All categories

2 years ago

. If CD = 4 and the perimeter of △ABC is 23, what is the perimeter of △ABE?

Mathematics

1 answer:

OlgaM077 [116]2 years ago

6 0

Answer:

Step-by-step explanation:

ABC has a perimeter of 23, so

AB + BC + AC = 23

The perimeter of ABE would be the sum of these sides:

AB + BC + CD + DE + AE

We also know:

AE = AC, CD=4 and DE=4 because ACD is congruent to AED

So, AB + BC + CD + DE + AE = AB + BC + 4 + 4+ AC = 23+8 = 31

You might be interested in

What is the commutative property in x+3y <br> PLEASE ANSWER QUICKLY

yarga [219]

Answer:

3y+x

Step-by-step explanation:

commutative property :

a+b=b+a

Similarly:

x+3y=3y+x

4 0

3 years ago

Read 2 more answers

Kayleigh is making goodie bags for people at school. She has 54 sour candy, 60 pieces of chocolate, and 12 fruit chews. She want

stellarik [79]

Answer:

Step-by-step explanation:

1 fruit chew

4 sour candy

5 chocolate

In each bag

6 0

3 years ago

I am thinking of a number. when it is increasd by 7, it equals 21. what is my number

andrey2020 [161]

If X is increased by 7 to make 21
X + 7 = 21
then rearrange the equation to get X
X = 21-7
X = 14(b)

6 0

3 years ago

Read 2 more answers

Two particles travel along the space curves r1(t) = t, t2, t3 r2(t) = 1 + 2t, 1 + 6t, 1 + 14t . Find the points at which their p

GuDViN [60]

Answer:

A) points at which paths intersect : (1,1,1) ; (2,4,8)

B) DNE

Step-by-step explanation:

A) To find the points in which the particle paths intersect, it is necessary to find the values of t for which the three components of both vectors are equal:

$t_1=1+2t_2\\\\t_1^2=1+6t_2\\\\t_1^3=1+14t_2$

you replace t1 from the first equation in the second equation:

$(1+2t_2)^2=1+6t_2\\\\1+4t_2+4t_2^2=1+6t_2\\\\4t_2^2-2t_2=0\\\\t_2(2t_2-1)=0\\\\t_2=0\\\\t_2=\frac{1}{2}$

Then, for t2 = 0 and t2=1/2 you obtain for t1:

$t_1=1+2(0)=1\\\\t_1=1+2(\frac{1}{2})=2$

Hence, for t1=1 and t2=0 the paths intersect. Furthermore, for t1=2 and t2=1/2 the paths also intersect.

The points at which the paths intersect are:

$r_1(1)=(1,1,1)=r_2(0)=(1,1,1)\\\\r_1(2)=(2,4,8)=r_2(\frac{1}{2})=(2,4,8)$

B) You have the following two trajectories of two independent particles:

$r_1(t)=(t,t^2,t^3)\\\\r_2(t)=(1+2t,1+6t,1+14t)$

To find the time in which the particles collide, it is necessary that both particles are in the same position on the same time. That is, each component of the vectors must coincide:

$t=1+2t\\\\t^2=1+6t\\\\t^3=1+14t$