All categories

3 years ago

DE is a Mid-segment of triangle ABC. How long is DE?

Mathematics

1 answer:

KIM [24]3 years ago

8 0

Answer:

DE = 14

Step-by-step explanation:

Since, DE is a Mid-segment of triangle ABC.

Therefore,

DE = 1/2 * AC

2 DE = AC

2(5x - 1) = 3x + 19

10x - 2 = 3x + 19

10x - 3x = 19 +2

7x = 21

x = 21/7

x = 3

DE = 5x - 1

DE = 5*3 - 1

DE = 15 - 1

DE = 14

You might be interested in

Peter wants to plot the point (-2, 3) on a coordinate plane. Which statement best describes how to plot this point starting from

nirvana33 [79]

Answer:

From the origin, go left 2 then up 3

Step-by-step explanation:

negative x value means move left, positive y value means move up

7 0

3 years ago

Read 2 more answers

Only need help on number 5 and 6

Alborosie

5) since (x-a)^2+(y-b)^2=r^2

plug in (-5,2) as (a,b) and r=3

Making the equation:

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

6) (x-0)^2+(y-0)^2=r^2

since we have (0,5)

0^2+5^2=r^2

Final equation: x^2+y^2=25

5 0

3 years ago

Rounded to the nearest whole number, how many yards is 32 kilometers?

earnstyle [38]

Answer:

B Is the correct answer

Step-by-step explanation:

4 0

4 years ago

Read 2 more answers

Line AB contains points A (0, 1) and B (1, 5). The slope of line AB is

Diano4ka-milaya [45]

(0,1)(1,5)
slope = (5 - 1) / (1 - 0) = 4/1 = 4 <==

8 0

3 years ago

Read 2 more answers

Determine the equation of the tangent of the hyperbola that is parallel to the given line:

marin [14]

The tangent line to the curve has slope equal to $\frac{\mathrm dy}{\mathrm dx}$ . Use implicit differentiation to find this derivative.

$9x^2-4y^2=32$

$\implies18x-16y\dfrac{\mathrm dy}{\mathrm dx}=0\implies\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{-18x}{-16y}=\dfrac{9x}{8y}$

The given line in slope-intercept form is

$9x+2y-1=0\implies y=\dfrac{1-9x}2$

and has slope -9/2. Any line parallel to this one has the same slope. The tangent to the hyperbola has this slope at points (<em>x</em>, <em>y</em>) such that

$-\dfrac92=\dfrac{9x}{8y}\implies x=-4y$

Find the points on the hyperbola where this condition is met.

$9(-4y)^2-4y^2=140y^2=32\implies y^2=\dfrac{32}{140}\implies y=\pm\sqrt{\dfrac8{35}}\implies x=\pm4\sqrt{\dfrac8{35}}$

Then use the point-slop formula to build the equations of these tangents:

$y\mp\sqrt{\dfrac8{35}}=-\dfrac92\left(x\pm4\sqrt{\dfrac8{35}}\right)$

8 0

3 years ago