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3 years ago

–2.5p – 20 = 9p + 37.5

Mathematics

2 answers:

Sever21 [200]3 years ago

7 0

Answer:

The answer is p = -5

Step-by-step explanation:

-2.5p - 20 = 9p + 37.5

(move constants to one side by adding 20 to both sides)

-2.5p = 9p + 57.5

(move all terms with the "p" variable to one side by subtracting 9 from both sides)

-11.5p = 57.5

(divide both sides by -11.5 to isolate the variable)

p = -5

Elenna [48]3 years ago

6 0

Answer:

p = - 5

Step-by-step explanation:

–2.5p – 20 = 9p + 37.5

combine like terms:

- 2.5p - 9p = 37.5 + 20

simplify:

- 11.5p = 57.5

p = 57.5 / -11.5

p = - 5

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Find the number of elements in A 1 ∪ A 2 ∪ A 3 if there are 200 elements in A 1 , 1000 in A 2 , and 5, 000 in A 3 if (a) A 1 ⊆ A

lina2011 [118]

Answer:

a. 4600

b. 6200

c. 6193

Step-by-step explanation:

Let $n(A)$ the number of elements in A.

Remember, the number of elements in $A_1 \cup A_2 \cup A_3$ satisfies

$n(A_1 \cup A_2 \cup A_3)=n(A_1)+n(A_2)+n(A_3)-n(A_1\cap A_2)-n(A_1\cap A_3)-n(A_2\cap A_3)-n(A_1\cap A_2 \cap A_3)$

Then,

a) If $A_1\subseteq A_2, n(A_1 \cap A_2)=n(A_1)=200$ , and if $A_2\subseteq A_3, n(A_2\cap A_3)=n(A_2)=1000$

Since $A_1\subseteq A_2\; and \; A_2\subseteq A_3, \; then \; A_1\cap A_2 \cap A_3= A_1$

$n(A_1 \cup A_2 \cup A_3)=\\=n(A_1)+n(A_2)+n(A_3)-n(A_1\cap A_2)-n(A_1\cap A_3)-n(A_2\cap A_3)-n(A_1\cap A_2 \cap A_3)=\\=200+1000+5000-200-200-1000-200=4600$

b) Since the sets are pairwise disjoint

$n(A_1 \cup A_2 \cup A_3)=\\n(A_1)+n(A_2)+n(A_3)-n(A_1\cap A_2)-n(A_1\cap A_3)-n(A_2\cap A_3)-n(A_1\cap A_2 \cap A_3)=\\200+1000+5000-0-0-0-0=6200$

c) Since there are two elements in common to each pair of sets and one element in all three sets, then

$n(A_1 \cup A_2 \cup A_3)=\\=n(A_1)+n(A_2)+n(A_3)-n(A_1\cap A_2)-n(A_1\cap A_3)-n(A_2\cap A_3)-n(A_1\cap A_2 \cap A_3)=\\=200+1000+5000-2-2-2-1=6193$

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3 years ago

Please Help! What is the distance between points (2, 8) and (7, 5)? Round to the nearest tenth of a unit.

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Distance formula is like pythagoren ahtoerem
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3 years ago

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creativ13 [48]

Answer: D

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Consider the second equation. Subtract x from both sides.

y−2−x=0

Add 2 to both sides. Anything plus zero gives itself.

y−x=2

To solve a pair of equations using substitution, first solve one of the equations for one of the variables. Then substitute the result for that variable in the other equation.

y−3x=−2,y−x=2

Choose one of the equations and solve it for y by isolating y on the left hand side of the equal sign.

y−3x=−2

Add 3x to both sides of the equation.

y=3x−2

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Add 3x to −x.

2x−2=2

Add 2 to both sides of the equation.

2x=4

Divide both sides by 2.

x=2

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3 years ago