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4 years ago

The difference between a number squared, and that same number is 30. What is the number

Mathematics

1 answer:

Volgvan4 years ago

6 0

Answer:

n^2 = n + 30

n^2 -n -30 = 0

Solving by quadratic formula

numbers equal 6 AND -5

DOUBLE -CHECK

6^2 -6 -30 = 0

36 -6 - 30 = 0 CORRECT

-5^2 - -5 -30 = 0

25 +5 -30 = 0 CORRECT

Two numbers 6 and -5

Step-by-step explanation:

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9.45 x 10^-5 In standard form

scoundrel [369]

The answer is : 0.0000945
Hope this helped have a good day!

4 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

NEED HELP ASAP!!!!!!

avanturin [10]

commitative property and

distributive property

5 0

3 years ago

Read 2 more answers

A. Consider the following algorithm segment: for i := 1 to 4, for j := 1 to i, [Statements in body of inner loop. None contain b

yuradex [85]

Answer:

(a) 4i times

(b) "i × n" times

Step-by-step explanation:

(a) Given the algorithm segment;

for i := 1 to 4, (Outer loop)

for j := 1 to i (Inner loop)

next j,

next i

The inner loop runs for i times, while the outer loop runs for 4 times.

The total times the inner loop would run when the entire algorithm is run is:

= i × 4

= 4i times

(b) Given the algorithm segment;

for i := 1 to n, (Outer loop)

for j := 1 to i (Inner loop)

next j,

next i

Where n is a set of positive integers.

The inner loop runs for "i" times, while the outer loop runs for "n" times.

The total times the inner loop would run when the entire algorithm is run is:

= i × n

= "i × n" times

6 0

3 years ago

Someone please help??

irga5000 [103]

Answer:

Im not 100% sure but i can tell you it is (D)

Step-by-step explanation:

8 0

3 years ago