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aleksley [76]
3 years ago
8

Solve the system by substitution. 5x+2y= 5 y=-2x + 3

Mathematics
1 answer:
Mamont248 [21]3 years ago
3 0
5x+2(-2x+3)=5
5x-4x+6=5
x=-1

Then plug back into either to solve for y.

y=-2(-1)+3
y=2+3
y=5

Substitution means you just replace one variable with the other equation so only one variable is showing. That’s how you solve for the first variable.
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ILL MARK BRAINLYEST
belka [17]

Step-by-step explanation:

Supplementary angles add up to 180 degrees

Complementary angles add up to 90 degrees

It is just a rule that is important to geometry

I hope that helps :)

4 0
3 years ago
Read 2 more answers
A number line goes from negative 5 to positive 5. Point D is at negative 4 and point E is at positive 5. A line is drawn from po
Ronch [10]

<em>Directed numbers</em> are numbers that have either a <u>positive</u> or <u>negative </u>sign, which can be shown on a <em>number line</em>. Therefore, point F is Fifteen-halves of line <em>segment</em> DE.

A <u>number line</u> is a system that can show the positions of <em>positive</em> or <em>negative</em> numbers. It has its <em>ends</em> ranging from <em>negative infinity</em> to <em>positive infinity</em>. Thus any <em>directed</em> number can be located on the line.

Directed numbers are numbers with either a <u>negative</u> or <u>positive </u>sign, which shows their direction with respect to the <em>number line.</em>

In the given question, the <u>distance</u> between points D and E is <em>9 units</em>. So that <em>dividing</em> 9 units in the ratio of 5 to 6, we have;

\frac{5}{6} x 9   = \frac{45}{6}

          = \frac{15}{2}

Therefore, the <em>location</em> of point F, which <u>partitions</u> the directed line segment from d to E into a 5:6 ratio is    \frac{15}{2}. Thus the<em> answer</em> is <u>Fifteen-halves.</u>

For further clarifications on a number line, visit: brainly.com/question/23379455

#SPJ1

7 0
2 years ago
What is the derivative of ln(lnx^3)
chubhunter [2.5K]

the derivative of ln(lnx^3) is \frac{1}{x(lnx)} .

<u>Step-by-step explanation:</u>

Here we have to find the derivative of ln(lnx^3) , Let's find out:

We have , ln(lnx^3) , Let's differentiate it w.r.t x :

⇒ \frac{d(ln(lnx^3))}{dx}

Let lnx^3 = u

⇒ \frac{d(ln(u))}{dx}

⇒ \frac{1}{u} (\frac{d(u)}{dx} )

⇒ \frac{1}{u} (\frac{d(ln(x^3))}{dx} )

Let x^3=v

⇒ \frac{1}{u} (\frac{d(ln(v))}{dx} )

⇒ \frac{1}{u}(\frac{1}{v} ) (\frac{d(v)}{dx} )

⇒ \frac{1}{u}(\frac{1}{v} ) (\frac{d(x^3)}{dx} )            

⇒ 3x^2(\frac{1}{u})(\frac{1}{v} )

Putting value of u & v we get:

⇒ 3x^2(\frac{1}{lnx^3})(\frac{1}{x^3} )

⇒ \frac{3}{x(lnx^3)}    { lnx^n = n(lnx)  }

⇒ \frac{3}{3x(lnx)}

⇒ \frac{1}{x(lnx)}

Therefore , the derivative of ln(lnx^3) is \frac{1}{x(lnx)} .

6 0
3 years ago
Units. Find the other
Tcecarenko [31]

Answer:

L = 12

Step-by-step explanation:

I didn't really know how to do it I just looked it up lol

6 0
3 years ago
Help me please,Thanks
alukav5142 [94]
I have not done measurements in a really long time and i am most likely wrong but i think its 45
8 0
4 years ago
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