Which of the following statements are true?

Mathematics

1 answer:

lana [24]3 years ago

7 0

Answer:

B, C, E, & F

Step-by-step explanation:

Option A is incorrect because the equation Ax = b is referred to as a matrix equation, not a vector equation.

Option B is correct. If Ax = b has a solution, vector b will a linear combination of columns of matrix A.

Option C is correct. In a matrix equation, product Ax when defined, is a sum of products.

Option D is incorrect. If an augmented matrix [Ab] had a pivot position in every row, there could be a pivot in the last column which would make it inconsistent.

Option E is correct. If the columns of an m×n matrix A span $R^m$ , then the equation Ax=b is consistent for each b in

Option F is correct. IfA is an m x n matrix whose columns do not span, then the equation Ax = b is inconsistent for some b in $R^m$

Options B, C, E, and F are correct.

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Is 10/3 equivalent to 70/21

slava [35]

Yes because if you divide both, the numerator and denominator, by 7 it would simplify to 10/3, so they are equivalent.

7 0

3 years ago

The length of a rectangle is 5 in longer than its width. if the perimeter of the rectangle is 34 in , find its area.

shtirl [24]

Area(A)= lw
perimeter(P)= 2w+2l= 34in
length(l)= w+5
width(w)= w

1/2 P= 17= w+l
17= w+(w+5)
17= 2w+5
2w=12
w=6 in

l= w+5
l= 6+5
l=11 in

A= lw= (11)(6)= 66in²

Answer:
Area of rectangle is 66 in².

3 0

3 years ago

1. (a) Solve the differential equation (x + 1)Dy/dx= xy, = given that y = 2 when x = 0. (b) Find the area between the two curves

erastova [34]

(a) The differential equation is separable, so we separate the variables and integrate:

$(x+1)\dfrac{dy}{dx} = xy \implies \dfrac{dy}y = \dfrac x{x+1} \, dx = \left(1-\dfrac1{x+1}\right) \, dx$

$\displaystyle \frac{dy}y = \int \left(1-\frac1{x+1}\right) \, dx$

$\ln|y| = x - \ln|x+1| + C$

When x = 0, we have y = 2, so we solve for the constant C :

$\ln|2| = 0 - \ln|0 + 1| + C \implies C = \ln(2)$

Then the particular solution to the DE is

$\ln|y| = x - \ln|x+1| + \ln(2)$

We can go on to solve explicitly for y in terms of x :

$e^{\ln|y|} = e^{x - \ln|x+1| + \ln(2)} \implies \boxed{y = \dfrac{2e^x}{x+1}}$

(b) The curves y = x² and y = 2x - x² intersect for

$x^2 = 2x - x^2 \implies 2x^2 - 2x = 2x (x - 1) = 0 \implies x = 0 \text{ or } x = 1$

and the bounded region is the set

$\left\{(x,y) ~:~ 0 \le x \le 1 \text{ and } x^2 \le y \le 2x - x^2\right\}$

The area of this region is

$\displaystyle \int_0^1 ((2x-x^2)-x^2) \, dx = 2 \int_0^1 (x-x^2) \, dx = 2 \left(\frac{x^2}2 - \frac{x^3}3\right)\bigg|_0^1 = 2\left(\frac12 - \frac13\right) = \boxed{\frac13}$