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Sveta_85 [38]
3 years ago
9

syeda has finished 20% of her math assignment. The assignment includes 50 problems how many problems has syeda completed?

Mathematics
1 answer:
Nostrana [21]3 years ago
5 0

Answer:

10 problems

Step-by-step explanation:

This is the answer because:

1) We have to solve for what 20% of 50 is since she finished 20% of her test:

20% of 50 = 10

\frac{2}{10} × \frac{50}{1} = \frac{100}{10} = 10

Therefore, the answer is 10 problems.

Hope this helps! :D

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Which of the following points are more than 5 units from the point P(−2, −2)? Select all that apply. A A (2, 1) B B (4, −1) C C
netineya [11]

The distance between any 2 points P(a,b) and Q(c,d) in the coordinate plane, is given by the formula:<span>

<span> |PQ|= \sqrt{ (a-c)^{2} + (b-d)^{2}}</span></span>


Using this formula we calculate the distances |PA|, |PB|, |PC|, |PD| and |PE| and compare to 5.


|PA|= \sqrt{ (-2-2)^{2} + (-2-1)^{2}}= \sqrt{16+9}= \sqrt{25}=5

|PB|= \sqrt{ (-2-4)^{2} + (-2+1)^{2}}= \sqrt{36+1}= \sqrt{37} \approx 6

|PC|= \sqrt{ (-2-2)^{2} + (-2+3)^{2}}= \sqrt{16+1}= \sqrt{17}\approx4

|PD|= \sqrt{ (-2+6)^{2} + (-2+6)^{2}}= \sqrt{16+16}= \sqrt{32}\ \textgreater \  \sqrt{25}=5

|PE|= \sqrt{ (-2+5)^{2} + (-2-1)^{2}}= \sqrt{9+9}= \sqrt{16}=4


Answer: B and D





3 0
3 years ago
Maya has 5 over 6 cup of ice cream. How many 1 over 4 -cup servings are in 5 over 6 cup of ice cream?
Drupady [299]
The answer is c because it’s not a or d and maybe not b so ye c
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3 years ago
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Please help with this
klemol [59]

Answer:

(h+k)(2)=4+1+2-2=5

(h-k)(3)=9+1-3-2=3+1-1=9

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7 0
3 years ago
To hang lights up on his house.Garrett place is a 14 foot ladder 4 feet from the base of the house. How high up the house will t
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This seems like a right triangle problem.

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So, the equation is 4^{2} +x^{2} =14^{2}.

Solve for x:

16+x^{2}= 196

x^{2}= 196-16

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x = 6√5 feet


Hope this helps!

5 0
3 years ago
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According to the graph of H(w) below, what happens when w gets very large?
GREYUIT [131]

Answer:

B. H(w) gets very small.

Step-by-step explanation:

This should be the answer you're looking for

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