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GalinKa [24]
3 years ago
14

Evaluate the determinant for the following matrix: [- 5 1 1 4] A. –21 B. 19 C. –17 D. –24

Mathematics
2 answers:
mariarad [96]3 years ago
6 0

Answer:

-21

Step-by-step explanation:

I assume this is a 2 by 2 matrix.

\begin{bmatrix} -5 & 1\\1 & 4 \end{bmatrix}

determinant = -5 * 4  - 1 * 1 = -20 - 1 = -21

lara [203]3 years ago
6 0

Answer:

Option A.

Step-by-step explanation:

If a matrix is defined as

A=\begin{bmatrix}a\:&\:b\:\\ c\:&\:d\:\end{bmatrix}\:

then its determinant is

|A|=\:ad-bc

Let the given matrix is

B=\begin{bmatrix}-5&1\\ \:1&4\end{bmatrix}

We need to find the determinant for the given matrix.

Using the above formula we get

|B|=\left(-5\right)\cdot \:4-1\cdot \:1

|B|=-21

The determinant of given matrix is -21. Therefore, the correct option is A.

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4−(2x+4)=5 solve for x
makvit [3.9K]

Answer:

-5/2

Step-by-step explanation:

4-2x-4=5

-2x=5+4-4

-2x=5

x=-5/2/-2 1/2/ -2.5

8 0
1 year ago
Is this solution true <br> 2<br> — 36+h=8,when h=4<br> 6
Crazy boy [7]

Answer:

yes

Step-by-step explanation:

8 0
3 years ago
Edgar collects sports T-shirt.He has 16 socoT-shirts and 24 rugby T- shirts. He has 12 hats. Haw many T -shirt does Edgar have i
Marianna [84]
Edgar has 40 shirts. The equation you could write is 24+16=40.  Hope this helps!!
7 0
3 years ago
Explain why it is correct to claim that A/B &lt; B/A is true.
PolarNik [594]

Answer:

Think of it as a small number and a large number. If the large number is being divided by a small number, of course that would be larger than the small number being divided by the large number.

3 0
2 years ago
What two numbers add to 8 and multiply to 28
anygoal [31]
Find numbers that multiply to 28 and add them to see if they add to 8
28=
1 and 28=29 not 8
2 and 14=16 not 8
4 and 7=11 not 8
that's it'
no 2 numbers
we must use quadratic formula

x+y=8
xy=28

x+y=8
subtract x fromb oths ides
y=8-x
subsitute
x(8-x)=28
distribute
8x-x^2=28
add x^2 to both sides
8x=28+x^2
subtract 8x
x^2-8x+28=0

if you have
ax^2+bx+c=0 then x=\frac{-b+/- \sqrt{b^{2}-4ac} }{2a}
so if we have
1x^2-8+28=0 then
a=1
b=-8
c=28

x=\frac{-(-8)+/- \sqrt{(-8)^{2}-4(1)(28)} }{2(1)}
x=\frac{8+/- \sqrt{-48} }{2}
x=\frac{8+/- 4 \sqrt{-3} }{2}
x=4+/- 2 \sqrt{-3}
x=4+/- 2i \sqrt{3}
there are no real numbers that satisfy this
5 0
3 years ago
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