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3 years ago

5

Choose the scenario that could be represented by 2/5

1 answer:

Gemiola [76]3 years ago

5 0

As far I as know
It should be A

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Use properties to compare expression 5 x ( 6 x 3 )and ( 5 x 6) x 3

nadya68 [22]

5(6*3) = 5(18) = 90

(5*6)3 = (30)3 = 90

The answer is the same by the associative law.

6 0

3 years ago

* One endpoint of a line segment has coordinates represented by (x + 2, 5y): The midpoint of the line segment is (6, –3).

kirill [66]

Answer:

$m = \frac{y2 - y1}{x2 - x1} \\ = \frac{ - 3 - 5y}{6 - (x + 2)} \\ = \frac{ - 3 - 5y}{6 - x - 2} \\ m = \frac{ - 3 - 5y}{4 - x}$

3 0

2 years ago

Please help. im lost. Question is in the picture

navik [9.2K]

In triangle QRT and triangle RTS , we have;
1.QR =SR (given)
2.RT bisects QS at T ( given )
3.T is the midpoint of QS (since RT is the bisector of QS i.e. QT=TS)
4.QT=TS (mid Point theorem)
5.TR=TR (common side)
6.triangle QRT congruent to triangle RTS (SAS criterion rule)
Hope this helps u....!!!

6 0

3 years ago

The figure will be dilated by a scale factor of 1.2. Find the new length. W: 25 in L: 40 in

Ghella [55]

The new length and width is 48in and 40in respectively

<h3>Dilation</h3>

Given the following information

Width = 25in
Length = 40in

If the length and the width are dilated by a factor of 1.2, the new length and width is expressed as;

W = 25(1.2) = 30in

L = 40(1.2)

L = 48 in

Hence the new length and width is 48in and 40in respectively

Learn more on dilation here: brainly.com/question/10253650

3 0

2 years ago

Rita is selecting a sandwich at the deli. The deli has four types of meat, three types of cheese and two types of bread. A delux

zalisa [80]

Answer:

24 possible dishes

Step-by-step explanation:

Given

Total

$Meat = 4$

$Cheese = 3$

$Bread = 2$

Selection

$Meat = 1$

$Cheese =2$

$Bread = 1$

Required

Determine the number of possible combo

Rita wants to select 1 of 4 meat types,

So, the number of selections is: $^4C_1$

Rita wants to select 2 of 3 cheese types,

So, the number of selections is: $^3C_2$

Rita wants to select 1 of 2 cheese types,

So, the number of selections is: $^2C_1$

The total number of selections is:

$Total = ^4C_1 * ^3C_2 * ^2C_1$

Using

$^nC_r = \frac{n!}{(n-r)!r!}$

We have:

$Total = \frac{4!}{(4-1)!1!} * \frac{3!}{(3-2)!2!} * \frac{2!}{(2-1)!1!}$

$Total = \frac{4!}{3!1!} * \frac{3!}{1!2!} * \frac{2!}{(1!1!}$

$Total = \frac{4*3!}{3!*1} * \frac{3*2!}{1*2!} * \frac{2*1}{1*1}$

$Total = 4* 3 * 2$

$Total = 24$

<em>Hence, there are 24 possible dishes</em>

4 0

3 years ago

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