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2 years ago

Describe and correct the error a student made in finding the midpoint of CD with C(4,5) and D(1,4).

Mathematics

1 answer:

Andru [333]2 years ago

6 0

Using the given information, the midpoint of CD is (2.5, 4.5)

<h3>Midpoint of a line </h3>

From the question, we are to determine the midpoint of line CD

The midpoint of a line is given by the formula,

[(x₁ + x₂)/2 , (y₁ + y₂)/2]

From the given information, we are to find the midpoint of line CD with coordinates C(4,5) and D(1,4)

∴ x₁ = 4

x₂ = 1

y₁ = 5

y₂ = 4

Putting the parameters into the formula for midpoint, we get

[(x₁ + x₂)/2 , (y₁ + y₂)/2]

[(4 + 1)/2 , (5 + 4)/2]

(5/2 , 9/2)

(2.5, 4.5)

Hence, the midpoint of CD is (2.5, 4.5)

Learn more on Midpoint of a line here: brainly.com/question/896396

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Step-by-step explanation:

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3 years ago

A trough has ends shaped like isosceles triangles, with width 5 m and height 7 m, and the trough is 12 m long. Water is being pu

Svet_ta [14]

Answer:

$\dfrac{dh}{dt}=21 \text{m/min}$

the rate of change of height when the water is 1 meter deep is 21 m/min

Step-by-step explanation:

First we need to find the volume of the trough given its dimensions and shape: (it has a prism shape so we can directly use that formula OR we can multiply the area of its triangular face with the length of the trough)

$V = \dfrac{1}{2}(bh)\times L$

here L is a constant since that won't change as the water is being filled in the trough, however 'b' and 'h' will be changing. The equation has two independent variables and we need to convert this equation so it is only dependent on 'h' (the height of the water).

As its an isosceles triangle we can find a relationship between b and h. the ratio between the b and h will be always be the same:

$\dfrac{b}{h} = \dfrac{5}{7}$

$b=\dfrac{5}{7}h$ this can be substituted back in the volume equation

$V = \dfrac{5}{14}h^2L$

the rate of the water flowing in is:

$\dfrac{dV}{dt} = 6$

The question is asking for the rate of change of height (m/min) hence that can be denoted as: $\frac{dh}{dt}$

Using the chainrule:

$\dfrac{dh}{dt}=\dfrac{dh}{dV}\times \dfrac{dV}{dt}$

the only thing missing in this equation is dh/dV which can be easily obtained by differentiating the volume equation with respect to h

$V = \dfrac{5}{14}h^2L$

$\dfrac{dV}{dh} = \dfrac{5}{7}hL$

reciprocating

$\dfrac{dh}{dV} = \dfrac{7}{5hL}$

plugging everything in the chain rule equation:

$\dfrac{dh}{dt}=\dfrac{dh}{dV}\times \dfrac{dV}{dt}$

$\dfrac{dh}{dt}=\dfrac{7}{5hL}\times 6$

$\dfrac{dh}{dt}=\dfrac{42}{5hL}$

L = 12, and h = 1 (when the water is 1m deep)

$\dfrac{dh}{dt}=\dfrac{42}{5(1)(12)}$

$\dfrac{dh}{dt}=21 \text{m/min}$

the rate of change of height when the water is 1 meter deep is 21 m/min

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4 years ago

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2 years ago

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Find the distance between each pair of points. Round your answer to the nearest tenth, if necessary. Hint: Use the Pythagorean T

Paul [167]

The distance between two points on the plane is given by the formula below

$\begin{gathered} A=(x_1,y_1),B=(x_2,y_2) \\ \Rightarrow d(A,B)=\sqrt[]{(x_1-x_2)^2+(y_1-y_2)^2} \end{gathered}$

Therefore, in our case,

$A=(-1,-3),B=(5,2)$

Thus,

$\begin{gathered} \Rightarrow d(A,B)=\sqrt[]{(-1-5)^2+(-3-2)^2}=\sqrt[]{6^2+5^2}=\sqrt[]{36+25}=\sqrt[]{61} \\ \Rightarrow d(A,B)=\sqrt[]{61} \end{gathered}$

Therefore, the answer is sqrt(61)

In general,

$-(-n)=n$

Remember that

$-n=(-1)\cdot n$

Therefore,

$\begin{gathered} a-(-n)=a+(-1)(-n)=a+(-1)(-1\cdot n)=a+(-1)^2\cdot n=a+1\cdot n=a+n \\ \Rightarrow a-(-n)=a+n \end{gathered}$

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2 years ago

Need help.Pleaseeeee

Marina CMI [18]

Answer:

The first one and the last one

Step-by-step explanation:

To sub x>0 and x<0 into every functions.

Then, u will get some function that y doesn't have a negative value or positive value.

Then, these function doesn't fulfill the range.

Then, u will get the answer.

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3 years ago

Describe and correct the error a student made in finding the midpoint of CD with ​C(​4,5) and ​D(​1,​4).

Describe and correct the error a student made in finding the midpoint of CD with C(4,5) and D(1,4).