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2 years ago

Please Help. Will give Brainliest.

Mathematics

1 answer:

mixas84 [53]2 years ago

4 0

Answer:

translation.........

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WILL GIVE BRAINLIESTTTTT!! What is the equation for the line?

AVprozaik [17]

Y=3x+4.5 is NOT the equation the correct answer is what the person under me typed

5 0

2 years ago

Read 2 more answers

8) Find the endpoint Cif M is the midpoint of segment CD and M (2, 4) and D (5,7)

Elenna [48]

Answer:

8. c. (-1, -1)

9. a. (-6, -1)

b. True

Step-by-step Explanation:

8. Given the midpoint M(2, 4), and one endpoint D(5, 7) of segment CD, the coordinate pair of the other endpoint C, can be calculated as follows:

let $D(5, 7) = (x_2, y_2)$

$C(?, ?) = (x_1, y_1)$

$M(2, 4) = (\frac{x_1 + 5}{2}, \frac{y_1 + 7}{2})$

Rewrite the equation to find the coordinates of C

$2 = \frac{x_1 + 5}{2}$ and $4 = \frac{y_1 + 7}{2}$

Solve for each:

$2 = \frac{x_1 + 5}{2}$

$2*2 = \frac{x_1 + 5}{2}*2$

$4 = x_1 + 5$

$4 - 5 = x_1 + 5 - 5$

$-1 = x_1$

$x_1 = -1$

$4 = \frac{y_1 + 7}{2}$

$4*2 = \frac{y_1 + 7}{2}*2$

$8 = y_1 + 7$

$8 - 7 = y_1 + 7 - 7$

$1 = y_1$

$y_1 = 1$

Coordinates of endpoint C is (-1, 1)

9. a.Given segment AB, with midpoint M(-4, -5), and endpoint A(-2, -9), find endpoint B as follows:

let $A(-2, -9) = (x_2, y_2)$

$B(?, ?) = (x_1, y_1)$

$M(-4, -5) = (\frac{x_1 + (-2)}{2}, \frac{y_1 + (-9)}{2})$

$-4 = \frac{x_1 - 2}{2}$ and $-5 = \frac{y_1 - 9}{2}$

Solve for each:

$-4 = \frac{x_1 - 2}{2}$

$-4*2 = \frac{x_1 - 2}{2}*2$

$-8 = x_1 - 2$

$-8 + 2 = x_1 - 2 + 2$

$-6 = x_1$

$x_1 = -6$

$-5 = \frac{y_1 - 9}{2}$

$-5*2 = \frac{y_1 - 9}{2}*2$

$-10 = y_1 - 9$

$-10 + 9 = y_1 - 9 + 9$

$-1 = y_1$

$y_1 = -1$

Coordinates of endpoint B is (-6, -1)

b. The midpoint of a segment, is the middle of the segment. It divides the segment into two equal parts. The answer is TRUE.

4 0

3 years ago

Factor f(x) = 15x^3 - 15x^2 - 90x completely and determine the exact value(s) of the zero(s) and enter them as a comma separated

Illusion [34]

Answer:

$x=-2,0,3$

Step-by-step explanation:

We have been given a function $f(x)=15x^3-15x^2-90x$ . We are asked to find the zeros of our given function.

To find the zeros of our given function, we will equate our given function by 0 as shown below:

$15x^3-15x^2-90x=0$

Now, we will factor our equation. We can see that all terms of our equation a common factor that is $15x$ .

Upon factoring out $15x$ , we will get:

$15x(x^2-x-6)=0$

Now, we will split the middle term of our equation into parts, whose sum is $-1$ and whose product is $-6$ . We know such two numbers are $-3\text{ and }2$ .

$15x(x^2-3x+2x-6)=0$

$15x((x^2-3x)+(2x-6))=0$

$15x(x(x-3)+2(x-3))=0$

$15x(x-3)(x+2)=0$

Now, we will use zero product property to find the zeros of our given function.

$15x=0\text{ (or) }(x-3)=0\text{ (or) }(x+2)=0$

$15x=0\text{ (or) }x-3=0\text{ (or) }x+2=0$

$\frac{15x}{15}=\frac{0}{15}\text{ (or) }x-3=0\text{ (or) }x+2=0$

$x=0\text{ (or) }x=3\text{ (or) }x=-2$

Therefore, the zeros of our given function are $x=-2,0,3$ .

7 0

3 years ago

Find the output, b, when the input, a, is 6. b = -1-7a b=

natka813 [3]

The answer is -43
B=-1-7(6)
B=-1-42
B=-43

7 0

3 years ago

Look at picture. Does anybody know the answer I’m lost?

masha68 [24]

Let $x,y$ be the dimensions of the rectangle. We know the equations for both area and perimeter:

$A=xy=36$

$P=2(x+y)=36 \iff x+y=18$

So, we have the following system:

$\begin{cases}xy=36\\x+y=18\end{cases}$

From the second equation, we can deduce

$y=18-x$

Plug this in the first equation to get

$xy=x(18-x)=-x^2+18=36$

Refactor as

$x^2-18x+36=0$

And solve with the usual quadratic formula to get

$x=9\pm3\sqrt{5}$

Both solutions are feasible, because they're both positive.

If we chose the positive solution, we have

$x=9+3\sqrt{5} \implies y=18-x=18-9-3\sqrt{5}=9-3\sqrt{5}$

If we choose the negative solution, we have

$x=9-3\sqrt{5} \implies y=18-x=18-9+3\sqrt{5}=9+3\sqrt{5}$

So, we're just swapping the role of $x$ and $y$ . The two dimensions of the rectangle are $9+3\sqrt{5}$ and $9-3\sqrt{5}$

6 0

2 years ago