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3 years ago

Can someone please help me, I don’t get this

Mathematics

1 answer:

leonid [27]3 years ago

4 0

Answer:

Sure Amaya , the grouping is the parenthesis, right?

Step-by-step explanation:

so just add up.. the things outside of the parenthesis

- $4^{2}$ * 4

so exponents come first in the order of what to do... right?

$4^{2}$ =16 okay , got that part, the next part is

16 * 4 = 64 okay got that part, now

- 64 is what is subtracted from the grouping , got it?

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(15 points!!) please help!! If you can answer the other questions on my page I will Venmo u!

Anon25 [30]

Answer:

y = (x - 3)² - 4 Vertex form

(3, -4) Vertex

Step-by-step explanation:

f(x) = x² - 6x + 5

Complete the square

y = (x - 3)² + 5 - (-3)²

y = (x - 3)² + 5 - 9

y = (x - 3)² - 4 Vertex form

(3, -4) Vertex

7 0

3 years ago

Solve the system of equations using substitution y=x-1 and y=-2x+5

pav-90 [236]

Since the two equations equal y, set them equal to each other.

x-1=-2x+5

From there, solve for x.

First get x on one side, by using the addition property of equality.

x-1=-2x+5
3x-1=5

Isolate x by adding 1.
3x=6

Lastly get x all by itself by dividing each side by 3.
x=2

You can now substitute your x-value, 2, into one of the equations (or both, if you wish; either one will result in the same answer.)

y=x-1
y=2-1
y=1

OR

y=-2x+5
y=-2(2)+5
y=-4+5
y=1

Final answer:
x=2 and y=1

Any questions or anything you would like me to clarify, feel free to ask :)

7 0

3 years ago

Can someone please help me with this geometry question

goldenfox [79]

Answer:

A. q = 39

Step-by-step explanation:

Since the lines are parallel, their sides will be proportional,

So,

Taking their proportion

=> $\frac{60}{40} = \frac{q}{26}$

Cross Multiplying

q × 40 = 26 × 60

q = $\frac{1560}{40}$

q = 39

7 0

2 years ago

Will mark as brainliest plz helpp

almond37 [142]

x - √3y - 4 = 0 → Choice A

Step-by-step explanation:

x - 4 = √3y

x - 4 - √3y = √3y - √3y

x - 4 - √3y = 0

x - √3y - 4 = 0

4 0

3 years ago

Find the limit

Lana71 [14]

Step-by-step explanation:

<h3>Appropriate Question :-</h3>

Find the limit

$\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right]$

$\large\underline{\sf{Solution-}}$

Given expression is

$\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right]$

On substituting directly x = 1, we get,

$\rm \: = \: \sf \dfrac{1-2}{1 - 1}-\dfrac{1}{1 - 3 + 2}$

$\rm \: = \sf \: \: - \infty \: - \: \infty$

which is indeterminant form.

Consider again,

$\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right]$

can be rewritten as

$\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x( {x}^{2} - 3x + 2)}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x( {x}^{2} - 2x - x + 2)}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x( x(x - 2) - 1(x - 2))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x(x - 2) \: (x - 1))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ {(x - 2)}^{2} - 1}{x(x - 2) \: (x - 1))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ (x - 2 - 1)(x - 2 + 1)}{x(x - 2) \: (x - 1))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ (x - 3)(x - 1)}{x(x - 2) \: (x - 1))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ (x - 3)}{x(x - 2)}\right]$

$\rm \: = \: \sf \: \dfrac{1 - 3}{1 \times (1 - 2)}$

$\rm \: = \: \sf \: \dfrac{ - 2}{ - 1}$

$\rm \: = \: \sf \boxed{2}$

Hence,

$\rm\implies \:\boxed{ \rm{ \:\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right] = 2 \: }}$

$\rule{190pt}{2pt}$

7 0

2 years ago

Read 2 more answers