Find the missing term and the missing coefficient. (6a − )5a = a2 − 35a

Answers

Answer 1

Answer: So, (6a - X)*5a = Y*(a^2) - 35a
=> 6a*5a - X*5a = Y*(a^2) - 35a
=> 30(a^2) - X*5a = Y(a^2) - 35a

30 = Y and
X*5 = 35 or X = 7

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Need the answer in under 5 minutes
Divide(6a^4+2a^3) divide 2a A) 3a^3+a^2 b)3a^4+a^3+2a c)3a^2+2a^3 d)3a^4+a^3-2a

Help with line equations?

Answers

First question

$line:~~~~y=x+3$

$curve:~~~~x^2+y^2=29$

now we have to replace the y of curve by the y of line, therefore

$x^2+(x+3)^2=29$

$x^2+x^2+6x+9=29$

$2x^2+6x+9-29=0$

$2x^2+6x-20=0$

we can multiply each member by $(1)/(2)$

$\boxed{\boxed{x^2+3x-10=0}}$

Now we have to find the roots of this funtion

$x^2+3x-10=0$

Sum and produc or Bhaskara

Then we find two axis

$\boxed{x_1=2~~and~~x_2=-5}$

now we replace this values to find the Y, we can replace in curve or line equation, I'll prefer to replace it in line equation.

$y=x+3$

$y_1=x_1+3$

$y_1=2+3$

$\boxed{y_1=5}$

$y_2=x_2+3$

$y_2=-5+3$

$\boxed{y_2=-2}$

Therefore

$\boxed{\boxed{P_1(2,5)~~and~~P_2(-5,-2)}}$
_______________________________________________________________

The second question give to us

$y=ax+b$

$P_1(2,13)$

$P_2(-1,-11)$

We just have to replace the value then we'll get a linear system.

point 1

$13=2a+b$

point 2

$-11=-a+b$

then our linear system will be

$\begin{Bmatrix}2a+b&=&13\n-a+b&=&-11\end{matrix}$

I'll multiply the second line by -1 and I'll add to first one

$\begin{Bmatrix}2a+(a)+b+(-b)&=&13+(-11)\n-a+b&=&-11\end{matrix}$

$\begin{Bmatrix}3a&=&24\n-a+b&=&-11\end{matrix}$

$\begin{Bmatrix}a&=&8\n-a+b&=&-11\end{matrix}$

therefore we can replace the value of a, at second line

$\begin{Bmatrix}a&=&8\n-8+b&=&-11\end{matrix}$

$\begin{Bmatrix}a&=&8\nb&=&-11+8\end{matrix}$

$\boxed{\boxed{\begin{Bmatrix}a&=&8\nb&=&-3\end{matrix}}}$

then our function will be

$\boxed{\boxed{y=8x-3}}$

_________________________________________________________________

The third one, we have

$line:~~~~y=3x-4$

$curve:~~~~y=x^2-2x-4$

This resolution will be the same of our first question.

Let's replace the y of curve by the y of line

$3x-4=x^2-2x-4$

$0=x^2-2x-4-3x+4$

therefore

$\boxed{\boxed{x^2-5x=0}}$

now we have to find the roots of this function.

$x^2-5x=0$

put x in evidence

$x*(x-5)=0$

$\boxed{x_1=0~~and~~x_2=5}$

then

$y=3x-4$

$y_1=3x_1-4$

$y_1=3*0-4$

$\boxed{y_1=-4}$

$y_2=3x_2-4$

$y_2=3*5-4$

$y_2=15-4$

$\boxed{y_2=11}$

our points will be

$\boxed{\boxed{P_1(0,-4)~~and~~P_2(5,11)}}$

I hope you enjoy it ;)

ActivityThe equation y = x + 1 represents plane A and y = -x + 21 represents plane B, where x is the time in minutes and y is the fuel in tons.

Part A
Go to your math tools and open the Graph tool to graph the two sets of equations. To see where the two lines intersect, change the scale so that the x-axis goes from 0 to 30 and the y-axis goes from 0 to 12. Paste a screenshot of the resulting graph in the answer space.

Part B
At which point do the lines intersect?

Part C
Do the coordinates of the point of intersection satisfy both equations simultaneously?

Answers

Answer:

Part-A: refer to the attachment

Part-B: (10,11)

Part-C: yes

step-by-step explanation:

Part-A:

refer to the attachment

(I used a online graphing calculator to graph the equations which made the work easy)

Part-B:

When two lines share exactly one common point, they are called the intersecting lines and thepointis called thepointof interception

Looking at the graph,we can understand that the two lines share a common point at (10,11),

hence,

The lines intercept at the point (10,11)

PartC:

well, to find the answer of this part, we can consider doing equality check by substituting the value of the point we got.

The point (10,11) means that the left and right hand side of both of the equations i.e $\text{y=x+1 and y=-x+21}$ are equal when x and y equal to 10 and 11 respectively.

So let's justify the points:

equation-1:

y = x + 1

substitute the value of x and y respectively:

$11\stackrel{?}{=}10+1$

simplify addition:

$11\stackrel{\checkmark}{=}11$

equation-2:

y = -x + 21

substitute the value of x and y respectively:

$11\stackrel{?}{=}-10+21$

simplify addition:

$11\stackrel{\checkmark}{=}11$

so,

Yes,the coordinates of the point of intersection satisfy both equations simultaneously

Answer:

Step-by-step explanation:

Transform the polynomial into a perfect square trinomial to solvex^2-12x=13
a){46,55}
b){-1,13}
c){-15,27}
d){-21,33}

Answers

If you would like to solve x^2 - 12 * x = 13, you can do this using the following steps:

x^2 - 12 * x = 13
x^2 - 12 * x - 13 = 0
(x - 13) * (x + 1) = 0
1. x = 13
2. x = -1

The correct result would be b) {-1, 13}.

Need a little help on this :)

Answers

Answer:

Step-by-step explanation:

Have A Great DAY!!!

Answer:

D. Charles, John, Thomas, Yurly

Step-by-step explanation:

I started by finding the greatest jump- Yurly. Since you have to order the names from least to greatest, Yurly would be last. Option D is the only option where Yurly is last, therefore option D would be the only possible correct option.

PLEASEEE ITS EASY ALGEBRA 1 ):

Answers

Answer:

6, 7, and 8

Step-by-step explanation:

2x/3 + 7 > 11

Subtract 7 from both sides

2x/3 > 4

Multiply both sides by 3

2x > 12

Divide both sides by 2

x > 6

6, 7, and 8 are the ones that work

The focus of a parabola is (3,-7) and the directrix is y = -4.What is an equation of the parabola?

Answers

Answer:

(a) (x -3)^2 = -6(y +5.5)

Step-by-step explanation:

The equation of a parabola can be written as ...

(x -h)^2 = 4p(y -k)

where (h, k) is the vertex, and p is the distance from the focus to the vertex.

The vertex is half-way between the focus and directrix, so is ...

(h, k) = (1/2)((3, -7) +(3, -4)) = (3, -5.5)

The focus is at y=-7, and the vertex is at y=-5.5, so the distance between them is ...

-7 -(-5.5) = -1.5

Then the equation for the parabola is ...

(x -3)^2 = 4(-1.5)(y -(-5.5))

(x -3)^2 = -6(y +5.5) . . . . matches the first choice

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A conference room is in the shape of a rectangle. Its floor has a length of (x − 4) meters and a width of (3x − 1) meters. The expression below represents the area of the floor of the room in square meters:(x − 4)(3x − 1) Which of the following simplified expressions represents the area of the floor of the conference room in square meters? (5 points) A.x2 − 13x + 4 B.3x2 − 13x + 4 C.3x2 − 11x + 4 D.x2 − 12x + 4

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The cost of a telephone call is 75 cents plus 25 cents per minute. Write an algebraic expression that models the cost of a telephone call that lasts t minutes.