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2 years ago

Whats the value of x?

Mathematics

1 answer:

Anvisha [2.4K]2 years ago

5 0

Answer:

x=3

Step-by-step explanation:

both of the two sides equal the same so it would be 3x=9x-18

solve you get x=3

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The difference of a number and 6 is the same as 5 times the sum of the number and 2. What is the number?

lutik1710 [3]

answer:

B -2

Step-by-step explanation:

equation is x-6=5x+2 and you some it like you would for a regular linear expression

6 0

2 years ago

Write the equation that the phrase represents, then solve the equation. The difference of a number and 1 2/7 is 5 1/2 ( with exp

BaLLatris [955]

Answer:

n = 95/14, or 6 11/14

Step-by-step explanation:

Let the number be n. Then:

n - 1 2/7 = 5 1/2.

Multiplying each term by the LCD (which is 14), we get:

14n - 18 = 77

Then 14n = 95

and so the desired result is n = 95/14, or 6 11/14

8 0

3 years ago

Work out the area of a rectangle with base, b = 36mm and perimeter, P = 94mm.

Alenkasestr [34]

Answer:

396

Step-by-step explanation:

A = bw

A = 36w

P = 2w + 2b

94 = 2w + 2(36)

94 = 2w + 72

-72 -72

--------------------

22 = 2w

---- ----

2 2

11 = w

A = 36(11)

A = 396

The area of the rectangle will be 396.

7 0

2 years ago

Whats the vertex-? Would it be -1???

Ahat [919]

Answer:

last option : vertex is 2, -1 . axis of symmetry is x = 2

3 0

1 year ago

Read 2 more answers

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

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

8 0

2 years ago