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Fudgin [204]
3 years ago
8

Add using a number line. +(-11)

Mathematics
1 answer:
Nikolay [14]3 years ago
4 0

Answer:

-11 is the anwer.

As it is present in left side, so -11 will be in left part of number line.

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Detailed solution is given below in the attached image

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Find the area of a triangle with a base length of 3 units and a height of 4 units.
svetlana [45]

Answer:

6 square units

Step-by-step explanation:

The area of a triangle can be calculated by multiplying the triangle's base by its height, and then dividing the result by 2. As an algebraic expression, this would be \frac{bh}{2}, where b is the triangle's base and h is the triangle's height. In this case, we know that b=3 and h=4. Therefore, the area of this triangle is \frac{3*4}{2} =\frac{12}{2} =6 square units. Hope this helps!

6 0
2 years ago
Read 2 more answers
A school administrator bought 6 desk sets. Each set consists of one table and one chair. He paid $48.50 for each chalr. If he pa
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The answer is $62.00

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3 years ago
Let n be a positive integer. We sample n numbers a1, a2,..., an from the set {1,...,n} uniformly at random, with replacement. We
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4 0
3 years ago
A motorboat takes 5 hours to travel 100mi going upstream. The return trip takes 2 hours going downstream. What is the rate of th
Temka [501]
The boat went 5hours upstream, let's say it has a "still water" speed rate of "b", it went 100miles... however, going upstream is going against the current, let's say the current has a speed rate of "c"

so, when the boat was going up, it wasn't really going "b" fast, it was going " b - c " fast, because the current was eroding speed from it

now, when coming down, the return trip, well the length is the same, so the distance is also 100miles, it only took 2hrs though, because, the boat wasn't coming down  "b" fast, it was coming down " b + c " fast, because the current was adding speed to it, so it came down quicker

now, recall your d = rt, distance = rate * time

\bf \begin{array}{lccclll}
&distance&rate&time\\
&-----&-----&-----\\
upstream&100&b-c&5\\
downstream&100&b+c&2
\end{array}
\\\\\\
\begin{cases}
100=(b-c)5\\
\qquad \frac{100}{5}=b-c\\
\qquad 20=b-c\\
\qquad 20+c=\boxed{b}\\
100=(b+c)2\\
\qquad 50=b+c\\
\qquad 50=\boxed{20+c}+c
\end{cases}

solve for "c", to see what's the current's speed

what's "b"?  well 20+c = b
4 0
3 years ago
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