Ask question

Ask question

All categories

3 years ago

15

Find two integers whose sum is -23 and product is 132

1 answer:

IRISSAK [1]3 years ago

6 0

Step-by-step explanation:

Let x and y be two integers

x * y = 132

x + y = -23

x= -23 - y

y(-23 - y) = 132

-23y - y^2 = 132

y^2 +23y + 132 = 0

(y + 11) (y + 12) = 0

y = -11 or - 12

If y = -11 then x + (-11) = -23 -> x = -12

If y = -12 then x + (-12) = -23 -> x = -11

So the two integers are -11 and -12

You might be interested in

Hi guys can you please help me with this test.

Anika [276]

Sure! but unfortunately the picture is blank. not sure if it’s just my screen

4 0

3 years ago

*WILL GIVE BRAINLIST TO WHO EVER IS RIGHT!*

Scorpion4ik [409]

Answer:g(X)=2x^2-10

Step-by-step explanation:

If a function f(x) is translated 'c' units down then the equation of the translated function be f(x)-c

Here, f(x) is translated 2 units down to obtain g(x).

Hence, we have

g(x)=f(x)-2

=2x^2-8-2

=2x^2-10

8 0

3 years ago

Solve for x 49 = 2592/x

adoni [48]

ok so basically x=1/2 which is the exact form. and the decimal form is x=0.5 Step-by-step explanation:

6 0

2 years ago

Arrange the entries of matrix A in increasing order of their cofactors values

givi [52]

To find the cofactor of

$A=\left[\begin{array}{ccc}7&5&3\\-7&4&-1\\-8&2&1\end{array}\right]$

We cross out the Row and columns of the respective entries and find the determinant of the remaining $2\times 2$ matrix with the alternating signs.

$Ac_{11}=\left|\begin{array}{ccc}4&-1\\2&1\end{array}\right|$

$Ac_{11}=4\times 1- -1\times 2$

$Ac_{11}=4+ 2$

$Ac_{11}=6$

$Ac_{12}=-\left|\begin{array}{ccc}-7&-1\\-8&1\end{array}\right|$

$Ac_{12}=-(-7\times 1- -1\times -8)$

$Ac_{12}=-(-7- 8)$

$Ac_{12}=15$

$Ac_{21}=-\left|\begin{array}{ccc}5&3\\2&1\end{array}\right|$

$Ac_{21}=-(5\times 1- 3\times 2)$

$Ac_{21}=-(5-6)$

$Ac_{21}=1$

$A_c{23}=-\left|\begin{array}{ccc}7&5\\-8&2\end{array}\right|$

$Ac_{23}=-(7\times 2 -8\times 5)$

$Ac_{23}=-(14-40)$

$Ac_{23}=26$

$A_c{31}=\left|\begin{array}{ccc}5&3\\4&-1\end{array}\right|$

$Ac_{31}=5\times -1 -4\times 3$

$Ac_{31}=-5-12$

$Ac_{31}=-17$

$A_c{33}=\left|\begin{array}{ccc}7&5\\-7&4\end{array}\right|$

$Ac_{33}=7\times 4- -7\times 5$

$Ac_{33}=28+35$

$Ac_{33}=63$

Therefore in increasing order, we have;

$Ac_{31}=-17,Ac_{21}=1,Ac_{11}=6,Ac_{23}=26,Ac_{12}=15, Ac_{33}=63$

7 0

3 years ago

Read 2 more answers

(Worth 30 points) What are the explicit equation and domain for an arithmetic sequence with a first term of 5 and a second term

kolbaska11 [484]

Answer:

$a_n=5-2(n-1)$ , all integers where n≥1

Step-by-step explanation:

we know that

The explicit equation for an arithmetic sequence is equal to

$a_n=a_1+d(n-1)$

a_n is the th term

a_1 is the first term

d is the common difference

n is the number of terms

we have

$a_1=5\\a_2=3$

Remember that

In an Arithmetic Sequence the difference between one term and the next is a constant, and this constant is called the common difference.

To find out the common difference subtract the first term from the second term

$d=a_2-a_1=3-5=-2$

substitute the given values in the formula

$a_n=5-2(n-1)$

The domain is all integers for $n\geq 1$

3 0

3 years ago