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3 years ago

14

The world's largest glacier is Lambert-

1 answer:

KiRa [710]3 years ago

8 0

15000 miles long. So length times width

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Prove or disprove that the point lies on the circle centered at the origin

tigry1 [53]

Answer:

Step-by-step explanation:

5 0

2 years ago

there are 2.54 centimeters in 1 inch. how many centimeters are there in 1 foot? in 1 yard? explain your reasoning

AVprozaik [17]

Answer:

1 foot = 30.48 centimeters

1 yard = 91.44 centimeters

Step-by-step explanation:

If there are 2.54 centimeters in an inch, and 12 inches in a foot, then you need to multiply 2.54 by 12.

Their are 36 inches in a yard so you need to multiply 2.54 by 36

3 0

3 years ago

Simplify, following the order of operations: 5 + 102 – 3 × (12 – 8) Question 6 options: 213 61 45 93

Pani-rosa [81]

Answer:

93

Step-by-step explanation:

Given in the question an expression

5 + 10² – 3 × (12 – 8)

To solve this we will use DMAS rule is followed when multiple arithmetic operations are there in a given problem like addition, subtraction, multiplication and division. It tells they should be performed in order of Division, Multiplication, Addition and Subtraction

Step 1

10²

100

5 + 100 – 3 × (12 – 8)

Step 2

(12 – 8)

(4)

5 + 100 – 3 × (4)

Step 2

3 × (4)

12

5 + 100 – 12

Step 3

5 + 100

105

105 - 12

Step 4

105 - 12

93

5 0

3 years ago

The area of ABED is 49 square units. Given AGequals9 units and ACequals10 units, what fraction of the area of ACIG is represent

Mrac [35]

Answer:

The fraction of the area of ACIG represented by the shaped region is 7/18

Step-by-step explanation:

see the attached figure to better understand the problem

step 1

In the square ABED find the length side of the square

we know that

AB=BE=ED=AD

The area of s square is

$A=b^{2}$

where b is the length side of the square

we have

$A=49\ units^2$

substitute

$49=b^{2}$

$b=7\ units$

therefore

$AB=BE=ED=AD=7\ units$

step 2

Find the area of ACIG

The area of rectangle ACIG is equal to

$A=(AC)(AG)$

substitute the given values

$A=(9)(10)=90\ units^2$

step 3

Find the area of shaded rectangle DEHG

The area of rectangle DEHG is equal to

$A=(DE)(DG)$

we have $DE=7\ units$

$DG=AG-AD=9-7=2\ units$

substitute $A=(7)(2)=14\ units^2$

step 4

Find the area of shaded rectangle BCFE

The area of rectangle BCFE is equal to

$A=(EF)(CF)$

we have

$EF=AC-AB=10-7=3\ units$

$CF=BE=7\ units$

substitute

$A=(3)(7)=21\ units^2$

step 5

sum the shaded areas

$14+21=35\ units^2$

step 6

Divide the area of of the shaded region by the area of ACIG

$\frac{35}{90}$

Simplify

Divide by 5 both numerator and denominator

$\frac{7}{18}$

therefore

The fraction of the area of ACIG represented by the shaped region is 7/18

5 0

3 years ago

Five times the larger of two

Scilla [17]

Answer:

-3, -1

Step-by-step explanation:

Smaller is x

larger is x+2

5x+10=19+8x

subtract 10 and 5x

0=9+3x

subtract 9

-9=3x

divide 3

-3=x

3 0

2 years ago