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3 years ago

Solve the following question

Mathematics

2 answers:

Anna11 [10]3 years ago

7 0

Answer:

they are 16 and 4

Step-by-step explanation:

We can call the numbers x and y and we can write:

x - y = 12

x + y = 20

Adding these equations gives us 2x = 32 which means x = 16 and substituting this value into the first equation gives us y = 4.

Yuki888 [10]3 years ago

6 0

Answer:

The numbers are 16 and 4

Step-by-step explanation:

Let the two numbers be x and y

x-y = 12

x+y = 20

Add the two equations together

x-y = 12

x+y = 20

-------------------

2x = 32

Divide by 2

2x/2 =32/2

x = 16

Now find y

x+y =20

16+y =20

Subtract 16

y = 20-16

y = 4

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Step-by-step explanation:

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Margo drove for 3.85 at an average speed of 61.4. Estimate the total distance that she drove.

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4 years ago

X/3+9=17 simplify your answer

frosja888 [35]

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3 years ago

A polynomial with rational coefficients has roots 5 and -6i. (The i is an imaginary number not variable). What is the polynomial

mars1129 [50]

Answer:

The answer is { $x^{3}$ -(5) $x^{2}$ (36)× $x$ -(180)

Step-by-step explanation:

A polynomial with rational coefficients has roots 5 and -6i (There i is the

imaginary number not variable )

As we knew that imaginary no comes in pair . This means that if -6i is

one root then the other root will be 6i

So if we assume the lowest polynomial that is possible is given as

{ $x^{3}$ -(sum of all roots) $x^{2}$ +(roots taken two at a time)

$x$ -(product of all roots)

{ $x^{3}$ -(5+6i+(-6i)) $x^{2}$ +(5×6i +5×(-6i) +6i×(-6i))

$x$ -(5×6i×(-6i))

$i^2$ = -1

{ $x^{3}$ -(5) $x^{2}$ (36)× $x$ -(180)

The general case we assume that the three previous roots and rest roots

a4,a5,a6 ...............an

$x^n$ -(sum of all roots)× $x^{n-1}$ +(roots taken two at a

time) × $x^{n-2}$ - .......................................... (product of all roots)

3 0

3 years ago

AMSWER ASAP PLEASE PLEASE PLEASE

Marysya12 [62]

Answer:

(2,14)

Step-by-step explanation:

so i solved this with substitution but there are other methods to solve this!

equation 1 : y = -x + 16
equation 2 : y = x + 12

solve for y in equation 1 !!
y = -x + 16
(add x on both sides)
y + x = 16
(subtract y on both sides)
x = -y + 16

equation 1 now equals : x = -y + 16

substitute equation 1 into equation 2 :)

y = (-y + 16) + 12
(add 16 + 12)
y = -y + 28
(add y on both sides)

2y = 28
(divide 2 on both sides)
y = 14

now to solve for x you can insert y !
you can use equation 1 or equation 2 to solve for x.

i'm using equation 2 :
y = x + 12
(14) = x + 12
(subtract 12 on both sides)
x = 2

therefore, the answer is (2, 14).

hope this helps :p

3 0

2 years ago