Find AB if BC = 4, BD = 5, and AD = 3.

I will give out BRAINLIEST for the correct answer!

earnstyle [38]

Answer:

2

Step-by-step explanation:

8 0

3 years ago

Determine the mean of the numbers 6, 17, 22, 28, and 37.

Kipish [7]

To find the mean or the avg you add up all the numbers you have and then divide by how many numbers you added together in this case 5

$6+ 17+22+28+37=110$

$\frac{110}{5} =22$

so the mean is 22, hope this helped you!

4 0

3 years ago

Which of the following sets are subspaces of R3 ?

Ratling [72]

Answer:

The following are the solution to the given points:

Step-by-step explanation:

for point A:

$\to A={(x,y,z)|3x+8y-5z=2} \\\\\to for(x_1, y_1, z_1),(x_2, y_2, z_2) \varepsilon A\\\\ a(x_1, y_1, z_1)+b(x_2, y_2, z_2) = (ax_1+bx_2,ay_1+by_2,az_1+bz_2)$

$=3(aX_l +bX_2) + 8(ay_1 + by_2) — 5(az_1+bz_2)\\\\=a(3X_l+8y_1- 5z_1)+b (3X_2+8y_2—5z_2)\\\\=2(a+b)$

The set A is not part of the subspace $R^3$

for point B:

$\to B={(x,y,z)|-4x-9y+7z=0}\\\\\to for(x_1,y_1,z_1),(x_2, y_2, z_2) \varepsilon B \\\\\to a(x_1, y_1, z_1)+b(x_2, y_2, z_2) = (ax_1+bx_2,ay_1+by_2,az_1+bz_2)$

$=-4(aX_l +bX_2) -9(ay_1 + by_2) +7(az_1+bz_2)\\\\=a(-4X_l-9y_1+7z_1)+b (-4X_2-9y_2+7z_2)\\\\=0$

$\to a(x_1,y_1,z_1)+b(x_2, y_2, z_2) \varepsilon B$

The set B is part of the subspace $R^3$

for point C: $\to C={(x,y,z)|x$

In this, the scalar multiplication can't behold

$\to for (-2,-1,2) \varepsilon C$

$\to -1(-2,-1,2)= (2,1,-1)$ ∉ C

this inequality is not hold

The set C is not a part of the subspace $R^3$

for point D:

$\to D={(-4,y,z)|\ y,\ z \ arbitrary \ numbers)$

The scalar multiplication s is not to hold

$\to for (-4, 1,2)\varepsilon D\\\\\to -1(-4,1,2) = (4,-1,-2)$ ∉ D

this is an inequality, which is not hold

The set D is not part of the subspace $R^3$

For point E:

$\to E= {(x,0,0)}|x \ is \ arbitrary) \\\\\to for (x_1,0 ,0) ,(x_{2},0 ,0) \varepsilon E \\\\\to a(x_1,0,0) +b(x_{2},0,0)= (ax_1+bx_2,0,0)\\$

The $x_1, x_2$ is the arbitrary, in which $ax_1+bx_2$ is arbitrary

$\to a(x_1,0,0)+b(x_2,0,0) \varepsilon E$

The set E is the part of the subspace $R^3$

For point F:

$\to F= {(-2x,-3x,-8x)}|x \ is \ arbitrary) \\\\\to for (-2x_1,-3x_1,-8x_1),(-2x_2,-3x_2,-8x_2)\varepsilon F \\\\\to a(-2x_1,-3x_1,-8x_1) +b(-2x_1,-3x_1,-8x_1)= (-2(ax_1+bx_2),-3(ax_1+bx_2),-8(ax_1+bx_2))$

The $x_1, x_2$ arbitrary so, they have $ax_1+bx_2$ as the arbitrary $\to a(-2x_1,-3x_1,-8x_1)+b(-2x_2,-3x_2,-8x_2) \varepsilon F$

The set F is the subspace of $R^3$

5 0

3 years ago

If f(x) = –x2 + 3x + 5 and g(x) = x2 + 2x, which graph shows the graph of (f + g)(x)? On a coordinate plane, a straight line wit

Mamont248 [21]

The graph of (f+g)(x) is on the coordinate plane, a straight line with a positive slope crosses the x-axis at (negative 1, 0) and crosses the y-axis at (0, 5).

<h3>What is the graph of a parabolic equation?</h3>

The graph of a parabolic equation is a curved U-shape graph from where the domain, the range, x-intercept(s), and the y-intercept(s) can be determined.

Given that:

f(x) = -x² + 3x + 5

g(x) = x² + 2x

To find (f + g)(x), we have:

(f + g)(x) = (-x² + 3x + 5)+(x² + 2x)

(f + g)(x) = -x² + 3x + 5 +x² + 2x

(f + g)(x) = 3x + 2x + 5

(f + g)(x) = 5x + 5

Use the Slope-intercept form: We are to find the graph of y = 5x + 5

y = 5x + 5

Here:

Slope = 5
x-intercept = (-1, 0)
y-intercept = (0,5)

Therefore, we can conclude that on the coordinate plane, a straight line with a positive slope crosses the x-axis at (negative 1, 0) and crosses the y-axis at (0, 5)

Learn more about parabolic equation here:

brainly.com/question/4061870

#SPJ1

4 0

2 years ago

Read 2 more answers

Write an equation for each condition listed.

Svetlanka [38]

Look for the y-intercept
since we know the slope is 3 we just going to plug in
-5=3(1)+y
-5-3=y
so the y-intercept is 8 now you can find the equation
y=3x-8

6 0

3 years ago