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SVEN [57.7K]
3 years ago
14

What is 9 + 10? A. 21 B. 19 C. 12 D. 16

Mathematics
2 answers:
lisabon 2012 [21]3 years ago
6 0

answer = 19 Haha.......

MAXImum [283]3 years ago
4 0

19

\mathtt \red{Hope \:  it  \: helps \:  uhh..}

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NEED HELP WITH NUMBER 4!!!?
Gnoma [55]

on the number line the numbers to the left have least value

4 0
3 years ago
In the figure below, ∠t= 133° . Find the measures of all remaining angle.
barxatty [35]

t is vertical to s because they share a point and are directly across from each other. vertical angles are congruent. therefore s=113 degrees

t and p are corresponding angles, which means they are congruent. this means p is also 113 degrees.

s and o are corresponding angles, so o is also 113 degrees.

t and u are supplementary. they share a straight line, so they add up to 180. you can find u by subtracting 113 from 180. 180-113= 67, so u is 67 degrees.

u and r are vertical angles, so r is 67 degrees.

r and m are corresponding, as well as u and q. therefore m and q both equal 67 degrees.

hope this helped :)

5 0
2 years ago
Hi does anyone know what m+20=6, please show work with it
erik [133]
So your equation is m + 20 = 6

The first think you do is subtract 20 from both sides in order to get “m” alone, since that’s the variable you’re solving for. So what you’ll you have next is
m + 20 - 20 = 6-20

Since 20 subtracted by itself is zero, you’ll be left with
m + 0 = -14

Or more simply

m = -14
8 0
3 years ago
Consider the following problem: a farmer with 750 feet of fencing wants to enclose a rectangular area and then divide it into fo
nlexa [21]

Answer:

14,062.5ft^2

Step-by-step explanation:

A farmer has a given length of fencing. Since this fencing is used to enclose a rectangular area, before it is then divided it into four pens using the same provided fencing material. This will definitely lead to five lengths of fencing which come along one side of the rectangle to form four pens.

Let us assume that the length of the each of the five sides be xft.

Then let the length of each of the two sides in the be yft.

Since the total length of fencing available is 750 feet.

Then we can have an equation of such:

5x+2y=750

Making y the subject

2y = 750-5x

y = 750-5x/2. ........1

The area of the rectangular region is given by:

A = xy. ......... 2

Combining (1) in (2) to find the area

A(x)= x. (750−5x)/2

=750x−5x^2/2 ........... (*)

Going by the means of first diffential calculus.

A(x)=750x−5x^2/2

A'(x) = 750-10x/2

Setting A'(x) = 0

750-10x/2 = 0

750-10x = 2×0

750-10x = 0

750= 10x

x = 750/10

x = 75

Now that we have x = 75

We can now substitute into equation (*)

A(x) = 750×75-5×75^2/2

= (56250-28125)/2

= 28126/2

= 14062.5

Hence the rectangular region of maximum area is

14062.5ft^2

5 0
3 years ago
Randy is walking home from school. According to the diagram above, what is his total distance from school to home? Show your wor
In-s [12.5K]

Answer:

First, when he walks, we can see in the image that between the school and his house he must walk 4 times a distance of 0.5km, so this is a total of 4¨*0.5km = 2km.

Then he needs to walk 2km.

Now if he has a jet-pack, he can ignore the buildings and just take the shorter path, here we can draw a triangle rectangle, in such a way that the hypotenuse of this triangle is the distance between the home and the school.

One of the catheti is the vertical distance (two blocks of 0.5km, so this catheti has a length of 2*0.5km = 1km), and the other one is the horizontal distance, also 1km.

The actual distance of this path is given by the Pythagorean's theorem:

A^2 + B^2 = H^2

Where A and B are the cathetus, and H is the hypotenuse, then:

H^2 = 1km^2 + 1km^2

H = (√2)km = 1.41km.

Now, in the case that he has a jet-pack, he can actually go to the school using this hypotenuse line as his path, in this case the distance and the displacement would be the same.

This is because the definitions of distance and displacement are:

Distance: "how much ground an object has covered"

Displacement: "Difference between the final position and the initial position"

When he walks, the distance is 2km and the displacement is 1.41km , but when he uses the jet pack, the distance is equal to the displacement, both are 1.41km.

7 0
4 years ago
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