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1 year ago

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Linear Relationships RetakeConsider the function f(x) and its transformation.f(x) = 5x + 1g(x) = f(x) + 6Write the equation in s

lope-intercept form for g(x).PLEASE HELP

1 answer:

zmey [24]1 year ago

5 0

Answer:

The equation in slope-intercept form is;

$g(x)=5x+7$

Explanation:

Given the function f(x);

$f(x)=5x+1$

And its transformation g(x);

$g(x)=f(x)+6$

To get the equation for g(x), let us substitute f(x) into the function g(x);

$\begin{gathered} g(x)=f(x)+6 \\ g(x)=5x+1+6 \\ g(x)=5x+7 \end{gathered}$

Therefore, the equation in slope-intercept form is;

$g(x)=5x+7$

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Simplify the expression. -2/3c -9/5 + 14c + 3/10

Andrew [12]

Answer:

$13 \frac{1}{3} c - 1 \frac{1}{2}$

Step-by-step explanation:

We want to combine alike terms so

-2/3c+14c & -9/5+3/10

14c-2/3

$\frac{14}{1} c - \frac{2}{3} c$

Multiply 14/1 by 3/3 to get the denominators the same

$\frac{42}{3} c - \frac{2}{3} c$

Subtract

$\frac{40}{3} c$

simplify

$13 \frac{1}{3} c$

now

$\frac{3}{10} - \frac{9}{5}$

Multiply 9/5 by 2/2

$\frac{3}{10} - \frac{ 18}{10}$

Subtract

$- \frac{15}{10}$

Simplify

$- 1 \frac{1}{2}$

put both together

$13 \frac{1}{3}c - 1 \frac{1}{2}$

Hope this helps! If you have any questions on how I got my answer feel free to ask. Stay safe!

5 0

2 years ago

Read 2 more answers

Find the quotient of these Complex Numbers. <br><br><br> (5-i) / (3+2i)

JulsSmile [24]

Let the given complex number

z = x + ix = $\dfrac{5-i}{3+2i}$

We have to find the standard form of complex number.

Solution:

∴ x + iy = $\dfrac{5-i}{3+2i}$

Rationalising numerator part of complex number, we get

x + iy = $\dfrac{5-i}{3+2i}\times \dfrac{3-2i}{3-2i}$

⇒ x + iy = $\dfrac{(5-i)(3-2i)}{3^2-(2i)^2}$

Using the algebraic identity:

(a + b)(a - b) = $a^{2}$ - $b^{2}$

⇒ x + iy = $\dfrac{15-10i-3i+2i^2}{9-4i^2}$

⇒ x + iy = $\dfrac{15-13i+2(-1)}{9-4(-1)}$ [ ∵ $i^{2} =-1$ ]

⇒ x + iy = $\dfrac{15-2-13i}{9+4}$

⇒ x + iy = $\dfrac{13-13i}{13}$

⇒ x + iy = $\dfrac{13(1-i)}{13}$

⇒ x + iy = 1 - i

Thus, the given complex number in standard form as "1 - i".

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3 years ago

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sergij07 [2.7K]

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2 years ago

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Studentka2010 [4]

Answer:

Step-by-step explanation:

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2 years ago

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True or false: some polynomial functions of degree 2 or higher have breaks in their graphs

aalyn [17]

The answer is false I hope this helps

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3 years ago