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3 years ago

Help,anyone can help me do quetion,I will mark brainlest.

Mathematics

2 answers:

Brilliant_brown [7]3 years ago

8 0

Answer:

Step-by-step explanation:

I can help you

2. Area of a rectangle 40 cm2

So. We know the area AND one of the sides of the rectangle.

We can calculate the area.

We know that area=length times width

A=lw

Rearrange that for w:

w=A/l

w=40:8

w=5

Now we know both l and w

Now we can calculate the perimeter

P=2a+2b

P=2(a+b)

P=2(8+5)

P=2(13)

P=26

I hope it is correct

fomenos3 years ago

5 0

Answer:

I have absolutely no clue sorry buddy

Step-by-step explanation:

There is no step-by-step explanation

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Coordinates will be fine PLs ASAP cause I need this rn<br> 13 POINTS!!!

grigory [225]

If the vertices of the triangle are J(-3,-5), K(-2,2) and L(1,-4) then its reflection in y axis are J(3,-5), K(2,2) and L(-1,-4)

Given,

The vertices of the triangle = J(-3,-5), K(-2,2) and L(1,-4)

We know,

In reflection in y axis (x,y) will become (-x,y)

Then the vertices of the reflection in y axis = J(3,-5), K(2,2) and L(-1,-4)

Draw the triangle using the vertices.

Hence, if the vertices of the triangle are J(-3,-5), K(-2,2) and L(1,-4) then its reflection in y axis are J(3,-5), K(2,2) and L(-1,-4)

Learn more about reflection here

brainly.com/question/15487308

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3 0

1 year ago

Trini rode her bike 12 miles on Friday. She rode 14 miles on Saturday and 15 miles on Sunday. How many miles did she ride in all

loris [4]

Answer: 41

Step-by-step explanation:

12+14+15= 41

4 0

2 years ago

An elevator containing five people can stop at any of seven floors. What is the probability that no two people exit at the same

elena-s [515]

Answer:

Approximately $0.15$ ( $360 / 2401$ .) (Assume that the choices of the $5$ passengers are independent. Also assume that the probability that a passenger chooses a particular floor is the same for all $7$ floors.)

Step-by-step explanation:

If there is no requirement that no two passengers exit at the same floor, each of these $5$ passenger could choose from any one of the $7$ floors. There would be a total of $7 \times 7 \times 7 \times 7 \times 7 = 7^{5}$ unique ways for these $5\!$ passengers to exit the elevator.

Assume that no two passengers are allowed to exit at the same floor.

The first passenger could choose from any of the $7$ floors.

However, the second passenger would not be able to choose the same floor as the first passenger. Thus, the second passenger would have to choose from only $(7 - 1) = 6$ floors.

Likewise, the third passenger would have to choose from only $(7 - 2) = 5$ floors.

Thus, under the requirement that no two passenger could exit at the same floor, there would be only $(7 \times 6 \times 5 \times 4 \times 3)$ unique ways for these two passengers to exit the elevator.

By the assumption that the choices of the passengers are independent and uniform across the $7$ floors. Each of these $7^{5}$ combinations would be equally likely.

Thus, the probability that the chosen combination satisfies the requirements (no two passengers exit at the same floor) would be:

$\begin{aligned}\frac{(7 \times 6 \times 5 \times 4 \times 3)}{7^{5}} \approx 0.15\end{aligned}$ .

5 0

2 years ago

Can someone please help me with my maths question

DIA [1.3K]

Answer:

$a. \ \dfrac{625 \cdot m}{27 \cdot n^{11}}$

$b. \ \dfrac{x^{3 \cdot m - 2}}{y^{ 3 + n}}$

Step-by-step explanation:

The question relates with rules of indices

(a) The give expression is presented as follows;

$\dfrac{m^3 \times \left (n^{-2} \right )^4 \times (5 \cdot m)^4}{\left (3 \cdot m^2 \cdot n \right )^3}$

By expanding the expression, we get;

$\dfrac{m^3 \times n^{-8} \times 5^4 \times m^4}{\left 3^3 \times m^6 \times n^3}$

Collecting like terms gives;

$\dfrac{m^{(3 + 4 - 6)} \times 5^4}{ 3^3 \times n^{3 + 8}} = \dfrac{625 \cdot m}{27 \cdot n^{11}}$

$\dfrac{m^3 \times \left (n^{-2} \right )^4 \times (5 \cdot m)^4}{\left (3 \cdot m^2 \cdot n \right )^3}= \dfrac{625 \cdot m}{27 \cdot n^{11}}$

(b) The given expression is presented as follows;

$x^{3 \cdot m + 2} \times \left (y^{n - 1} \right )^3 \div (x \cdot y^n)^4$

Therefore, we get;

$x^{3 \cdot m + 2} \times \left (y^{n - 1} \right )^3 \times x^{-4} \times y^{-4 \cdot n}$

Collecting like terms gives;

$x^{3 \cdot m + 2 - 4} \times \left (y^{3 \cdot n - 3 -4 \cdot n}} \right ) = x^{3 \cdot m - 2} \times \left (y^{ - 3 -n}} \right ) = x^{3 \cdot m - 2} \div \left (y^{ 3 + n}} \right )$