What is the graph function f(x)= x^2-7x+12/x-3

Answer:

There would be 20 ounces of almonds in the bag

Step-by-step explanation:

System of Equations

To solve the problem, we'll set up these two variables:

x = ounces of almonds

y = ounces of raisins

The ratio of almonds to raisins in the trail mix is 5:7, thus:

[tex]\displaystyle \frac{x}{y}=\frac{5}{7}[/tex]

Cross-multiplying:

7x=5y         [1]

A 3-pound bag of this trail mix contains 3*16=48 ounces that include almonds and raisins, thus:

x + y = 48      [2]

Multiplying [2] by 5:

5x + 5y = 240

Substituting from [1]:

5x + 7x = 240

12x = 240

x = 240/12

x = 20

There would be 20 ounces of almonds in the bag

Answer:

A

Step-by-step explanation:

Answer:

what is the y-coordinate for A?

Step-by-step explanation:

Answer:

[tex]\boxed {\boxed {\sf E. \ y=x-3}}[/tex]

Step-by-step explanation:

To find the equation of the line, we must first find the slope, then use the point-slope formula.

1. Find the Slope

The slope formula is the change in y over the change in x, or:

[tex]m=\frac{y_2-y_1}{x_2-x_1}[/tex]

Where (x₁, y₁) and (x₂, y₂) are the points the line passes through.

The points given in the problem are (6,3) and (4,1). Therefore:

[tex]x_1=6 \\y_1=3 \\x_2=4 \\y_2=1[/tex]

Substitute the values into the formula.

[tex]m=\frac{1-3}{4-6}[/tex]

Solve the numerator.

[tex]m=\frac{-2}{4-6}[/tex]

Solve the denominator.

[tex]m=\frac{-2}{-2}[/tex]

Divide.

[tex]m=1[/tex]

2. Find the Equation of the Line

We have the slope and a point, so we can use the point-slope formula.

[tex]y-y_1=m(x-x_1)[/tex]

We know the slope is 1 and we can pick either point to use for (x₁, y₁). Let's use (4,1).

[tex]y-1=1(x-4)[/tex]

Distribute the 1.

[tex]y-1=(1*x)+(1*-4)\\y-1=x-4[/tex]

We want to find the equation in y=mx+b, so we must isolate the variable on one side of the equation.

1 is being subtracted from y and the inverse of subtraction is addition. Add 1 to both sides of the equation.

[tex]y-1+1=x-4+1\\y=x-4+1\\y=x-3[/tex]

In slope-intercept form, the equation of the line is y=x-3

Answer:c

Step-by-step explanation:

C

Answer:

13.48

Step-by-step explanation:

Answer:

B=10

Step-by-step explanation:

worked it out on a graph on the internet

Step-by-step explanation:

4y(3x+2)–2(3x+2)

(4y–2)(3x+2)

2(2y–1)(3x+2)

The Factor of given equation  [tex]4y(3x+2)-2(3x+2)[/tex]   is    [tex](2y-1) (6x+4)[/tex] .

Factor in an equation is something which is multiplied by something else or we can say that numbers or the variables that are multiplied together.

We have,

 [tex]4y(3x+2)-2(3x+2)[/tex]  

So, first simplify the expression,

      [tex]12xy+8y -6x-4[/tex]

⇒    [tex]12xy-6x+8y-4[/tex]

⇒     [tex]6x(2y-1)+4(2y-1)[/tex]

From we can see that [tex](2y-1)[/tex] is common in both terms,

Therefore,

⇒    [tex](2y-1) (6x+4)[/tex]

So,    [tex](2y-1) (6x+4)[/tex]  is the factor of the given expression.

Hence, we can say that the Factor of given equation [tex]4y(3x+2)-2(3x+2)[/tex]   is    [tex](2y-1) (6x+4)[/tex] .

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Answer:

4, see below :)

Step-by-step explanation:

We can rewrite the equation into:

3.5 + 2 - 1.5

First we add 3.5 and 2

3 and 2 are in the ones place so:

5, and then you added the .5

5.5

Then:

5.5 - 1.5

First we subtract the tenths place so:

.5 - .5 = 0

Then the ones:

5 - 1 = 4

Answer:

4 all i did was, i used a calculator

Answer:

This question is so bad but it is most likely 25% or 0. this has no context in the question

Step-by-step explanation:

The matrix is not properly formatted.

However, I'm able to rearrange the question as:

[tex]\left[\begin{array}{ccc}1&1&1|-1\\-2&3&5|3\\3&2&4|1\end{array}\right][/tex]

Operations:

[tex]2R_1 + R_2 ->R_2[/tex]

[tex]-3R_1 +R_3 ->R_3[/tex]

Please note that the above may not reflect the original question. However, you should be able to implement my steps in your question.

Answer:

[tex]\left[\begin{array}{ccc}1&1&1|-1\\0&5&7|1\\0&-1&1|4\end{array}\right][/tex]

Step-by-step explanation:

The first operation:

[tex]2R_1 + R_2 ->R_2[/tex]

This means that the new second row (R2) is derived by:

Multiplying the first row (R1) by 2; add this to the second row

The row 1 elements are:

[tex]\left[\begin{array}{ccc}1&1&1|-1\end{array}\right][/tex]

Multiply by 2

[tex]2 * \left[\begin{array}{ccc}1&1&1|-1\end{array}\right] = \left[\begin{array}{ccc}2&2&2|-2\end{array}\right][/tex]

Add to row 2 elements are: [tex]\left[\begin{array}{ccc}-2&3&5|3\end{array}\right][/tex]

[tex]\left[\begin{array}{ccc}2&2&2|-2\end{array}\right] + \left[\begin{array}{ccc}-2&3&5|3\end{array}\right][/tex]

[tex]\left[\begin{array}{ccc}0&5&7|1\end{array}\right][/tex]

The second operation:

[tex]-3R_1 +R_3 ->R_3[/tex]

This means that the new third row (R3) is derived by:

Multiplying the first row (R1) by -3; add this to the third row

The row 1 elements are:

[tex]\left[\begin{array}{ccc}1&1&1|-1\end{array}\right][/tex]

Multiply by -3

[tex]-3 * \left[\begin{array}{ccc}1&1&1|-1\end{array}\right] = \left[\begin{array}{ccc}-3&-3&-3|3\end{array}\right][/tex]

Add to row 2 elements are: [tex]\left[\begin{array}{ccc}3&2&4|1\end{array}\right][/tex]

[tex]\left[\begin{array}{ccc}-3&-3&-3|3\end{array}\right] + \left[\begin{array}{ccc}3&2&4|1\end{array}\right][/tex]

[tex]\left[\begin{array}{ccc}0&-1&1|4\end{array}\right][/tex]

Hence, the new matrix is:

[tex]\left[\begin{array}{ccc}1&1&1|-1\\0&5&7|1\\0&-1&1|4\end{array}\right][/tex]

Answer:

Step-by-step explanation:

:)

Answer:

3 days

Step-by-step explanation: (See Annex)

According to the problem statement, both readings (Isabela and Angel) are models for a straight line:

10 and 20 are the slopes of I and A respectively

I = 35 + 10*t

A = 5 + 20*t

The common point of the two lines will be the day when they both will be on the same page ( see annex)

then:

35 + 10*t  =  5 +20*t

35 - 5  =  20*t - 10*t

10*t  =  30

t  = 3

Answer:

Option C is correct

Step-by-step explanation:

Plz help answer my question my lattes question plz

Answer:

C

Step-by-step explanation:

Answer:

3x+4=3*2+4

3*2+4=6+4

6+4=10

10=10

Answer:

B. 68pi

Step-by-step explanation:

to find the surface area of a right cone given radius and slant height:

let A = surface area

let r = radius

let l = slant height.

A=pi*r*l + pi* r^2

plug in the values

A=pi*52+pi*16= about 213.62

The surface area of the cone is 68π sq. units.

The surface area of a figure is the number of unit squares it takes to cover the entire surface without gaps or overlaps. If a three-dimensional figure has flat sides, the sides are called faces. The surface area is the total of the areas of the faces.

Given that, the slant height and the radius of a cone is 13 units and 4 units respectively,

We need to find the surface area of the cone,

SA = π×r×l + π×r²

SA = π(4×13+16)

SA = 68π

Hence, the surface area of the cone is 68π sq. units.

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Answer:

x = 20

Step-by-step explanation:

The sum of angle of a triangle is 180. Taking the sum of the angles and equating to 180 will give:

75 + 63 + 4x+2 = 180

138+2+4x = 180

140+2x = 180

2x = 180=140

2x = 40

x = 20

Hence the value of x is 20

Answer:

Solution in photo

Step-by-step explanation:

Answer:

b.c

Step-by-step explanation:

Answer:

3) Because the run go by 4 and the rise go up by 2

Step-by-step explanation:

Answer:

y = 80

Step-by-step explanation:

the angles of a triangle = 180. the top angle is 80 and the bottom is 20. 180 - 20 - 80 = 80

Answer:

Following are the solution to this question:

Step-by-step explanation:

[tex]\to tellurium-131\\\\ \to T(t) =1000 \times e^{\frac{\log(2)}{25}t}[/tex]

where t = in miles

In part A:

When scientists got t=0, therefore the tellurium quantity.

[tex]\to T(t) = 1000 \times e^{0}[/tex]

           [tex]= 1000 \times 1 \\\\= 1000 \ grams[/tex]

In part B:

When 1 hours =  60 min  

[tex]T(t) = 1000 \times e^{\frac{\log 2}{25} \times 60} \\\\[/tex]

      [tex]= 1000 \times e^{\frac{\log 2}{5} \times 12}\\\\= 1000 \times e^{0.72247199}\\\\\= 1000 \times 2.05951803\\\\=2059.51803\\\\=2059.51 \ grams[/tex]

Task Radioactive isotopes decay by emitting radiation is that :

A) The tellurium-131 were in the sample the moment the scientist received are 1000 grams.

B)The grams of tellurium-131 will remain in the sample after one hour is 2059.51.

Given :

Part A:

When scientists got t=0,

t=1000*[tex]e^{o}[/tex]

t=1000*1

t=1000 grams

The tellurium-131 were in the sample the moment the scientist received are 1000 grams.

Part B:

The grams of tellurium-131 will remain in the sample after one hour is :

The grams of tellurium-131 will remain in the sample after one hour is 2059.51.

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Answer:

a. 100

b. 156

Step-by-step explanation:

a.

(n - 2)180 = (6 - 2)180 = 4(180) = 720

? + 166 + 108 + 121 + 135 + 90 = 720

? = 100

b.

(n - 2)180 = (8 - 2)180 = 6(180) = 1080

? + 128 + 99 + 161 + 113 + 142 + 146 + 135 = 1080

? = 156

Answer:

51 months

explanation

Answer:

The right answer is

[tex]y = 2x[/tex]