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3 years ago

Help Me With This Please #Geometry

Mathematics

1 answer:

arsen [322]3 years ago

7 0

First, you find the area of one triangle.

Area of Triangle= 1/2 base x height

here we plug in

(1/2)5 x 7
2.5 x 7
17.5

So 17.5 is the area of one triangle, you have 4
so we multiply 17.5 x 4= 70

Now we need to find the area of the base which is a square so the area = length x width or side x side

plug in 5 x 5= 25

so we add everything up 70+25= 95

Answer is choice c 95 ft sq.

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Point slope and slope intercept

tino4ka555 [31]

Answer:

Step-by-step explanation:

Observe what happens to y if we move from x = 1 to x = 9: y decreases by 2. Thus, the slope of the line in question is m = rise / run = -2/8, or -1/4.

Writing the equation in point-slope form, y - k = m(x - h), we get:

y - 7 = (-1/4)(x - 1) (point-slope form)

Solving for y results in the slope-intercept form:

y = 7 - x/4 + 1, or

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Answer:

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4 years ago

25. Paulo's family arrived at the reunion at

Rom4ik [11]

Answer:

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Step-by-step explanation:

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3 years ago

The following describes a linear function?

Mrrafil [7]

Answer:

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Step-by-step explanation:

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3 years ago

(a) Show that a differentiable function f decreases most rapidly at x in the direction opposite the gradient vector, that is, in

Sophie [7]

Answer:

Step-by-step explanation:

$\text{Show that a differentiable function f decreases most rapidly at x in the }$

$\text{direction opposite the gradient vector, that is, in the direction of}$ $-\bigtriangledown f(x)$ $\text{. Let}\ \theta \ \text{be the angle between} \bigtriangledown f(x) \ \text{and unit vector u. Then } D_u f = \mathbf{|\bigtriangledown f| \ cos \ \theta }}$

$\text{Since the minimum value of} \ \ \mathbf{cos \ \theta} \ \ is \mathbf{-1} \ \text{occuring \ for \ 0} \le \ \theta \ < 2x, \\ \\ when \ \theta = \mathbf{\pi} , \text{the mnimum value of} \ D_uf \ is} -|\bigtriangledown f|, \text{occuring when the direction of u is } \\ \\ \ \mathbf{the \ opposite \ of} \ \text{the direction of } \ \bigtriangledown f (assuming \ \bigtriangledown f\ is \ not \ zero)$

b) $\text{From part A:}$

$If \ f(x,y) = x^4y -x^2y^2 \ \ decreases \ fastest \ at \ the \point \ (2,-5)\\ \\ F(x,y) = x^4y -x^2y^3 \\ \\ f_x = \dfrac{df}{dx}= \dfrac{d}{dx}(x^4y-x^2y^3) \\ \\ f_x = \dfrac{df}{dx}= y4x^3 -2y^3x \\ \\ For(2,-5) \\ \\ f_x = (-5)4(2)^3 -2(-5)^3(2) \\ \\ \mathbf{ f_x = 340}$

$However; f_y = \dfrac{df}{dy} = \dfrac{d}{dy}(x^4y - x^2y^3) \\ \\ f_y = x^4 -3x^2y^2 \\ \\ Now, for (2, -5)\\ \\f_y = (2)^4 -3(2)^2(-5)^2 \\ \\ f_y = -284$

$So; \bigtriangledown = < 340,-284> \text{this is the direction of fastest decrease}$

6 0

3 years ago