What type of association is there ??

You’ve seen two methods for finding the area of ΔABC—using coordinate algebra (by hand) and using geometry software. How are the

Phantasy [73]

AnsweYou’ve seen two methods for finding the area of ΔABC—using coordinate algebra (by hand) and using geometry software. How are the two methods similar? How are they different? Why might coordinate algebra be important in making and using geometry software?

r:

Step-by-step explanation:

8 0

2 years ago

Can anyone help? <br><br>Give the answer in four significant figures.

Llana [10]

X + 0.5y = 3
x^2 - y = 15

from the first equation:-
0.5y = 3 - x
y = 6 - 2x

Substitute for y in second equation:-
x^2 - (6 - 2x) = 15
x^2 + 2x - 21 = 0
x = [-2 +/- sqrt((4 - 4*-21)] / 2
= 3.6904 , -5.6904

Plugging in these into first equation we get

3.6904 + 0.5y = 3 so y = 2(3-3.6904) = -1.3808
-5.6904 + 0.5y = 3 so y = 2(3 + 5.6904) = 17.3808

Solution is x = 3.690, y = -1.381 ; x = -5.690, y = 17.38 ( to 4 s.f's)

5 0

3 years ago

What diagonals does a kite intersect at?

Romashka [77]

The diagonals of a kite meet at a right angle.

3 0

3 years ago

What is the sum of the

Elena-2011 [213]

Answer:

C

Hope This Helps! :) Please Give Brainliest if this does!

5 0

3 years ago

Write the equation of the quadratic function whose graph passes through <img src="https://tex.z-dn.net/?f=%28-3%2C2%29" id="TexF

blagie [28]

Answer:

$f(x)=x^2+3x+2$

Step-by-step explanation:

We want to write the equation of a quadratic whose graph passes through (-3, 2), (-1, 0), and (1, 6).

Remember that the standard quadratic function is given by:

$f(x)=ax^2+bx+c$

Since it passes through the point (-3, 2). This means that when $x=-3$ , $f(x)=f(-3)=2$ . Hence:

$f(-3)=2=a(-3)^2+b(-3)+c$

Simplify:

$2=9a-3b+c$

Perform the same computations for the coordinates (-1, 0) and (1, 6). Therefore:

$0=a(-1)^2+b(-1)+c \\ \\0=a-b+c$

And for (1, 6):

$6=a(1)^2+b(1)+c\\\\ 6=a+b+c$

So, we have a triple system of equations:

$\left\{ \begin{array}{ll} 2=9a-3b+c &\\ 0=a-b+c \\6=a+b+c \end{array} \right.$

We can solve this using elimination.

Notice that the b term in Equation 2 and 3 are opposites. Hence, let's add them together. This yields:

$(0+6)=(a+a)+(-b+b)+(c+c)$

Compute:

$6=2a+2c$

Let's divide both sides by 2:

$3=a+c$

Now, let's eliminate b again but we will use Equation 1 and 2.

Notice that if we multiply Equation 2 by -3, then the b terms will be opposites. So:

$-3(0)=-3(a-b+c)$

Multiply:

$0=-3a+3b-3c$

Add this to Equation 1:

$(0+2)=(9a-3a)+(-3b+3b)+(c-3c)$

Compute:

$2=6a-2c$

Again, we can divide both sides by 2:

$1=3a-c$

So, we know have two equations with only two variables:

$3=a+c\text{ and } 1=3a-c$

We can solve for a using elimination since the c term are opposites of each other. Add the two equations together:

$(3+1)=(a+3a)+(c-c)$

Compute:

$4=4a$

Solve for a:

$a=1$

So, the value of a is 1.

Using either of the two equations, we can now find c. Let's use the first one. Hence:

$3=a+c$

Substitute 1 for a and solve for c:

$\begin{aligned} c+(1)&=3 \\c&=2 \end{aligned}$

So, the value of c is 2.

Finally, using any of the three original equations, solve for b:

We can use Equation 3. Hence:

$6=a+b+c$

Substitute in known values and solve for b:

$6=(1)+b+(2)\\\\6=3+b\\\\b=3$

Therefore, a=1, b=3, and c=2.

Hence, our quadratic function is:

$f(x)=x^2+3x+2$

5 0

3 years ago