Please help fast this is the unit test

itzel noticed that the distance from her house to the library, which is 40 miles, was one fifth the distance from her house to h

Maksim231197 [3]

Answer:

200 miles

Step-by-step explanation:

40 x 5 = 200

5 0

3 years ago

a confectioner has 500 mint, 500 orange, and 500 strawberry flavored sweets. He wishes to make packets containing 10 mint, 5 ora

Readme [11.4K]

$Inside\ each\ packet\ he\ used\ the\ greatest\ amount\ of\ mint\ sweets,\\so\ we\ can\ focus\ only\ on\ it.\\\\We\ should\ divide\ all\ of\ mint\ sweets\ by\ one\ packet\ amount:\\\\500:10=50\\\\He\ can\ make\ 50\ packets\ of\ this\ type.$

8 0

3 years ago

12x^2-11x+2 sho work

Scrat [10]

Answer: (3x−2)(4x−1)

Step-by-step explanation:

Start factoring with -11x.

12x^(2)-11x+2

12x2+(−8−3)x+2

12x2+(−8x−3x)+2

12x2−8x−3x+2

Factor the highest common denominator of each group.

Group the first two terms and the last two.

(12x2−8x)+(−3x+2)

Factor the highest common denominator (MCD) of each group.

4x(3x−2)−(3x−2)

Factoring the highest common denominator 3x−2.

(3x−2)(4x−1)

4 0

3 years ago

What is the equation of the line whose y-Intercept is 3 and slope is 1?

Arada [10]

y = x + 3

as x increases by 1, y increases by 1 (Slope)

when x equals 0, y equals 3 (y-intercept)

5 0

3 years ago

Compute the line integral with respect to arc length of the function f(x, y, z) = xy2 along the parametrized curve that is the l

SIZIF [17.4K]

Answer:

$\displaystyle\frac{15\sqrt{3}}{4}-90\sqrt{146}$

Step-by-step explanation:

The line integral with respect to arc length of the function f(x, y, z) = xy2 along the parametrized curve that is the line segment from (1, 1, 1) to (2, 2, 2) followed by the line segment from (2, 2, 2) to (−9, 6, 5) equals the sum of the line integral of f along each path separately.

Let

$C_1,C_2$

be the two paths.

Recall that if we parametrize a path C as $(r_1(t),r_2(t),r_3(t))$ with the parameter t varying on some interval [a,b], then the line integral with respect to arc length of a function f is

$\displaystyle\int_{C}f(x,y,z)ds=\displaystyle\int_{a}^{b}f(r_1,r_2,r_3)\sqrt{(r'_1)^2+(r'_2)^2+(r'_3)^2}dt$

Given any two points P, Q we can parametrize the line segment from P to Q as

r(t) = tQ + (1-t)P with 0≤ t≤ 1

The parametrization of the line segment from (1,1,1) to (2,2,2) is

r(t) = t(2,2,2) + (1-t)(1,1,1) = (1+t, 1+t, 1+t)

r'(t) = (1,1,1)

and

$\displaystyle\int_{C_1}f(x,y,z)ds=\displaystyle\int_{0}^{1}f(1+t,1+t,1+t)\sqrt{3}dt=\\\\=\sqrt{3}\displaystyle\int_{0}^{1}(1+t)(1+t)^2dt=\sqrt{3}\displaystyle\int_{0}^{1}(1+t)^3dt=\displaystyle\frac{15\sqrt{3}}{4}$

The parametrization of the line segment from (2,2,2) to

(-9,6,5) is

r(t) = t(-9,6,5) + (1-t)(2,2,2) = (2-11t, 2+4t, 2+3t)

r'(t) = (-11,4,3)

and

$\displaystyle\int_{C_2}f(x,y,z)ds=\displaystyle\int_{0}^{1}f(2-11t,2+4t,2+3t)\sqrt{146}dt=\\\\=\sqrt{146}\displaystyle\int_{0}^{1}(2-11t)(2+4t)^2dt=-90\sqrt{146}$

Hence

$\displaystyle\int_{C}f(x,y,z)ds=\displaystyle\int_{C_1}f(x,y,z)ds+\displaystyle\int_{C_2}f(x,y,z)ds=\\\\=\boxed{\displaystyle\frac{15\sqrt{3}}{4}-90\sqrt{146}}$

8 0

3 years ago