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Andrei [34K]
3 years ago
8

Graph the equation y = 4x + 6.

Mathematics
1 answer:
LenKa [72]3 years ago
3 0
That’s how you graph it

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I really need someone to answer this, QUICK!
Elodia [21]

Answer:

A. LCM=72   B. GCF=7    C. 7(5+9)

Step-by-step explanation:

A. 8/2=4/2=2  9/3=3   Use the denominators to find the LCM. In this case, the LCM is 72.

B.  35/5=7   63/3=21/3=7    The GCF for these numbers is 7 because it is the only factor that they have in common.

C.   7 x 14=88     Since 5 and 9 don't have common factors, they work perfectly for this problem.

6 0
2 years ago
Help fast !!!! I need to get this done before I go to school ixl AA.1
Lorico [155]

Answer:

slope = - \frac{7}{5}

Step-by-step explanation:

Calculate the slope m using the slope formula

m = \frac{y_{2}-y_{1}  }{x_{2}-x_{1}  }

with (x₁, y₁ ) = (- 5, 0) and (x₂, y₂ ) = (0, - 7 ) ← 2 points on the line

m = \frac{-7-0}{0-(-5)} = \frac{-7}{0+5} = \frac{-7}{5} = - \frac{7}{5}

4 0
2 years ago
What is the slope of the line given by the equation y = -3x?
s2008m [1.1K]
The answer to this question would be B, -3. The reason for this is because when you have an equation in slope intercept form, it would be y = mx + b. In this case your b is equal to 0 so you have only y = mx. The variable m represents the slope, which is in this case, equal to -3. Hope this helps. Please rate, leave a thanks, and mark a brainiest answer. (Not necessarily mine). Thanks, it really helps! :D
7 0
3 years ago
Read 2 more answers
What is the answer and how do you solve this??
OLga [1]
Since they are both equal to y, you can set -7x and x+8 equal to each other.
- 7x = x + 8 and solve for x then use what you got for x to solve for y
5 0
2 years ago
Read 2 more answers
Please help me answer this question
ratelena [41]

By the definition of the hyperbolic function tanh x, we have proven that \frac{1-tanh \ x}{1 + tanh \ x}=e^{-2x}

<h3>Hyperbolic functions & Proof of identities </h3>

By definition

tanh \ x=\frac{e^{x} -e^{-x}}{e^{x} +e^{-x}}

Then,

\frac{1-tanh \ x}{1 + tanh \ x}=\frac{1-\frac{e^{x} -e^{-x}}{e^{x} +e^{-x}} }{1+ \frac{e^{x} -e^{-x}}{e^{x} +e^{-x}} }

=1-\frac{e^{x} -e^{-x}}{e^{x} +e^{-x}} \div (1+ \frac{e^{x} -e^{-x}}{e^{x} +e^{-x}} })

=\frac{e^{x} +e^{-x}-(e^{x} -e^{-x})}{e^{x} +e^{-x}} \div (\frac{e^{x} +e^{-x}+e^{x} -e^{-x}}{e^{x} +e^{-x}} })

=\frac{e^{x} +e^{-x}-e^{x} +e^{-x}}{e^{x} +e^{-x}} \div \frac{e^{x} +e^{-x}+e^{x} -e^{-x}}{e^{x} +e^{-x}} }

=\frac{e^{-x}+e^{-x}}{e^{x} +e^{-x}} \div \frac{e^{x} +e^{x} }{e^{x} +e^{-x}} }

=\frac{2e^{-x}}{e^{x} +e^{-x}} \div \frac{2e^{x} }{e^{x} +e^{-x}} }

=\frac{2e^{-x}}{e^{x} +e^{-x}} \times \frac{e^{x} +e^{-x}}{2e^{x}}

=\frac{2e^{-x}}{1} \times \frac{1}{2e^{x}}

=\frac{2e^{-x}}{2e^{x}}

=\frac{e^{-x}}{e^{x}}
=e^{-x} \times \frac{1}{e^{x}}

= e^{-x} \times e^{-x}

= e^{-x+-x}

= e^{-x-x}

= e^{-2x}

Hence, we have proven that \frac{1-tanh \ x}{1 + tanh \ x}=e^{-2x}

Learn more on Proof of Identities here: brainly.com/question/2561079

#SPJ1

6 0
2 years ago
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