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murzikaleks [220]
2 years ago
9

I need help with this please help asap

Mathematics
2 answers:
dusya [7]2 years ago
5 0

Answer: 5-3

Step-by-step explanation:

According to the law Pemdas, you do parenthesis first.

Inside the parenthesis is 5-3+6.

It's not answer choice d, since it would be -3+6 and not 3+6.

MrMuchimi2 years ago
4 0

Answer:

5-3

Step-by-step explanation:

PEMDAS

Parenthesis

Exponents

Multiplication

Division

Addition

Subtraction

(And left to right)

In this case the parenthesis are first, and 5-3 is the first operation from left to right

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2 years ago
A rectangle has a length 6 more than it's width if the width is decreased by 2 and the length decreased by 4 the resulting has a
Rashid [163]

Answer:

Length of original rectangle: 11 units.

\frac{\text{Area of original rectangle}}{\text{Area of new rectangle}}=\frac{55}{21}

\text{Perimeter of new rectangle}=20

Step-by-step explanation:

Let x represent width of the original rectangle.  

We have been given that a rectangle has a length 6 more than it's width. S the length of the original rectangle would be x+6.

We have been given that when the width is decreased by 2 and the length decreased by 4 the resulting has an area of 21 square units.

The width of new rectangle would be x-2.

The length of new rectangle would be x+6-4=x+2.

The area of new rectangle would be (x+2)(x-2).

Now we will equate area of new rectangle with 21 and solve for x as:

(x+2)(x-2)=21

Applying difference of squares, we will get:

x^2-2^2=21

x^2-4=21

x^2-4+4=21+4

x^2=25

Since width cannot be negative, so we will take positive square root of both sides.

\sqrt{x^2}=\sqrt{25}

x=5

Therefore, the width of original rectangle is 5 units.

Length of the original rectangle would be x+6\Rightarrow x+5=11.

Therefore, the length of original rectangle is 11 units.

\text{Area of original rectangle}=5\times 11

\text{Area of original rectangle}=55    

Therefore, area of the original rectangle is 55 square units.

Now we will find ratio of the original rectangle area to the new rectangle area as:

\frac{\text{Area of original rectangle}}{\text{Area of new rectangle}}=\frac{55}{21}

We know that perimeter of rectangle is two times the sum of length and width.

\text{Perimeter of new rectangle}=2((x+2)+(x-2))

\text{Perimeter of new rectangle}=2((5+2)+(5-2))

\text{Perimeter of new rectangle}=2(7+3)

\text{Perimeter of new rectangle}=2(10)

\text{Perimeter of new rectangle}=20

Therefore, the perimeter of the new rectangle is 20 units.

7 0
3 years ago
You want to wrap a gift shaped like the regular triangular prism shown. How many
kotegsom [21]

Answer:

697.6 \ in^2

Step-by-step explanation:

-Given the dimensions of the box is 8.7 in, 14 in and 10 in.

Let length=14 in, width=10 in and height =8.7 in

#The surface area of a rectangular prism is calculated as:

Surface \ Area=2(wl+hl+hw)\\\\l=length, w=width, h=height\\\\\therefore Surface \ Area=2(14\times 10+10\times 8.7+14\times 8.7)\\\\=697.6 \ in^2

Hence, the wrapping paper needed has an area of 697.6 \ in^2

8 0
3 years ago
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