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3 years ago

Where can we put parenthesis in 21 + 21 divided by 7 to make it equivalent to 6?

Mathematics

2 answers:

timurjin [86]3 years ago

7 0

Answer:

1.) $\frac{(21+21)}{7}$

2.) $(\frac{21+21}{7} )$

Step-by-step explanation:

$\frac{21+21}{7}$

If you just solve it like this, you get 6.

So, the parenthesis would go on the whole thing, or where 21+21 is.

1.) $\frac{(21+21)}{7}$

2.) $(\frac{21+21}{7} )$

DiKsa [7]3 years ago

4 0

(21+21)÷7

Step-by-step explanation:

21+21=42 42÷7=6 I'm not good at explaining lol

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There are 5 less oranges on the tree today than there was yesterday Plz help

topjm [15]

That’s all ? is there like no other explanation to make it understood better ?

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3 years ago

Read 2 more answers

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Goryan [66]

Answer:

a. Jared and Ali

Step-by-step explanation:

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3 years ago

A rectangular swimming pool is 15 meters long, 10 1/2

daser333 [38]

V = lwh

15(10.5)(4.5) = 708.75

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3 years ago

Find the area of the composite figure.

Roman55 [17]

Answer:

The Area of the composite figure would be 76.26 in^2

Step-by-step explanation:

According to the Figure Given:

Total Horizontal Distance = 14 in

Length = 6 in

To Find :

The Area of the composite figure

Solution:

Firstly we need to find the area of Rectangular part.

So We know that,

$\boxed{ \rm \: Area \: of \: Rectangle = Length×Breadth}$

Here, Length is 6 in but the breadth is unknown.

To Find out the breadth, we’ll use this formula:

$\boxed{\rm \: Breadth = total \: distance - Radius}$

According to the Figure, we can see one side of a rectangle and radius of the circle are common, hence,

$\longrightarrow\rm \: Length \: of \: the \: circle = Radius$

Since Length = 6 in ;

$\longrightarrow \rm \: 6 \: in = radius$

Hence Radius is 6 in.

So Substitute the value of Total distance and Radius:

Total Horizontal Distance= 14
Radius = 6

$\longrightarrow\rm \: Breadth = 14-6$

$\longrightarrow\rm \: Breadth = 8 \: in$

Hence, the Breadth is 8 in.

Then, Substitute the values of Length and Breadth in the formula of Rectangle :

Length = 6
Breadth = 8

$\longrightarrow\rm \: Area \: of \: Rectangle = 6 \times 8$

$\longrightarrow \rm \: Area \: of \: Rectangle = 48 \: in {}^{2}$

Then, We need to find the area of Quarter circle :

We know that,

$\boxed{\rm Area_{(Quarter \; Circle) } = \cfrac{\pi{r} {}^{2} }{4}}$

Now Substitute their values:

r = radius = 6
π = 3.14

$\longrightarrow\rm Area_{(Quarter \; Circle) } = \cfrac{3.14 \times 6 {}^{2} }{4}$

Solve it.

$\longrightarrow\rm Area_{(Quarter \; Circle) } = \cfrac{3.14 \times 36}{4}$

$\longrightarrow\rm Area_{(Quarter \; Circle) } = \cfrac{3.14 \times \cancel{{36} } \: ^{9} }{ \cancel4}$

$\longrightarrow\rm Area_{(Quarter \; Circle)} =3.14 \times 9$

$\longrightarrow\rm Area_{(Quarter \; Circle) } = 28.26 \: {in}^{2}$

Now we can Find out the total Area of composite figure:

We know that,

$\boxed{ \rm \: Area_{(Composite Figure)} =Area_{(rectangle)}+ Area_{ (Quarter Circle)}}$

So Substitute their values:

$\rm Area_{(rectangle)}$ = 48
$\rm Area_{(Quarter Circle)}$ = 28.26

$\longrightarrow \rm \: Area_{(Composite Figure)} =48 + 28 .26$

Solve it.

$\longrightarrow \rm \: Area_{(Composite Figure)} =\boxed{\tt 76.26 \:\rm in {}^{2}}$

Hence, the area of the composite figure would be 76.26 in² or 76.26 sq. in.

$\rule{225pt}{2pt}$

I hope this helps!

3 0

2 years ago

Three stamps can be attached to each other in various ways.how many ways might three stamps be attached?

ivanzaharov [21]

Let's call the stamps A, B, and C. They can each be used only once. I assume all 3 must be used in each possible arrangement.
There are two ways to solve this. We can list each possible arrangement of stamps, or we can plug in the numbers to a formula.
Let's find all possible arrangements first. We can easily start spouting out possible arrangements of the 3 stamps, but to make sure we find them all, let's go in alphabetical order. First, let's look at the arrangements that start with A:
ABC
ACB
There are no other ways to arrange 3 stamps with the first stamp being A. Let's look at the ways to arrange them starting with B:
BAC
BCA
Try finding the arrangements that start with C:
C_ _
C_ _
Or we can try a little formula; y×(y-1)×(y-2)×(y-3)...until the (y-x) = 1 where y=the number of items.
In this case there are 3 stamps, so y=3, and the formula looks like this: 3×(3-1)×(3-2).
Confused? Let me explain why it works.
There are 3 possibilities for the first stamp: A, B, or C.
There are 2 possibilities for the second space: The two stamps that are not in the first space.
There is 1 possibility for the third space: the stamp not used in the first or second space.
So the number of possibilities, in this case, is 3×2×1.
We can see that the number of ways that 3 stamps can be attached is the same regardless of method used.

8 0

3 years ago