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2 years ago

I'm struggling in this subject so if you'd give a step-by-step explanation that'd be great thx

Mathematics

2 answers:

defon2 years ago

8 0

A. First you must start by plugging in your equation solve Y = 2(5) + 1 then you must simplify by solving your 2×5+ 1 and then you end up with Y equals 11

B. Again plug in your equation Y = 2(-5) + 1 then you must simplify by solving your 2×-5+ 1 and then you end up with Y equals -10+ one which simplifies to y = -9

I hope this helps

spayn [35]2 years ago

6 0

Here the equation given to us is,

${\blue{\boxed{\pink{y = 2x + 1}}}}$

Now what we have to do is, we have to put the value of x in brackets () in both the equations . Why we have to put value of x in a bracket?

This is because 2 is being multiplied by the variable x. So let's solve the equations

a) When x = 5, what is the value of y?

Put the value of x here

y = 2x + 1

y = 2(5) + 1

y = 10 + 1

y = 11

b) When x = -5, what is the value of y?

y = 2x + 1

y = 2(-5) + 1

y = -10 + 1

y = -9

ANSWERS :-

a) y = 11

b) y = -9

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When given the graph of a function, the domain would include all the points that there is a graph. The strategy is to find what <em>is not</em> included.

What we are looking for are points of discontinuity. Think of it as when you remove your pencil from the paper.

From left to right, the graph stops at x = -3. So anything less than -3 is in the domain. Next, the graph starts up again at x =-1 after an asymptote (the vertical dashed lines). This piece goes to x = 4. So our domain is from -1 to 4.

Lastly, there's a jump from 4 to 5 and the graph goes on again. After 5, we take all the stuff more than it. So x > 5 is in the domain.

So x < -3, - 1 < x < 4, and x > 5 appears to be our domain. However, end points needed to be checked to see if we include them or not. Again we go left to right.

At x = -3 there is a filled (or closed) circle and that means we include -3.

At x = -1 there is an asymptote. Asymptotes are things you get close to but don't get to. (Think of it as the "I'm Not Touching" game you play on car trips.) So we exclude -1.

At x = 4 there is an unfilled (or open) circle and that means we exclude 4.

At x = 5 there is a filled circle so we include 5.

Now we refine our domain for the endpoints.

x ≤ -3, -1 < x < 4, x ≥ 5 is our domain.

The problem gives us intervals, and we gave it in inequalities. When we include an endpoint we use brackets - [ and } and when we exclude and endpoint we use parentheses - ( and ). Let's go back to x ≤ -3. Anything less works, and -3 is included (closed circle). That interval is (-∞, -3]. Next is the piece between -1 and 4. Since both are excluded, (-1,4) is our interval. We include 5 to write x ≥5 as the interval [5,∞).

Put the bolded ones all together and use the union, ∪, symbol to connect them, since something on the graph could be in any piece.

Our domain is (-∞, -3] ∪(-1,4) ∪ [5,∞).

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3 years ago

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solong [7]

Answer:

$(x+\frac{b}{2a})^2+(\frac{4ac}{4a^2}-\frac{b^2}{4a^2})=0$

$(x+\frac{b}{2a})^2=\frac{b^2-4ac}{4a^2}$

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Step-by-step explanation:

$x^2+\frac{b}{a}x+\frac{c}{a}=0$

They wanted to complete the square so they took the thing in front of x and divided by 2 then squared. Whatever you add in, you must take out.

$x^2+\frac{b}{a}x+(\frac{b}{2a})^2+\frac{c}{a}-(\frac{b}{2a})^2=0$

Now we are read to write that one part (the first three terms together) as a square:

$(x+\frac{b}{2a})^2+\frac{c}{a}-(\frac{b}{2a})^2=0$

I don't see this but what happens if we find a common denominator for those 2 terms after the square. (b/2a)^2=b^2/4a^2 so we need to multiply that one fraction by 4a/4a.

$(x+\frac{b}{2a})^2+\frac{4ac}{4a^2}-\frac{b^2}{4a^2}=0$