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3 years ago

9 to 2

Mathematics

1 answer:

Arlecino [84]3 years ago

5 0

Answer:

$\frac { 4 } { 5 }$

Step-by-step explanation:

We are given the following two points and we are to find the slope of the line which passes through these points:

$(3.-1)$ and $(-2,-5)$

The formula of the slope is given by:

Slope = $\frac { y _ 2 - y _ 1 } { x _ 2 - x _ 1 }$

Substituting the given coordinates in the above formula:

Slope = $\frac { - 5 - ( - 1 ) } { - 2 - 3 } = \frac { 4 } { 5 }$

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Can anyone help me find the coordinates (?,?) given that the shape is a parallelogram? (15 points)

xxTIMURxx [149]

The missing coordinates of the parallelogram is (m + h, n).

Solution:

Diagonals of the parallelogram bisect each other.

Solve using mid-point formula:

$$\text{Midpoint} =\left( \frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right )$

Here $x_1=m, y_1=n, x_2=h, y_2=0$

$$=\left( \frac{m+h}{2}, \frac{n+0}{2}\right )$

$$\text{Midpoint} =\left( \frac{m+h}{2}, \frac{n}{2}\right )$

To find the missing coordinate:

Let the missing coordinates by x and y.

Here $x_1=0, y_1=0, x_2=x, y_2=y$

$$\text{Midpoint}=\left( \frac{0+x}{2}, \frac{0+y}{2}\right )$

$$\left( \frac{m+h}{2}, \frac{n}{2}\right )=\left( \frac{0+x}{2}, \frac{0+y}{2}\right )$

$$\left( \frac{m+h}{2}, \frac{n}{2}\right )=\left( \frac{x}{2}, \frac{y}{2}\right )$

Now equate the x-coordinate.

$$ \frac{m+h}{2}=\frac{x}{2}$

Multiply by 2 on both sides of the equation, we get

m + h = x

x = m + h

Now equate the y-coordinate.

$$\frac{n}{2} = \frac{y}{2}$

Multiply by 2 on both sides of the equation, we get

n = y

y = n

Hence the missing coordinates of the parallelogram is (m + h, n).

5 0

3 years ago

Use the remainder theorem to divide 5x^2+9x-2 by x+3. what is the remainder?

Anna [14]

Answer:

A)16

Step-by-step explanation:

Given

f(x)=5x^2+9x-2

Remainder theorem states that when f(x) is divided by x-a then the remainder can be calculated by calculating f(a).

Now Using the remainder theorem to divide 5x^2+9x-2 by x+3 to find the remainder:

f(x)=5x^2+9x-2

f(-3) = 5(-3)^2 +9(-3) -2

=5(9) - 27 -2

= 45-29

= 16 !

5 0

2 years ago

A circle with radius of 6 cm sits inside a circle with radius of 9 cm. What is

laiz [17]

Answer:

area of Circle = (pi)r^2

3.14×4

12.56

7 0

3 years ago

Find the first five terms of the sequence in which a1 = –10 and an = 4an – 1 + 7, if n ≥ 2.

Naddika [18.5K]

Answer:

-10, -33, -125, -493, -1965

Step-by-step explanation:

a1 is -10, meaning the first term is -10. therefore the first term of the sequence has to be -10

7 0

3 years ago

Find a and b such that gcd(a, b) = 14, a > 2000, b > 2000, and the only prime divisors of a and b are 2 and 7.

choli [55]

Answer:

$a = 2^\alpha.14 \ ,\alpha\geq 8\\b = 7^\delta.14 \ , \delta\geq 3$

Step-by-step explanation:

As 2 and 7 are the only prime divisors of both $a$ and $b$ we know that both can be written as:

$a = 2^\alpha . 7^\beta \ , \alpha ,\beta\ \in \mathbb{N}_{0}\\b = 2^\gamma . 7^\delta \ , \gamma ,\delta\ \in \mathbb{N}_{0}\\$

Where $\mathbb{N}_{0}$ is the set of all natural numbers adding the zero (careful because this part is important as I'll explain next).

We also know that 14 divides both numbers and that is actually the greatest common divisor between them. So we can rewrite a and b as follows:

$a = 2^\alpha . 7^\beta.14 \ , \alpha ,\beta\ \in \mathbb{N}_{0}\\b = 2^\gamma . 7^\delta.14 \ , \gamma ,\delta\ \in \mathbb{N}_{0}\\$

Why do I write them like this? Because this way is easier to observe that if $\alpha$ and $\gamma$ were both greater than zero, then 28 would divide both hence 14 wouldn't be their g.c.d.. Likewise, if $\beta$ and $\delta$ were both greater than zero, then 98 would divide both and once again, 14 wouldn't be their g.c.d.

So either of them has to be equal to zero. And then we have that

$a = 2^\alpha.14 \\b = 7^\delta.14$

All we have left to do is find the possible values for $\alpha$ and $\delta$ so that $a>2000 \ , b>2000$ and that only happens if $\alpha\geq 8$ and $\delta\geq 3$

8 0

3 years ago