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7nadin3 [17]
3 years ago
5

The product of 4 and a number ,increased by 22 is 146. What is the number ?

Mathematics
1 answer:
maks197457 [2]3 years ago
7 0
Not absolutely sure I understand the question, but should be 120
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What are the solutions to
Svetllana [295]

Answer:

B and F

Step-by-step explanation:

Given

x² + 4x + 4 = 12 ( subtract 4 from both sides )

x² + 4x = 8

Using the method of completing the square to solve for x

add ( half the coefficient of the x- term )² to both sides

x² + 2(2)x + 4 = 8 + 4, that is

(x + 2)² = 12 ( take the square root of both sides )

x + 2 = ± \sqrt{12} ( subtract 2 from both sides )

x = ± \sqrt{12} - 2

  = ± 2\sqrt{3} - 2

Hence

x = 2\sqrt{3} - 2 → B

x = - 2\sqrt{3} - 2 → F

4 0
4 years ago
Find the distance between the following coordinates: I(-a,b), J(a,b)
olchik [2.2K]

Answer:

2a

Step-by-step explanation:

using distance formula:

√(x₂-x₁)² + (y₂-y₁)²

putting values,

√{a-(a)}² + {b-b}² = √ {a+a}² + 0²

= √(2a)² = √4a² = 2a

Hope this helps:)

8 0
3 years ago
Four friends go to the movies and pay $11 per ticket. They also each buy a snack
Setler79 [48]

Answer:

Each snack combo costed $8

Step-by-step explanation:

4 friends* 11 per ticket = $44 for four tickets

76 - 44 = $32 spent on all the snack combos

$32 ÷ 4 friends = $8 spent on each snack combo

So, each snack combo costed $8

Hope this helps!

6 0
2 years ago
Read 2 more answers
If cot theta= 4/3, find csc theta
Marrrta [24]

Answer: Option d.

Step-by-step explanation:

The trigonometric identity needed is:

csc^2\theta=cot^2\theta+1

Knowing that cot\theta=\frac{4}{3}:

Substitute it into csc^2\theta=cot^2\theta+1:

csc^2\theta=(\frac{4}{3})^2+1

Simplify the expression:

csc^2\theta=(\frac{4}{3})^2+1\\\\csc^2\theta=\frac{16}{9}+1\\\\csc^2\theta=\frac{25}{9}

Solve for csc\theta. Apply square root at both sides of the expression:

\sqrt{csc^2\theta}=\±\sqrt{\frac{25}{9}}

csc\theta=\frac{5}{3}

8 0
3 years ago
The measure of the exterior angle of the triangle is
Nookie1986 [14]

Answer:

jsjsjsisusbbdbdjsjcjdjsjjsjsjsjsjdjxjx

4 0
3 years ago
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