What has for congruent sides?

Answers

Answer 1
Answer: Every square and rhombus has, and any other polygon
with more than four sides may have.

Also, many shipping and packing cartons and crates have.

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2x + y = 5 , x + y = 4Solve the system of linear equations.
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A swimming pool holds 900 cubic meters of water. If its length is 20 meters and its height is 3 meters, find its width

Answers

V=900m³

V=L*w*h

V=20*3*w
V=60*w

60*w=900

w=900/60
w=15m
Volume \ of \ a \ Cuboid: \n \n V=9000 m^3 \nthe \ length : \ l=20 \ m \n the \ height: \ h=3 m \n the \ width : \ w= ? \n \nV= l\cdot w \cdot h \n \n900 =20 \cdot w \cdot 3 \n \n900=60w \ \ / :60\n \n w=15 \ m \n\nAnswer : \ A \ swimming \ holds \ 15 \ meters\ the \ width .

Solve for x₁ and x₂
6x₁+3x₂=60
6x₁-2x₂=80

Answers

\left \{ {\big{6x_1+3x_2=60\ \ \ \ \ \ \ \ \ \ \ } \atop \big{6x_1-2x_2=80\ /\cdot(-1)}} \right. \n\n \left \{ {\big{6x_1+3x_2=60\ \ \ \ } \atop \big{-6x_1+2x_2=-80}} \right. \n-----------\n3x_2+2x_2=60-80\n5x_2=-20\ /:5\n\nx_2=-4\n\n6x_1+3x_2=60\ /:3\n2x_1+x_2=20\ \ \ \Rightarrow\ \ \ 2x_1+(-4)=20\ \ \ \Rightarrow\ \ \ 2x_1=24\ /:2\n\nx_1=12\n\n \left \{ {\big{x+1=12} \atop\big {x_2=-4}} \right.
first let's change the "x1" to "A" Nd the "x2" to "B" __ so now the problem reads "6A + 3B = 60" And "6A - 2B= 80" we want to isolate variable. and the easiest one is the "B" by getting rid of the "A". set up a subtraction problem where you take one equation away from the other. 6a-6a=0. 3b - (-2b) = 5b and 60-80=-20. so your new equation reads 5B= -20. you wanna get B on one side so yu divide 5 on both sides giving you b= -4 or "x2"=-4. from there you can plug -4 into any one of the original equations. ex: 6A -2(-4)=80 then 6A + 8=80 then you subtract 8 from both sides because you're trying to isolate the "A" variable. which leaves you with 6A = 72. and now you wanna get"A" on one side so yu divide 6 from both sides and get A=12 or "x1"=12. do a quick check and plug in these numbers into any one of the equation.

2. What is the sample space for the chance experiment described on this scenario card?

Answers

Answer:18 outcomes

Step-by-step explanation:

There are 18 outcomes on this scenario card. Students can list all of the outcomes or describe them. Here is the list

(the first number is the outcome from Spinner 1, and the second number is the outcome from Spinner 2).

What is the vertex of the graph of f(x) = |x – 13| + 11?

Answers

Hello,

the graph of f(x) is joint here.
It has a form of "V" with (13,11) like vertex;
It is a minimum.
It is composed of 2 half straight lines y=-x-2 (for x<13) and y=x+2 (for x>13)

Answer:

x>13

Step-by-step explanation:

edge 2021

Can u help me with this please

Answers

1. Turkey trot
2. Pumpkin pie
3. Family dinner
4. Mashed potatoes
5. Football
6. Wishbone
7. Cut the turkey
8. Beans (I can’t tell what the first emoji is)
9. Cornucopia
10. Black Friday

Answer:

1. turkey trolley

2. pumpkin pie

3. family dinner

4. smashed potatoes

5. football

6. wishbones

7. kill the turkey

8. green beans

9. corn you policeman xD

10. black friday

some of the stuff i know but i tried

Step-by-step explanation:

cards are dealt one by one from a well shuffled pack of 52 cards. find the probability that exactly n cards are dealt before the first ace appears. if the cards are drawn further, then find the probability that exactly k cards are dealt in all before the second ace.

Answers

To find the probability that exactly n cards are dealt before the first ace appears, we can use the concept of a geometric distribution. In a geometric distribution, we're interested in the number of trials (in this case, card draws) required for a success to occur (in this case, drawing an ace) for the first time.

The probability of drawing an ace in a single draw from a well-shuffled pack of 52 cards is 4/52 because there are 4 aces out of 52 cards.

So, the probability of drawing a non-ace in a single draw is (52 - 4)/52 = 48/52.

Now, let X be the random variable representing the number of cards drawn before the first ace appears. X follows a geometric distribution with parameter p, where p is the probability of success on a single trial.

P(X = n) = (1 - p)^(n - 1) * p

In this case, p is the probability of drawing an ace on a single trial, which is 4/52, and n is the number of cards drawn before the first ace.

So, the probability that exactly n cards are dealt before the first ace appears is:

P(X = n) = (1 - 4/52)^(n - 1) * (4/52)

Now, to find the probability that exactly k cards are dealt in all before the second ace appears, we need to consider two scenarios:

1. The first ace appears on the nth card, and the second ace appears on the kth card after that. This is represented by P(X = n) * P(X = k).

2. The first ace appears on the kth card, and the second ace appears on the nth card after that. This is represented by P(X = k) * P(X = n).

So, the total probability that exactly k cards are dealt before the second ace appears is:

P(X = n) * P(X = k) + P(X = k) * P(X = n)

You can calculate this probability using the formula for the geometric distribution with p = 4/52 as mentioned earlier for both P(X = n) and P(X = k).