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2 years ago

The sum of x and y decreased by their product

Mathematics

1 answer:

motikmotik2 years ago

7 0

The sum of x and y=x+y
their product=xy
Therefore the expression is
(x+y)-(xy)

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What is the number writren in standard form six million, seven hundred ,twenty

enot [183]

6,720,000. I hope that helps!

4 0

2 years ago

Read 2 more answers

3p - 5 = 19<br> Answer when solved??

lianna [129]

Answer: P=8

Step-by-step explanation:

3p-5=19

U turn -5 to +5

Then u add 5 to both sides, -5 and 19

3p-5=19

+5=+5

-5 and +5 cancel each other out so know it’s

3p=24

Because 19 plus 5 is 24

Now u have to get the variable by itself by dividing 3 on both sides of the equal sign 3p and 24

3 and 3 cancel each other out so now you only have p=24 but then 24 divided 3 is 8

3 0

2 years ago

Helppppp pleaseeee!!!!!!

maks197457 [2]

I got 3.825

your equation would be
(2x-1)(x+7)=72

when you distribute youd get
2x^2 +13x -7 =72

move the 72 over giving you
2x^2 +13x -79=0

i used the quadratic formula and the answer would be 3.825 because a negative answer wouldn’t make sense in this situation

5 0

1 year ago

SOME ONE PLEASE HELP !!!!

Andrei [34K]

x = 20 - -5 = 20 + 5 = 25

y = 16 - 1 = 15

7 0

2 years ago

The given matrix is the augmented matrix for a linear system. Use technology to perform the row operations needed to transform t

shtirl [24]

Answer:

$x_{1} = \frac{176}{127} + \frac{71}{127}x_{4}\\\\ x_{2} = \frac{284}{127} + \frac{131}{254}x_{4}\\\\x_{3} = \frac{845}{127} + \frac{663}{254}x_{4}\\$

Step-by-step explanation:

As the given Augmented matrix is

$\left[\begin{array}{ccccc}9&-2&0&-4&:8\\0&7&-1&-1&:9\\8&12&-6&5&:-2\end{array}\right]$

Step 1 :

$r_{1}$ ↔ $r_{1} - r_{2}$

$\left[\begin{array}{ccccc}1&-14&6&-9&:10\\0&7&-1&-1&:9\\8&12&-6&5&:-2\end{array}\right]$

Step 2 :

$r_{3}$ ↔ $r_{3} - 8r_{1}$

$\left[\begin{array}{ccccc}1&-14&6&-9&:10\\0&7&-1&-1&:9\\0&124&-54&77&:-82\end{array}\right]$

Step 3 :

$r_{2}$ ↔ $\frac{r_{2}}{7}$

$\left[\begin{array}{ccccc}1&-14&6&-9&:10\\0&1&-\frac{1}{7} &-\frac{1}{7} &:\frac{9}{7} \\0&124&-54&77&:-82\end{array}\right]$

Step 4 :

$r_{1}$ ↔ $r_{1} + 14r_{2}$ , $r_{3}$ ↔ $r_{3} - 124r_{2}$

$\left[\begin{array}{ccccc}1&0&4&-11&:-8\\0&1&-\frac{1}{7} &-\frac{1}{7} &:\frac{9}{7} \\0&0&- \frac{254}{7} &\frac{663}{7} &:-\frac{1690}{7} \end{array}\right]$

Step 5 :

$r_{3}$ ↔ $\frac{r_{3}. 7}{254}$

$\left[\begin{array}{ccccc}1&0&4&-11&:-8\\0&1&-\frac{1}{7} &-\frac{1}{7} &:\frac{9}{7} \\0&0&1&-\frac{663}{254} &:-\frac{1690}{254} \end{array}\right]$

Step 6 :

$r_{1}$ ↔ $r_{1} - 4r_{3}$ , $r_{2}$ ↔ $r_{2} + \frac{1}{7} r_{3}$

$\left[\begin{array}{ccccc}1&0&0&-\frac{71}{127} &:\frac{176}{127} \\0&1&0&-\frac{131}{254} &:\frac{284}{127} \\0&0&1&-\frac{663}{254} &:\frac{845}{127} \end{array}\right]$

∴ we get

$x_{1} = \frac{176}{127} + \frac{71}{127}x_{4}\\\\ x_{2} = \frac{284}{127} + \frac{131}{254}x_{4}\\\\x_{3} = \frac{845}{127} + \frac{663}{254}x_{4}\\$

6 0

2 years ago