Please 10 points for one question Please don’t answer if you don’t know

Positive numbers are located to the right of zero on the number line. True False

Trava [24]

$\text {Hi!}$

$\text {The Answer to this Problem Is:}$

$\fbox {True!}$

$\text {When we look at a number line with positive and negative numbers we see}$ $\text {that positives are to the right and negatives are to the left.}$

$\text {As you can see the number line on the image that all}$ $\text {the negative numbers are to the left and positive numbers are to the right.}$

$\text {\underline {Remember}}$

$\text {Positive to the Right.}$

$\text {Negative to the Left.}$

$\text {Best of Luck!}$

7 0

3 years ago

Read 2 more answers

In shop a crunchy peanut butter costs $325 for 250g

FinnZ [79.3K]

Answer:

shop b

Step-by-step explanation:

We can find for example how much 100 grams of peanut butter cost in each shop using a proportion

shop A 100 : 250 = x : 325

x = (325 * 100) / 250 = $130

shop B 100 : 340 = x : 408

x = (408 * 100)/ 340 = $120

We discovered that shop B is cheaper

5 0

3 years ago

If a=2,b=-1,c=-3 and d=-2.find the value of 7a2-3ac+d2

hram777 [196]

Answer:

42

Step-by-step explanation:

It's just a distribution property where the given value of each variables (a, b, c, d) are being distributed to the given equation:

7a2-3ac+d2

7(2)(2) - 3(2)(-3) + (-2)(2)

28 + 18 - 4 = 42

3 0

3 years ago

Solve for x 2x^2-8x+5=0

ss7ja [257]

X=7 over 8 or 0.875 is the answer

8 0

3 years ago

Please determine whether the set S = x^2 + 3x + 1, 2x^2 + x - 1, 4.c is a basis for P2. Please explain and show all work. It is

ohaa [14]

The vectors in $S$ form a basis of $P_2$ if they are mutually linearly independent and span $P_2$ .

To check for independence, we can compute the Wronskian determinant:

$\begin{vmatrix}x^2+3x+1&2x^2+x-1&4\\2x+3&4x+1&0\\2&4&0\end{vmatrix}=4\begin{vmatrix}2x+3&4x+1\\2&4\end{vmatrix}=40\neq0$

The determinant is non-zero, so the vectors are indeed independent.

To check if they span $P_2$ , you need to show that any vector in $P_2$ can be expressed as a linear combination of the vectors in $S$ . We can write an arbitrary vector in $P_2$ as

$p=ax^2+bx+c$

Then we need to show that there is always some choice of scalars $k_1,k_2,k_3$ such that

$k_1(x^2+3x+1)+k_2(2x^2+x-1)+k_34=p$

This is equivalent to solving

$(k_1+2k_2)x^2+(3k_1+k_2)x+(k_1-k_2+4k_3)=ax^2+bx+c$

or the system (in matrix form)

$\begin{bmatrix}1&1&0\\3&1&0\\1&-1&4\end{bmatrix}\begin{bmatrix}k_1\\k_2\\k_3\end{bmatrix}=\begin{bmatrix}a\\b\\c\end{bmatrix}$

This has a solution if the coefficient matrix on the left is invertible. It is, because

$\begin{vmatrix}1&1&0\\3&1&0\\1&-1&4\end{vmatrix}=4\begin{vmatrix}1&2\\3&1\end{vmatrix}=-20\neq0$

(that is, the coefficient matrix is not singular, so an inverse exists)

Compute the inverse any way you like; you should get

$\begin{bmatrix}1&1&0\\3&1&0\\1&-1&4\end{bmatrix}^{-1}=-\dfrac1{20}\begin{bmatrix}4&-8&0\\-12&4&0\\-4&3&-5\end{bmatrix}$

Then

$\begin{bmatrix}k_1\\k_2\\k_3\end{bmatrix}=\begin{bmatrix}1&1&0\\3&1&0\\1&-1&4\end{bmatrix}^{-1}\begin{bmatrix}a\\b\\c\end{bmatrix}$

$\implies k_1=\dfrac{2b-a}5,k_2=\dfrac{3a-b}5,k_3=\dfrac{4a-3b+5c}{20}$

A solution exists for any choice of $a,b,c$ , so the vectors in $S$ indeed span $P_2$ .

The vectors in $S$ are independent and span $P_2$ , so $S$ forms a basis of $P_2$ .

5 0

3 years ago