B is the midpoint of AC. Solve for the length of AB if AB=5x+3 and BC=7x+9

Answer:

Graph of line going through (0, 8) with a slope of 1.5  represents Ariel's remaining entries

Step-by-step explanation:

Ariel wants to write 8 entries for her blog.

After 2 hours, she has 5 entries left. [tex](x_1,y_1)=(2,5)[/tex]

After 4 hours, she has 2 entries left. [tex](x_2,y_2)=(4,2)[/tex]

So, now using two point slope form:

[tex]y-y_1=m(x-x_1)[/tex] ---1

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

So , substituting the given points we get the equation that represents the given situation .

[tex]m = \frac{2-5}{4-2}[/tex]

[tex]m = \frac{-3}{2}[/tex]

[tex]m = -1.5[/tex]

Now,

[tex]y-5 = -1.5(x-2)[/tex]

[tex]y-5 = -1.5x+3[/tex]

Thus the slope of the given situation is 1.5

Now at x = 0

[tex]y-5 = -1.5*0+3[/tex]

[tex]y=8[/tex]

So, at 0 hour 8 entries left .

So, graph of line going through (0, 8) with a slope of 1.5  represents Ariel's remaining entries

Within a specific stretch spanning 5 miles, the roadway inclines approximately 0.53 miles or 2771.64 feet, rounded to the nearest hundredth of a mile.

Trigonometric principles aid in determining the elevation gained by the highway over this distance.

Given:

An angle of ascent (θ) of 6 degrees and a horizontal distance (adjacent side) of 5 miles, our objective is to ascertain the vertical distance (opposite side).

Utilizing the tangent function in trigonometry:

[tex]\[ \tan(\theta) = \frac{{\text{{opposite}}}}{{\text{{adjacent}}}} \][/tex]

[tex]\[ \tan(6^\circ) = \frac{{\text{{height}}}}{{5}} \][/tex]

The calculation for the height is:

height = 5 × tan(6°)

Plugging in the values:

height ≈ 5 × 0.1051

height ≈ 0.5255 miles

To convert this to feet (1 mile = 5280 feet):

height ≈ 0.5255 × 5280  

height ≈ 2771.64 feet

Thus, within this 5-mile segment, the roadway ascends approximately 0.53 miles or 2771.64 feet, rounded to the nearest hundredth of a mile.

Answer:

Option 1,3 and 4 are correct.

Step-by-step explanation:

Given is a quadratic equation

[tex]x^2-10x+25=0[/tex]

Marcus used quadratic formula to solve this

The correct use of formula is

[tex]\frac{-b+/\sqrt{b^2-4ac} }{2a}[/tex]

for the quadratic equation [tex]ax^2+bx+c=0[/tex]

Comparing we have a =1, b =-10: c=25

The student should have substituted -10 for b but he wrong did only 10

II option is wrong since 2 is substituted correctly

III option is right -4ac i.e. the product x 4 must have been subtracted

IV option is right since repeated root is there so only one solution

Answer:

Linear equation is V(t) = 1500 - 290t

Step-by-step explanation:

To find the linear equation representing the value of the robot we will calculate per year depreciation in the cost of the robot.

Rate of depreciation = [tex]\frac{\text{(Manufacturing cost in 2005-selling price in 2010)}}{5}[/tex]

                                 = [tex]\frac{[(1.5)(10^{6})-50000]}{5}[/tex]

                                 = [tex]\frac{(1500000-50000)}{5}[/tex]

                                 = [tex]\frac{1450000}{5}=290000[/tex]

If depreciation is represented by V(t) (in $000) then equation will be

V(t) = 1500 - 290t

Answer:

[tex]1+\sqrt{2}[/tex]

Step-by-step explanation:

We have been given a radical expression [tex]\sqrt{3+2\sqrt{2}}[/tex]. We are asked to simplify the given expression.

We will add [tex](\sqrt{2})^2-2[/tex] to our given expression as after adding and subtracting same quantity the value of our expression will be same.

[tex]\sqrt{3+2\sqrt{2}+(\sqrt{2})^2-2}[/tex]

[tex]\sqrt{3-2+2\sqrt{2}+(\sqrt{2})^2}[/tex]

[tex]\sqrt{1+2\sqrt{2}+(\sqrt{2})^2}[/tex]

Using perfect square formula [tex](a+b)^2=a^2+2ab+b^2[/tex] we can rewrite our expression as:

[tex]\sqrt{1^2+2\sqrt{2}\cdot 1+(\sqrt{2})^2}[/tex]

[tex]\sqrt{(1+\sqrt{2})^2}[/tex]

Applying radical rule [tex]\sqrt[n]{x^n} =x[/tex], we will get,

[tex](1+\sqrt{2})^{\frac{2}{2}}=1+\sqrt{2}[/tex]

Therefore, the simplified form of our given expression would be [tex]1+\sqrt{2}[/tex].

The converse of the statement 'If two lines are parallel, they do not intersect' is 'If two lines do not intersect, they are parallel', which is false. A counterexample to this converse is skew lines that do not intersect but are not parallel. Thus, we can't write both statements as a biconditional

The subject of your question is regarding conditional statements and their converses in mathematics. A conditional statement is often expressed in the form: 'If p, then q', where p is a hypothesis and q is a conclusion. The converse of a conditional statement switches the hypothesis and the conclusion, i.e., 'If q, then p'.

In the context of your question, the conditional statement given is 'If two lines are parallel, they do not intersect'. The converse of this statement is 'If two lines do not intersect, they are parallel'. However, this converse is not true. A counterexample can be two lines that do not intersect but are not parallel because they are skew lines, which exist in three dimensions and do not lie in the same plane. Thus, the statements can't be written as a biconditional as one is not a valid conditional.

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Answer:

Ryan use 4. 50 yards of cloth to make 6 backpack.

Step-by-step explanation:

Given : ryan makes 6 backpacks he uses 3/4 of cloth to make each backpack.

To find  : what is the total amount cloth ,in yards ryan use to make all 6 backpacks.

Solution : We have given that

Ryan make a backpack to use =  [tex]\frac{3}{4}[/tex] of a cloth.

Ryan make such 6 backpack = [tex]6*\frac{3}{4}[/tex]

                                                =   [tex]\frac{18}{4}[/tex] .

                                               = 4. 50 yards.

Therefore, Ryan use 4. 50 yards of cloth to make 6 backpack.

If the repair took 5 hours, the total cost would be $250.

To find the total cost if the repair took 5 hours, we need to evaluate the function[tex]\(C(x)\) at \(x = 5\).[/tex]

Given that [tex]\(C(x) = 35x + 75\),[/tex] we substitute [tex]\(x = 5\)[/tex] into the function:

[tex]\[C(5) = 35 \times 5 + 75\]\[C(5) = 175 + 75\]\[C(5) = 250\][/tex]

So, if the repair took 5 hours, the total cost would be $250.

Complete correct question:

The function C(x) = 35x + 75 represents the labor cost (in dollars) for Bob's Auto Repair to replace our alternator, where x is the number of hours of labor. What would be the total cost if the repair took 5 hours?

Answer with explanation:

The given equation which represents a circle having center at, (2,0) and radius =8 is,

 [tex](x - 2)^2 + y^2 = 64\\\\y^2=64 - (x-2)^2\\\\y^2=64 - (x^2-4 x +4)\\\\ y^2=64 -x^2+4 x-4\\\\y^2=60+4 x-x^2[/tex]

Result  of isolating y² in the equation below, gives

   Option A

            [tex]y^2=60+4 x-x^2[/tex]

Answer:

yes

Step-by-step explanation:

When performing certain statistical calculations, finding roots, or working with powers, calculators or computers are necessary for the calculations; otherwise, more traditional methods will be sufficient.

Answer:

The pressure of the gas 0°C is about 699.

Step-by-step explanation:

The given table of values is

Temperature      Pressure

     10                      723

     25                     763

     40                     800

     55                     845

     80                      900

The general form of a trend line is

[tex]y=a+bx[/tex]

where, b is slope and a is y-intercept.

[tex]b=\frac{\sum_{i=1}^nx_iy_i-n\overline{x}\overline{y}}{\sum_{i=1}^nx_i^2-n\overline{x}^2}[/tex]

[tex]a=\overline{y}-b\overline{x}[/tex]

Using graphing calculator, we get

[tex]a=699.006826[/tex]

[tex]b=2.552218[/tex]

The equation of line of best bit is

[tex]y = 2.552218x + 699.006826[/tex]

where, y is pressure at x°C.

We need to find the pressure of the gas at 0°C.

Substitute x=0 in the above equation.

[tex]y = 2.552218(0) + 699.006826[/tex]

[tex]y =699.006826[/tex]

[tex]y \approx 699[/tex]

Therefore the pressure of the gas 0°C is about 699.