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Oduvanchick [21]

2 years ago

11

which expression is equivalent to 6^3 x 6 1/4?

1 answer:

soldi70 [24.7K]2 years ago

7 0

Answer:

number 3<em><u> </u></em><em><u>I</u></em><em><u> </u></em><em><u>think</u></em><em><u> </u></em><em><u>I'm</u></em><em><u> </u></em><em><u>not</u></em><em><u> </u></em><em><u>sure</u></em>

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How many times greater is the value of the 2 in 204,936 than the value of the 2 in 124,936

miss Akunina [59]

10 times 20,000 equals 200,000. So the answer is C, the value of the 2 is 10 times greater

3 0

2 years ago

Linda bought a cube jewelry box. The volume of the jewelry box is 64 cubic inches. What is the edge length of Linda’s jewelry bo

yawa3891 [41]

Answer:

The edge length will be equal to 4 inches.

Step-by-step explanation:

The volume of a cube = s * s * s (in which s represent side of the cube)

V = s * s * s

V = s³

64 = s³

64 = 4³

So edge length of the cube will be 4 inches.

7 0

2 years ago

Use Euler's method with step size 0.2 to estimate y(1), where y(x) is the solution of the initial-value problem y' = x2y − 1 2 y

irina [24]

Answer:

Therefore the value of y(1)= 0.9152.

Step-by-step explanation:

According to the Euler's method

y(x+h)≈ y(x) + hy'(x) ....(1)

Given that y(0) =3 and step size (h) = 0.2.

$y'(x)= x^2y(x)-\frac12y^2(x)$

Putting the value of y'(x) in equation (1)

$y(x+h)\approx y(x) +h(x^2y(x)-\frac12y^2(x))$

Substituting x =0 and h= 0.2

$y(0+0.2)\approx y(0)+0.2[0\times y(0)-\frac12 (y(0))^2]$

$\Rightarrow y(0.2)\approx 3+0.2[-\frac12 \times3]$ [∵ y(0) =3 ]

$\Rightarrow y(0.2)\approx 2.7$

Substituting x =0.2 and h= 0.2

$y(0.2+0.2)\approx y(0.2)+0.2[(0.2)^2\times y(0.2)-\frac12 (y(0.2))^2]$

$\Rightarrow y(0.4)\approx 2.7+0.2[(0.2)^2\times 2.7- \frac12(2.7)^2]$

$\Rightarrow y(0.4)\approx 1.9926$

Substituting x =0.4 and h= 0.2

$y(0.4+0.2)\approx y(0.4)+0.2[(0.4)^2\times y(0.4)-\frac12 (y(0.4))^2]$

$\Rightarrow y(0.6)\approx 1.9926+0.2[(0.4)^2\times 1.9926- \frac12(1.9926)^2]$

$\Rightarrow y(0.6)\approx 1.6593$

Substituting x =0.6 and h= 0.2

$y(0.6+0.2)\approx y(0.6)+0.2[(0.6)^2\times y(0.6)-\frac12 (y(0.6))^2]$

$\Rightarrow y(0.8)\approx 1.6593+0.2[(0.6)^2\times 1.6593- \frac12(1.6593)^2]$

$\Rightarrow y(0.6)\approx 0.8800$

Substituting x =0.8 and h= 0.2

$y(0.8+0.2)\approx y(0.8)+0.2[(0.8)^2\times y(0.8)-\frac12 (y(0.8))^2]$

$\Rightarrow y(1.0)\approx 0.8800+0.2[(0.8)^2\times 0.8800- \frac12(0.8800)^2]$

$\Rightarrow y(1.0)\approx 0.9152$

Therefore the value of y(1)= 0.9152.

4 0

2 years ago

Plllzzz im new and i neeed help

ella [17]

Answer:

4082

Step-by-step explanation:

Given

The composite object

Required

The volume

The object is a mix of a cone and a hemisphere

Such that:

<u>Cone</u>

$r = 10cm$ ---- radius (r = 20/2)

$h = 19cm$

<u>Hemisphere</u>

$r=10cm$

The volume of the cone is:

$V_1 = \frac{1}{3}\pi r^2h$

$V_1 = \frac{1}{3}\pi * 10^2 * 19$

$V_1 = \frac{1900}{3}\pi$

The volume of the hemisphere is:

$V_2 = \frac{2}{3}\pi r^3$

$V_2 = \frac{2}{3}\pi 10^3$

$V_2 = \frac{2000}{3}\pi$

So, the volume of the object is:

$V = V_1 + V_2$

$V = \frac{1900}{3}\pi + \frac{2000}{3}\pi$

$V = \frac{3900}{3}\pi$

$V = 1300\pi$

$V = 1300 * 3.14$

$V = 4082$

8 0

2 years ago

Please help in Math!? What value of y makes the equation true?

Zepler [3.9K]

$\frac{-7}{12} = \frac{y}{-36}$

First, simplify $\frac{y}{-36}$ to $- \frac{y}{36}$ / Your problem should look like: $- \frac{7}{12} = -\frac{y}{36}$
Second, multiply both sides by 36. / Your problem should look like: $-21=-y$
Third, multiply both sides by -1. / Your problem should look like: $21 = y$
Fourth, switch your sides. / Your problem should look like: $y = 21$

Answer: C) 21

5 0

2 years ago