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sweet [91]
3 years ago
13

Identify the domain and range of y = 4x - 5 + 3

Mathematics
1 answer:
Zepler [3.9K]3 years ago
5 0

Answer:

\huge\boxed{\text{Domain:}\ D=\mathbb{R}=(-\infty;\ \infty)}\\\boxed{\text{Range:}\ R=\mathbb{R}=(-\infty;\ \infty)}

Step-by-step explanation:

y = 4x - 5 + 3 = 4x - 2

It's an equation of a line in the slope-intercept form.

y = 4x - 2 → f(x) = 4x - 2 - it's a lienar function.

The domain of a linear function is the set of all real numbers.

If a slope of linear function is different than 0, then the range is the set of all real numbers.

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Is it bad to have a foot fettish lol please be honest
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Step-by-step explanation:

8 0
2 years ago
A triangle has vertices at (3, 2), (-4,4), and (-2,5). What are the coordinates of the vertices of the image
MrMuchimi

Answer:

The triangle is translated 2 units down and 4 units to the right.

Step-by-step explanation:

7 0
3 years ago
Math math math math math <br> i need help<br> tysm :) &lt;3
stiv31 [10]

Answer:

C. x^14

Step-by-step explanation:

(x^9)^0 is 1

(x^7)^2 is x^14

x^14*1 is x^14

The Answer is C. x^14

Hope this helps!

5 0
2 years ago
A quadratic equation function f(x)=ax^2 passes through (0,0) and (5,5). write an equation for this function
Alex
Equation: f(x)=(1/5)x^2 or f(x)=(0.2)x^2

Check:
f(0)=(1/5)(0)^2
f(0)=(1/5)(0)
f(0)=0✓
f(5)=(1/5)(5)^2
f(5)=(1/5)(25)
f(5)=5✓
6 0
3 years ago
What is the slope of the line tangent to the curve y+2 = (x^2/2) - 2siny at the point (2,0)?
kirza4 [7]
<span>, y+2 = (x^2/2) - 2sin(y) so we are taking the derivative y in respect to x so we have dy/dx use chain rule on y so y' = 2x/2 - 2cos(y)*y'

</span><span>Now rearrange it to solve for y' y' = 2x/2 - 2cos(y)*y' 0 = x - 2cos(y)y' - y' - x = 2cos(y)y' - y' -x = y'(2cos(y) - 1) -x/(2cos(y) - 1) = y'
</span><span>we know when f(2) = 0 so thus y = 0 so when f'(2) = -2/(2cos(0)-1)
</span><span>2/2 = 1
</span><span>f'(2) = -2/(2cos(0)-1) cos(0) = 1 thus f'(2) = -2/(2(1)-1) = -2/-1 = 2 f'(2) = 2
</span>
6 0
2 years ago
Read 2 more answers
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