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3 years ago

WILL GIVE BRAINLIEST 15 POINTS!!! Consider the following pair of equations:

1 answer:

5 0

The two lines will intersect at the point (2, 3).

We can find this by setting the two equations equal to eachother, as the point at which they intersect is the point where they share the same x and y value. First, we have to get both equations in the form y = mx + c, so:
-x - y = -5
+ y
-x = -5 + y
+ 5
y = -x + 5

Now we can set -x + 5 equal to x + 1 and solve to find the x coordinate:

-x + 5 = x + 1
+ x
5 = 2x + 1
- 1
4 = 2x
÷ 2
2 = x

With the x coordinate, we can substitute it in to one of the equations and find our y coordinate:

y = 2 + 1 = 3

So the final answer is (2, 3).

I hope this helps!

-x + 5 = x + 1

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3 years ago

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svetoff [14.1K]

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4 years ago

Why does multiplying by 10 shift each digit one place to left

lesya692 [45]

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6 0

3 years ago

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Simplify the exspresion below <br><br> 3(7.1y + 4x)

givi [52]

Answer:

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Step-by-step explanation:

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3 years ago

Simplify the expression below and write it as a single logarithm:

OLEGan [10]

The simplification of 3log(x + 4) – 2log(x – 7) + 5log(x - 2) - log(x^2) is $\log \left(\frac{(x+4)^{3} \times(x-2)^{5}}{(x-7)^{2} \times x^{2}}\right)$

<u>Solution:</u>

Given, expression is $3 \log (x+4)-2 \log (x-7)+5 \log (x-2)-\log \left(x^{2}\right)$

We have to write in as single logarithm by simplifying it.

Now, take the given expression.

$\rightarrow 3 \log (x+4)-2 \log (x-7)+5 \log (x-2)-\log \left(x^{2}\right)$

Rearranging the terms we get,

$\left.\rightarrow 3 \log (x+4)+5 \log (x-2)-2 \log (x-7)+\log \left(x^{2}\right)\right)$

$\text { since a } \times \log b=\log \left(b^{a}\right)$

$\rightarrow \log (x+4)^{3}+\log (x-2)^{5}-\left(\log (x-7)^{2}+\log \left(x^{2}\right)\right)$

$\text { We know that } \log a \times \log b=\log a b$

$\rightarrow \log \left((x+4)^{3} \times(x-2)^{5}\right)-\left(\log \left((x-7)^{2} \times\left(x^{2}\right)\right)\right.$

$\text { We know that } \log a-\log b=\log \frac{a}{b}$

$\rightarrow \log \left(\frac{(x+4)^{3} \times(x-2)^{5}}{(x-7)^{2} \times x^{2}}\right)$

Hence, the simplified form $\rightarrow \log \left(\frac{(x+4)^{3} \times(x-2)^{5}}{(x-7)^{2} \times x^{2}}\right)$

4 0

3 years ago

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