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2 years ago

Help me pleaseee!! Would appreciate ^^ BRAINLIST, thanks and points*

Mathematics

1 answer:

harina [27]2 years ago

5 0

Answer:

1) 4900 is square

2) The greatest common factor 30 and 126 is 6. The GCF of 126 and 39204 is 18

3) i) 44100 ii) 48024900

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Evan received a box in the mail that was 37 inches long, 19 inches wide, and 15 inches tall. What is the volume of the box.

saul85 [17]

$v=lwh\\v=37 \times 19 \times 15\\v=10545$ inches

8 0

3 years ago

For what value of x does 64^3x = 512^2x+12

aleksandr82 [10.1K]

Answer:

<h2>no solution</h2>

Step-by-step explanation:

$Use:\\\\(a^n)^m=a^{nm}\\--------------\\64=2^6\\512=2^9\\\\\text{therefore}\\\\64^{3x}=512^{2x+12}\\\\(2^6)^{3x}=(2^9)^{2x+12}\\\\2^{(6)(3x)}=2^{(9)(2x+12)}\iff6(3x)=9(2x+12)\\\\18x=9(2x+12)\qquad\text{divide both sides by 9}\\\\\dfrac{18x}{9}=\dfrac{9(2x+12)}{9}\\\\2x=2x+12\qquad\text{subtract 2x from both sides}\\2x-2x=2x-2x+12\\\\0=12\qquad\bold{FALSE}$

3 0

3 years ago

Does anyone know the area?

Anestetic [448]

Answer:

192 $cm^{2}$

Step-by-step explanation:

We use the same area equation for a rectangle

A = b * l

A = 12 * 16

A = 192

5 0

3 years ago

Read 2 more answers

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

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