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Anestetic [448]
3 years ago
11

A coordinate plane. Which expression could help you find the distance between (10, 4) and (–6, 4)?

Mathematics
1 answer:
nirvana33 [79]3 years ago
3 0

Answer:

d=16

Step-by-step explanation:

d: distance (10,4),(-6,4)

d=√(x2-x1)^2+(y2-y1)^2

d=√(-6-10)^2+(4-4)^2

d=√16^2

d=16

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Please someone help me
Yuki888 [10]

Answer: H

Step-by-step explanation:

We can see this rectangular prism will have side lengths of 5, 3, 7.

So we can use the fact Surface Area = 2(lh+wh+lw) to get:

2(15 + 35 + 21) = 2(71) = 142

5 0
2 years ago
(02.02)A train can travel 744 miles in 3 hours. What is the unit rate that this train is traveling per hour? _____ miles per hou
Fynjy0 [20]

If the train travels 744 miles in 3 hours then to determine the unit rate the train is traveling per hour you have to divide the miles by the hours, in this case (744 miles)/(3 hours). The final result concludes as 248 miles per hour.

7 0
3 years ago
1. Express <img src="https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bx%282x%2B3%29%20%7D" id="TexFormula1" title="\frac{1}{x(2x+3) }" a
katovenus [111]

1. Let a and b be coefficients such that

\dfrac1{x(2x+3)} = \dfrac ax + \dfrac b{2x+3}

Combining the fractions on the right gives

\dfrac1{x(2x+3)} = \dfrac{a(2x+3) + bx}{x(2x+3)}

\implies 1 = (2a+b)x + 3a

\implies \begin{cases}3a=1 \\ 2a+b=0\end{cases} \implies a=\dfrac13, b = -\dfrac23

so that

\dfrac1{x(2x+3)} = \boxed{\dfrac13 \left(\dfrac1x - \dfrac2{2x+3}\right)}

2. a. The given ODE is separable as

x(2x+3) \dfrac{dy}dx} = y \implies \dfrac{dy}y = \dfrac{dx}{x(2x+3)}

Using the result of part (1), integrating both sides gives

\ln|y| = \dfrac13 \left(\ln|x| - \ln|2x+3|\right) + C

Given that y = 1 when x = 1, we find

\ln|1| = \dfrac13 \left(\ln|1| - \ln|5|\right) + C \implies C = \dfrac13\ln(5)

so the particular solution to the ODE is

\ln|y| = \dfrac13 \left(\ln|x| - \ln|2x+3|\right) + \dfrac13\ln(5)

We can solve this explicitly for y :

\ln|y| = \dfrac13 \left(\ln|x| - \ln|2x+3| + \ln(5)\right)

\ln|y| = \dfrac13 \ln\left|\dfrac{5x}{2x+3}\right|

\ln|y| = \ln\left|\sqrt[3]{\dfrac{5x}{2x+3}}\right|

\boxed{y = \sqrt[3]{\dfrac{5x}{2x+3}}}

2. b. When x = 9, we get

y = \sqrt[3]{\dfrac{45}{21}} = \sqrt[3]{\dfrac{15}7} \approx \boxed{1.29}

8 0
2 years ago
Find the lateral area of the regular pyramid<br><br> L.A. =
Arisa [49]

Answer:

FOR REGULAR PYRAMID with those dimension.

L.A = 96

FOR HEXAGONAL PYRAMID with those dimension

L.A = 171.71

Step-by-step explanation:

Please the question asked for L.A of a REGULAR PYRAMID, but the figure is a HEXAGON PYRAMID.

Hence I solved for both:

FOR REGULAR PYRAMID

Lateral Area (L.A) = 1/2* p * l

Where p = Perimeter of base

P = 4s

P = 4 * 6

P = 24cm

l = slanted height

l = 8cm

L.A = 1/2 * 24 * 8

L.A = 1/2 ( 192)

L.A = 96cm ^ 2

FOR AN HEXAGONAL PYRAMID

Lateral Area = 3a √ h^2 + (3a^2) / 4

Where:

a = Base Edge = 6

h = Height = 8

L.A = 3*6 √ 8^2 + ( 3*6^2) / 4

L.A = 18 √ 64 + ( 3 * 36) / 4

L.A = 18 √ 64 + 108/4

L.A = 18 √ 64+27

L.A = 18 √ 91

L.A = 18 * 9.539

L.A = 171.71

8 0
3 years ago
Help please! and explain if you can
EleoNora [17]

Answer:

4) y = 9

5) x = -1

6) y = -5x − 9

Step-by-step explanation:

4) x = 3 is a vertical line with an x-intercept of (3, 0).  It has an undefined slope.

A perpendicular line to that will be y = a, which is a horizontal line with a y-intercept of (0, a) and a slope of 0.  Since this line passes through (3, 9), the equation of the line must be y = 9.

5) The y-axis is x = 0.  A parallel line that includes the point (-1, -2) is x = -1.

6) Parallel lines have the same slope, so:

y = -5x + b

To find the value of b (the y-intercept), plug in the point (-2, 1):

1 = -5(-2) + b

1 = 10 + b

b = -9

y = -5x − 9

3 0
3 years ago
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