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Svetlanka [38]
3 years ago
10

If 80 inches is equivalent to 44 inches, then 28 inches would be equivalent to what?

Mathematics
1 answer:
harina [27]3 years ago
4 0

Answer:

Well no cheating not from me

Step-by-step explanation:

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Which segment is parallel to AD
Alex73 [517]

Answer:

BC

Parallel: Two sides or lines are parallel if they are lines that are always the same distance from each other and will never intersect or touch      

8 0
3 years ago
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Javion drank 7/8 quart of juice in two days. He drank 1/4 quart of juice yesterday. How much did he drink today?
o-na [289]

Step-by-step explanation:

7/8 - 1/4 = 5/8

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4 0
2 years ago
The monthly revenue, f(x) for the Sweet Treats Ice Cream truck
irakobra [83]

Answer:

The vertex is (-4, -52), and it represents the minimum for the function.

Step-by-step explanation:

3 0
3 years ago
Graph triangle RST with vertices R(3, 7), S(-5, -2), and T(3, -5) and its image after a reflection over x = -3.​
kirill115 [55]

Given:

The vertices of a triangle are R(3, 7), S(-5, -2), and T(3, -5).

To find:

The vertices of the triangle after a reflection over x = -3 and plot the triangle and its image on the graph.

Solution:

If a figure reflected across the line x=a, then

(x,y)\to (-(x-a)+a,y)

(x,y)\to (-x+a+a,y)

(x,y)\to (2a-x,y)

The triangle after a reflection over x = -3. So, the rule of reflection is

(x,y)\to (2(-3)-x,y)

(x,y)\to (-6-x,y)

The vertices of triangle after reflection are

R(3,7)\to R'(-6-3,7)

R(3,7)\to R'(-9,7)

Similarly,

S(-5,-2)\to S'(-6-(-5),-2)

S(-5,-2)\to S'(-6+5,-2)

S(-5,-2)\to S'(-1,-2)

And,

T(3,-5)\to T'(-6-3,-5)

T(3,-5)\to T'(-9,-5)

Therefore, the vertices of triangle after reflection over x=-3 are R'(-9,7), S'(-1,-2) and T'(-3,-5).

3 0
3 years ago
23 degrees 20’48 is the same as _degrees.
allochka39001 [22]

Answer:

\large\boxed{23^o20'48''=\left(23\dfrac{26}{75}\right)^o}

Step-by-step explanation:

\text{We know:}\\\\1^o=60'\to1'=\left(\dfrac{1}{60}\right)^o\\\\1'=60''\to1^o=(60)(60'')=3600''\to1''=\left(\dfrac{1}{3600}\right)^o\\\\23^o20'48''\\\\20'=\left(\dfrac{20}{60}\right)^o=\left(\dfrac{1}{3}\right)^o\\\\48''=\left(\dfrac{48}{3600}\right)^o=\left(\dfrac{1}{75}\right)^o\\\\\text{Therefore}\\\\23^o20'48''=23^o+20'+48''=23^o+\left(\dfrac{1}{3}\right)^o+\left(\dfrac{1}{75}\right)^o\\\\=\left(23+\dfrac{1\cdot25}{3\cdot75}+\dfrac{1}{75}\right)^o=\left(23+\dfrac{25}{75}+\dfrac{1}{75}\right)^o=\left(23\dfrac{26}{75}\right)^o

4 0
3 years ago
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