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3 years ago

WHY IS STEPHAN WRONG AND WHAT IS THE CORRECT ANSWER!!!!! PLEASE EXPLAIN ND WILL GIVE BRAINLIEST

Mathematics

1 answer:

Sliva [168]3 years ago

3 0

Answer:

IM NOT SURE I have not learned this yet... but i think its... B

Step-by-step explanation:

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What is the area of the figure?

zalisa [80]

Answer: 57 ft²

Step-by-step explanation:

Find the areas of the two individual rectangles and then add that up to find the area of the figure:

Rectangle A: Rectangle B:

= Length * Width = 11.5 * 3

= 7.5 * 3 = 34.5ft²

= 22.5 ft²

Area of figure:

= 22.5 + 34.5

= 57 ft²

3 0

3 years ago

Read 2 more answers

Determine all real values of p such that the set of all linear combination of u = (3, p) and v = (1, 2) is all of R2. Justify yo

Rama09 [41]

Answer:

p ∈ IR - {6}

Step-by-step explanation:

The set of all linear combination of two vectors ''u'' and ''v'' that belong to R2

is all R2 ⇔

$u\neq 0_{R2}$

$v\neq 0_{R2}$

And also u and v must be linearly independent.

In order to achieve the final condition, we can make a matrix that belongs to $R^{2x2}$ using the vectors ''u'' and ''v'' to form its columns, and next calculate the determinant. Finally, we will need that this determinant must be different to zero.

Let's make the matrix :

$A=\left[\begin{array}{cc}3&1&p&2\end{array}\right]$

We used the first vector ''u'' as the first column of the matrix A

We used the second vector ''v'' as the second column of the matrix A

The determinant of the matrix ''A'' is

$Det(A)=6-p$

We need this determinant to be different to zero

$6-p\neq 0$

$p\neq 6$

The only restriction in order to the set of all linear combination of ''u'' and ''v'' to be R2 is that $p\neq 6$

We can write : p ∈ IR - {6}

Notice that is $p=6$ ⇒

$u=(3,6)$

$v=(1,2)$

If we write $3v=3(1,2)=(3,6)=u$ , the vectors ''u'' and ''v'' wouldn't be linearly independent and therefore the set of all linear combination of ''u'' and ''b'' wouldn't be R2.