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iVinArrow [24]
3 years ago
12

Evaluate the expression 2x+3 if x= 3

Mathematics
2 answers:
ikadub [295]3 years ago
6 0

Answer:

9

Step-by-step explanation:

2(x=3)+3

6+3

9

It is nine

Citrus2011 [14]3 years ago
4 0

let the value be a

2x+3=a

2(3)+3=a

6+3=a

9=a

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A kite was flying 50 ft high from ground level. Suddenly it loses 20 ft in height due to wind.
goldfiish [28.3K]

Answer:

30ft

Step-by-step explanation:

50- 20 = 30

hope it helps!

5 0
3 years ago
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Find the value of each variable. If you give the decimal value, round to the nearest tenth.
DiKsa [7]

u = 10.4 and v = 12

Solution:

In the given 2 sides of a triangle are 60°, 60°.

Sum of all the angles of a triangle = 180°

60° + 60° + third angle = 180°

⇒ third angle = 180° – 60° – 60°

⇒ third angle = 60°

All angles are equal, therefore the given triangle is an equilateral triangle.

⇒ All sides are equal in length.

⇒ v = 12

The line drawn from the top angle divides the triangle into two equal parts

and the line is perpendicular.

12 ÷ 2 = 6

Using Pythagoras theorem,

\text{Base}^2+\text{Opposite}^2=\text{Hypotenuse}^2

⇒ 6^2+\text{opposite}^2=12^2

⇒ 36+\text{opposite}^2=144

⇒ \text{opposite}^2=108

⇒ \text{opposite}=6\sqrt3=10.4

⇒ u = 10.4

Hence, u = 10.4 and v = 12.

7 0
3 years ago
Camilla is 14 years old. She is<br> one-third of her teacher's age.<br> How old is her teacher?
Mamont248 [21]
Her teacher would be 42. you would take 14 * 3 and get the teachers age.
6 0
3 years ago
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What are the answer to these questions:
Tomtit [17]
1. 40/60 + 15/60 + 24/60 + 10/60 = 1.48

2.30/60 + 20/60 + 15/60 + 12/60 + 10/60 = 1.45
3 0
3 years ago
What is the smallest integer $n$, greater than $1$, such that $n^{-1}\pmod{130}$ and $n^{-1}\pmod{231}$ are both defined?
olasank [31]

First of all, the modular inverse of n modulo k can only exist if GCD(n, k) = 1.

We have

130 = 2 • 5 • 13

231 = 3 • 7 • 11

so n must be free of 2, 3, 5, 7, 11, and 13, which are the first six primes. It follows that n = 17 must the least integer that satisfies the conditions.

To verify the claim, we try to solve the system of congruences

\begin{cases} 17x \equiv 1 \pmod{130} \\ 17y \equiv 1 \pmod{231} \end{cases}

Use the Euclidean algorithm to express 1 as a linear combination of 130 and 17:

130 = 7 • 17 + 11

17 = 1 • 11 + 6

11 = 1 • 6 + 5

6 = 1 • 5 + 1

⇒   1 = 23 • 17 - 3 • 130

Then

23 • 17 - 3 • 130 ≡ 23 • 17 ≡ 1 (mod 130)

so that x = 23.

Repeat for 231 and 17:

231 = 13 • 17 + 10

17 = 1 • 10 + 7

10 = 1 • 7 + 3

7 = 2 • 3 + 1

⇒   1 = 68 • 17 - 5 • 231

Then

68 • 17 - 5 • 231 ≡ = 68 • 17 ≡ 1 (mod 231)

so that y = 68.

3 0
3 years ago
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