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3 years ago

What is the line of symmetry

Mathematics

1 answer:

Lerok [7]3 years ago

6 0

Answer:

is the imaginary line where you could fold the image and have both halves match exactly.

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Can someone please help me with this? Thank you <33

Mekhanik [1.2K]

Answer:

Step-by-step explanation:

4 0

3 years ago

Determine the side length.

White raven [17]

Answer:

Step-by-step explanation:

The given shape is a right angled triangle.

Here,

Hypotenuse = 2√3

Another side = √3

Let the missing side length be " x ".

By Hypotenuse theorem,

x² + ( √3 )² = ( 2√3 )²

x² + 3 = 12

x² = 12 - 3

x² = 9

x² = 3²

x = 3

Therefore, the side length is 3.

8 0

2 years ago

I really don't understand this I need help

Salsk061 [2.6K]

In looking at it I would guess it is asking which of the points is on the line? To answer that you would have to take each point an plug it into the equation and see if it is true.

3x - 7 = y

A(9,20)
3(9) - 7 = 20
27 - 7 = 20
20 = 20 True - this point is on the line

B(5,0)
3(5) - 7 = 0
15 - 7 = 0
8 = 0 False - this point is not on the line

C(3,2)
3(3) - 7 = 2
9 - 7 = 2
2 = 2 True - this point is on the line

I hope this helps

7 0

3 years ago

Find the midpoint of the segment with the following endpoints (-3,-9)(-7,-3)

barxatty [35]

Answer:

(-5,-6)

Step-by-step explanation:

Let(-3,-9)=(x1,y1)

(-7,-3)=(x2,y2)

Using midpoint formula,

Midpoint(x,y)=((x1+x2)/2,(y1+y2)/2)

=((-3+(-7))/2,(-9+(-3))/2))

=((-3-7)/2),(-9-3)/2))

=((-10/2),(-12/2))

=(-5,-6)

Thus midpoint(x,y)=(-5,-6)

8 0

4 years ago

Harry drove 2 hours on the freeway then decreased his speed by 20mph and drove 6 more hours. total trip was 352 what was his spe

poizon [28]

Recall your d = rt, distance = rate * time

we know the total trip was 352miles, so, let's say he was going for the first 2hrs at "r" speed, he went "d" miles, then he slowed down to "r - 20" and went 6 more hours, since the total trip is 352 miles, the 6hrs take the slack from 352 - d, and that's how many miles he covered on those 6hours

$\bf \begin{array}{lccclll}
&distance&rate&time\\
&-----&-----&-----\\
\textit{before slowing}&d&r&2\\
\textit{after slowing}&352-d&r-20&6
\end{array}
\\\\\\

\begin{cases}
\boxed{d}=2r\\
352-d=(r-20)6\\
----------\\
352-\boxed{2r}=(r-20)6
\end{cases}$

solve for "r".

7 0

4 years ago