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wolverine [178]
3 years ago
9

CAN SOMEONE HELP ME WITH THIS PROBLEM? WILL GIVE BRAINLEIST!!!

Mathematics
2 answers:
djyliett [7]3 years ago
6 0

the answer is c) 59.5

Ludmilka [50]3 years ago
6 0

Answer:

D because 84 of 50 is 168...

solutions:

84:50*100 =

(84*100):50 =

8400:50 = 168

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What is equation created from the sqrt(x+2)+2))^2
Naddika [18.5K]

Answer:

x + 4√(x + 2) + 6

Step-by-step explanation:

All we need to do is expand by using FOIL (First, Outside, Inside, Last):

x + 2 + 2√(x + 2) + 2√(x + 2) + 4

Then we combine like terms to get our final answer:

x + 2 + 4√(x + 2) + 4

x + 4√(x + 2) + 6

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3 years ago
Solve:<br><br> Picture Below
kkurt [141]

Answer: 8/3 or 2 2/3

4 0
2 years ago
Given: ABCD ∥gram, BK ⊥ AD , AB ⊥ BD AB=6, AK=3 Find: m∠A, BK, AABCD
jenyasd209 [6]

1. Consider right triangle ABK. The hypotenuse AB is 6 un. and the leg AK is 3 un. Since the leg is half of the hypotenuse, then the opposite to the leg angle is 30°. This means that m∠ABK=30°. Then m∠BAK=90°-30°=60°.

2. By the Pythagorean theorem,

AB^2=AK^2+BK^2,\\\\6^2=3^2+BK^2,\\\\BK^2=36-9,\\\\BK^2=27,\\\\BK=3\sqrt{3}\ un.

3. Consider right triangle ABD. In this triangle AD is hypotenuse and m∠ADB=90°-m∠BAD=90°-60°=30°. Then the leg AB opposite to the angle 30° is half of the hypotenuse and AD=12 un.

4. The area of parallelogram is

A_{ABCD}=AD\cdot BK=12\cdot 3\sqrt{3}=36\sqrt{3}\ sq. un.

5 0
3 years ago
Let n be a positive integer. (a) Prove that n^3 = n + 3n(n - 1) + 6 C(n, 3) by counting the number of ordered triples (a,b,c), w
Sophie [7]

(a) Let's call (a,b,c) a smiley face if b is less than a and b is less than c, because when we plot the graph, we get a happy face!  And if b is greater than a and b is greater than c, that's a frowny face, because we get a frowny face when we turn a smiely face up-side-down.

There are other kinds of faces like smirks (like a is less than b and b is less than c) and neutral faces (like when a is equal to b and b is equal to c).  If the face is neutral, then a equals b and b equals c, so when we choose a, b, and c are also chosen, and there are n choices for a, so there are n neutral faces.

Now we count the number of smirks.  There are n ways to choose a, and there are n - 1 ways to choose b.  We also multiply by 3, because the value that we chose for a could have also been the value of b, or the value of c.  So there are 3n(n - 1) smirks.

Now we count the number of smiley faces.  There are n ways to choose a, then n - 1 ways to choose b, then n - 2 ways to choose c.  So there are n(n - 1)(n - 2) = 3C(n,3) smiley faces.  By symmetry, there are 3C(n,3) frowny faces.

Therefore, the total number of faces is n^3 = n + 3n(n - 1) + 6C(n,3).

(b) To choose three numbers, we can choose two groups one group with two numbers and the other group has one number.  The total of n + 2 numbers can be separated into two groups.

In the first way, one group has 2 numbers, and the other group has n numbers.  They form a total of n + 2 numbers.  There are C(2,2) = 1 ways to choose two numbers from the 2 group.  There are C(n,1) = n ways to choose one number from the one group.  This gives us a first term of 1*n.

In the second way, one group has three numbers, and the other group has n - 1 numbers.  They form a total of n - 1 numbers.  There are C(3,2) = 2 ways to choose two numbers from the 2 group.  There are C(n - 1,1) = n - 1 ways to choose one number from the one group.  This gives us a second term of 2*(n - 1).  The pattern will continue until we reach n.  So the two sides are equal.

5 0
3 years ago
Data about cell phone usage was collected from 14 different people. The list shoes the number of incoming calls each person rece
Rudik [331]

Answer: 121

Step-by-step explanation:

I calculated, trust. Also took the test.

7 0
3 years ago
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