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3 years ago

6

One number is 5 greater than another. The product of the numbers are 84. Find both the two positive and two negative sets of num

bers.

2 answers:

Bond [772]3 years ago

6 0

GIVEN

one number is greater than the other by 5.

the product of both are 84.

find out the number.

To proof =

let assume that one number be x

other number be x+5

the equation becomes

⇒ x(x+8 ) = 84

⇒ x²+5x-84 =0

⇒ x²+ 12x-7x-84 = 0

⇒( x-7)(x+12)=0

⇒x=7, x=-12

now the positive numbers are 7 and 12

now the negative numbers are -12 and -7

hence proved

Pepsi [2]3 years ago

3 0

Let x,y be the two numbers.

Given that one number is 5 greater than another.

Let x be the smaller number ans y be the greater number.

That is y=x+5. Let this be the first equation.

And also given that product of the two numbers is 84.

That is x*y = 84, let us plugin y=x+5 here.

x*(x+5) = 84

x^2 + 5x -84 = 0.

x^2+12x-7x-84 = 0

x(x+12)-7(x+12) =0

(x-7)(x+12)=0

That is x= 7 or -12.

If x=7, y= 7+5=12.

If x=-12, y= -12+5 = -7.

Hence two positive numbers corresponding to given conditions are 7,12.

And two negative numbers corresponding to given conditions are -12,-7.

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Charra [1.4K]

Answer:

$2a^3b^2\sqrt[3]{3a}$

Step-by-step explanation:

Use the following rules for exponents:

$a^m*a^n=a^{m+n}\\\\\sqrt[3]{x^3}=x$

Simplify 24. Find two factors of 24, one of which should be a perfect cube:

$8*3=24\\\\2^3=8$

Insert:

$\sqrt[3]{2^3*3a^{10}b^6}$

Now split the exponents. Split 10 into as many 3's as possible:

$10=3+3+3+1$

Insert as exponents:

$\sqrt[3]{2^3*3*a^3*a^3*a^3*a^1*b^6}$

Split 6 into as many 3's as possible:

$6=3+3$

Insert as exponents:

$\sqrt[3]{2^3*3*a^3*a^3*a^3*a^1*b^3*b^3}$

Now simplify. Any terms with an exponent of 3 will be moved out of the radical (rule #2):

$2\sqrt[3]{3*a^3*a^3*a^3*a^1*b^3*b^3}\\\\\\2*a*a*a\sqrt[3]{3*a^1*b^3*b^3}\\\\\\2*a*a*a*b*b\sqrt[3]{3*a^1}$

Simplify:

$2a^3b^2\sqrt[3]{3a}$

:Done

6 0

3 years ago

Someone find x !!!!!!

BartSMP [9]

14. 128
15.25
16.90
those should be right

6 0

3 years ago

What is the GCF of 12 and 7

madam [21]

Answer:

The GCF of 7 and 12 is 1.

8 0

3 years ago

Read 2 more answers

Andre45 [30]

Answer:

$a_n=-3(3)^{n-1}$ ; {-3,-9, -27,- 81, -243, ...}

$a_n=-3(-3)^{n-1}$ ; {-3, 9,-27, 81, -243, ...}

$a_n=3(\frac{1}{2})^{n-1}$ ; {3, 1.5, 0.75, 0.375, 0.1875, ...}

$a_n=243(\frac{1}{3})^{n-1}$ ; {243, 81, 27, 9, 3, ...}

Step-by-step explanation:

The first explicit equation is

$a_n=-3(3)^{n-1}$

At n=1,

$a_1=-3(3)^{1-1}=-3$

At n=2,

$a_2=-3(3)^{2-1}=-9$

At n=3,

$a_3=-3(3)^{3-1}=-27$

Therefore, the geometric sequence is {-3,-9, -27,- 81, -243, ...}.

The second explicit equation is

$a_n=-3(-3)^{n-1}$

At n=1,

$a_1=-3(-3)^{1-1}=-3$

At n=2,

$a_2=-3(-3)^{2-1}=9$

At n=3,

$a_3=-3(-3)^{3-1}=-27$

Therefore, the geometric sequence is {-3, 9,-27, 81, -243, ...}.

The third explicit equation is

$a_n=3(\frac{1}{2})^{n-1}$

At n=1,

$a_1=3(\frac{1}{2})^{1-1}=3$

At n=2,

$a_2=3(\frac{1}{2})^{2-1}=1.5$

At n=3,

$a_3=3(\frac{1}{2})^{3-1}=0.75$

Therefore, the geometric sequence is {3, 1.5, 0.75, 0.375, 0.1875, ...}.

The fourth explicit equation is

$a_n=243(\frac{1}{3})^{n-1}$

At n=1,

$a_1=243(\frac{1}{3})^{1-1}=243$

At n=2,

$a_2=243(\frac{1}{3})^{2-1}=81$

At n=3,

$a_3=243(\frac{1}{3})^{3-1}=27$

Therefore, the geometric sequence is {243, 81, 27, 9, 3, ...}.

6 0

3 years ago

You wanna answer it you get SOME POINTS . AND YOU GET TO MARRY THE GIRL OF YOU'R DREAMS ! ! ! IK right . So do it .

uysha [10]

Answer:

The answer to your question is 20 feet

Step-by-step explanation:

Data

A = (-2, 1)

B = (4, 1)

C = (-2, -3)

D = (4, -3)

Process

1.- Calculate the distance from C and D

dCD = $\sqrt{(4 + 2)^{2} + (3 - 3)^{2}}$

dCD = $\sqrt{6^{2}}$

dCD = 6

2.- Calculate the distance from A to C

dAC = $\sqrt{(2 - 2)^{2} + (-3 - 1)^{2}}$

dAC = $\sqrt{4^{2}}$

dAC = 4

3.- Calculate the perimeter

Perimeter = 2(dCD) + 2(dAC)

-Substitution

Perimeter = 2(6) + 2(4)

-Simplification

Perimeter = 12 + 8

-Result

Perimeter = 20 ft

3 0

3 years ago

Read 2 more answers