Solve this linear equation using substitution. (Please show work!)

Mathematics

2 answers:

julsineya [31]3 years ago

7 0

The answer is 7=1-7x

Genrish500 [490]3 years ago

5 0

Answer:

y=1-7x

Step-by-step explanation:

7x+y=1

The first thing to do is get the y by its self. So u cancel the 7x out by-7x and bring it to the other side.

7x-7x = 0

Now all we have left is: y=1-7x

Hope it helps and is right.

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(PHOTO) Please help need this done by tomorrow thank you.

Andre45 [30]

Answer:

x = 8

Step-by-step explanation:

These are corresponding angles, and they are equal to one another.

$7x+3=3x+35\\4x+3=35\\4x=32\\x=8$

8 0

3 years ago

Evaluate the expression when x = –11.

Aleksandr [31]

2x - 19
Let's substitute the value of x in the equation

2 ( -11 ) - 19
-22 - 19
-41

So correct option is D.

Hope this helps you.. :)

5 0

3 years ago

Read 2 more answers

Root 3 cosec140° - sec140°=4 prove that

Lorico [155]

Answer:

Step-by-step explanation:

We are to show that $\sqrt{3} cosec140^{0} - sec140^{0} = 4\\$

Proof:

From trigonometry identity;

$cosec \theta = \frac{1}{sin\theta} \\sec\theta = \frac{1}{cos\theta}$

$\sqrt{3} cosec140^{0} - sec140^{0} \\= \frac{\sqrt{3} }{sin140} - \frac{1}{cos140} \\= \frac{\sqrt{3}cos140-sin140 }{sin140cos140} \\$

From trigonometry, 2sinAcosA = Sin2A

$= \frac{\sqrt{3}cos140-sin140 }{sin140cos140} \\\\= \frac{\sqrt{3}cos140-sin140 }{sin280/2}\\= \frac{4(\sqrt{3}/2cos140-1/2sin140) }{2sin280}\\\\= \frac{4(\sqrt{3}/2cos140-1/2sin140) }{sin280}\\since sin420 = \sqrt{3}/2 \ and \ cos420 = 1/2 \\ then\\\frac{4(sin420cos140-cos420sin140) }{sin280}$