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2 years ago

15

Which equation can be used to find the answer? Christina climbs to the top of Mount Graham in Arizona. On the first day, she cli

mbs to an elevation of 1029 ft. The elevation of Mount Graham is 10,724 ft. How many total feet of Mount Graham’s elevation does Christina climb on the other days?
A. e + 1029 = 10,724 e = 9695 9695 ft
B. e = 1029 · 2 e = 2058 2058 ft
C. e = 10,724 ÷ 2 e = 5362 5362 ft
D. e – 1029 = 10,724 e = 11,753 11,753 ft

1 answer:

ZanzabumX [31]2 years ago

5 0

It would be A because 10,724 minus 1029 equals 9,695

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I am not sure if I am correct.
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The Square sand pit in park lot has an area of 16 ft. There is a rope along one side of the pit. How long is the rope?

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3 years ago

Draw at least six different sized rectangles that have an area of 64 square units

olga_2 [115]

Let

x------> the length of the rectangle

y------> the width of the rectangle

we know that

The area of a rectangle is equal to

$A=x*y$

$A=64\ units^{2}$

so

$x*y=64$ --------> equation 1

let's assume different values of x to get the different values of y

<u>case 1)</u> For x=64 units

substitute in the equation 1

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$y=64/x$

$y=64/64$

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the dimensions of the rectangle are 64 units x 1 unit

see the draw in the attached figure N 1

<u>case 2)</u> For x=32 units

substitute in the equation 1

$x*y=64$

$y=64/x$

$y=64/32$

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the dimensions of the rectangle are 32 units x 2 units

see the draw in the attached figure N 2

<u>case 3)</u> For x=16 units

substitute in the equation 1

$x*y=64$

$y=64/x$

$y=64/16$

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see the draw in the attached figure N 3

<u>case 4)</u> For x=30 units

substitute in the equation 1

$x*y=64$

$y=64/x$

$y=64/30$

$y=\frac{32}{15} =2\frac{2}{15} \ units$

the dimensions of the rectangle are 30 units x 2 (2/15) units

see the draw in the attached figure N 4

<u>case 5)</u> For x=40 units

substitute in the equation 1

$x*y=64$

$y=64/x$

$y=64/40$

$y=1.60\ units$

the dimensions of the rectangle are 40 units x 1.60 units

see the draw in the attached figure N 5

<u>case 6)</u> For x=60 units

substitute in the equation 1

$x*y=64$

$y=64/x$

$y=64/60$

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3 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

3 years ago

Please prove the following trigonometric identity.<br><br>

Brums [2.3K]

Step-by-step explanation:

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∵ 1-cos²x = sin²x

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∵tanx = sinx×cosx

∴ sinx/ cosx × cosx = sinx

Since, LHS = RHS proved ___

3 0

2 years ago