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Feliz [49]
3 years ago
9

Which value is the solution to the open sentence? 2 + 3x= 5

Mathematics
2 answers:
3241004551 [841]3 years ago
3 0

Answer:

x = 1

Step-by-step explanation:

2 + 3*x = 5

3*x = 5 - 2

3*x = 3

x = 3/3

x = 1

OLEGan [10]3 years ago
3 0
Answer would be one
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Given that sin theta = 1/4, 0→theta→π/2, what<br>. what is the exact value of cos 8?​
natali 33 [55]
<h3>Answer: Choice B \frac{\sqrt{15}}{4}</h3>

===========================================================

Work Shown:

Angle theta is between 0 and pi/2, so this angle is in quadrant Q1.

Square both sides of the given equation

\sin \theta = \frac{1}{4}\\\\\sin^2 \theta = \left(\frac{1}{4}\right)^2\\\\\sin^2 \theta = \frac{1}{16}

Then use the pythagorean trig identity to get

\sin^2 \theta + \cos^2 \theta = 1\\\\\cos^2 \theta = 1-\sin^2 \theta\\\\\cos \theta = \sqrt{1-\sin^2 \theta} \ \ \ \text{cosine is positive in Q1}\\\\\cos \theta = \sqrt{1-\frac{1}{16}}\\\\\cos \theta = \sqrt{\frac{16}{16}-\frac{1}{16}}\\\\\cos \theta = \sqrt{\frac{16-1}{16}}\\\\\cos \theta = \sqrt{\frac{15}{16}}\\\\\cos \theta = \frac{\sqrt{15}}{\sqrt{16}}\\\\\cos \theta = \frac{\sqrt{15}}{4}\\\\

3 0
3 years ago
For each of the following functions, determine if they are injective. Also determine if theyare surjective. Also determine if th
timurjin [86]

Answer:

Check below

Step-by-step explanation:

1. Definition for intervals

(a,b)=\left \{ x\in\Re : a

2. Functions

1) \Re \rightarrow \Re \\ f(x)=x

Let's perform graph tests.

That's an one to one, injective function. Look how any horizontal line touches that only once. Also, It's a surjective and a bijective one.

2)\Re\geq0\rightarrow\Re , f(x)=x+1\\

Injective, surjective and bijective.

Injective: a horizontal line crosses the graph in one point.

3)f:\Re\geq 0\rightarrow\Re, f(x) = cos(x)

The cosine function is not injective, bijective nor surjective.

4)f:\Re\rightarrow\Re \:f(x)=ex

Since e, is euler number it's a constant. It's also injective, surjective and bijective.

5)  Quite unclear format

6) \:f:\Re\rightarrow(0,\infty), f(x) =ex

Despite the Restriction for the CoDomain, the function remains injective, surjective and therefore bijective.

7) f:\Re\geq 0 \rightarrow  \Re\geq0, f(x) =x^{4}

Not injective nor surjective therefore not bijective too.

8).f:\{-1,2,-3\}}\rightarrow \{1,4,9\}, f(x) =x^{2}

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Injective (one to one), Surjective,  and Bijective.

10) f:\Re\geq 0\rightarrow [-1,1], f(x)= cos(x)\\-1=cos(x) \therefore x=\pi,3\pi,5\pi,etc.

Surjective.

11.f:R\geq 0[-1,1], f(x) = 0\\

Surjective

12.f: US Citizens→Z, f(x) = the SSN of x.

General function

13.f: US Zip Codes→US States, f(x) = The state that x belongs to.

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7 0
3 years ago
Write an expression equivalent to (6x + 4y) – 2y by combining like terms. Enter the numbers in the boxes
raketka [301]

6x +2y

Solution:

Given expression is (6x + 4y) –2y.

To find the equivalent expression for the given expression.

⇒ (6x + 4y) –2y

⇒ 6x + 4y –2y

Combine like terms together.

⇒ 6x + (4y –2y)

⇒ 6x + 2y

6x +2y is equivalent to (6x + 4y) –2y.

Hence, the expression equivalent to (6x + 4y) –2y is 6x + 2y.

5 0
3 years ago
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