{16} * 16$

$16 * 160 = a$

$2560 = a$

Now, we can easily solve for the 6th term

Recall that the formula of a GP is

$T_n = a r^{n-1}$

Here, n = 6;

$T_6 = a r^{6-1}$

$T_6 = a r^5$

$T_6 = 2560 r^5$

r = ± $\frac{1}{2}$

So,

$T_6 = 2560( \frac{1}{2}^5)$ or $T_6 = 2560( \frac{-1}{2}^5)$

$T_6 = 2560( \frac{1}{32})$ or $T_6 = 2560( \frac{-1}{32})$

$T_6 = 80$ or $T_6 = -80$

$T_6 =$ ±80

Hence, the 6th term is positive/negative 80

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3 years ago

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podryga [215]

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3 years ago

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Answer is 20, hope this helps BRAINLIEST?

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3 years ago

Read 2 more answers