No termino de entender esta parte de la tarea me pueden ayudar porfabor

A person on tour has dollar 360 for his daily expenses. If he extends his tour for 4 days, he has to cut down his daily expenses

mash [69]

Answer:

The original duration of the tour = 20 days

Step-by-step explanation:

Solution:

Total expenses for the tour = $360

Let the original tour duration be for $x$ days.

So, for $x$ days the total expense = $360

Thus the daily expense in dollars can be given by = $\frac{360}{x}$

Tour extension and effect on daily expenses.

The tour is extended by 4 days.

Tour duration now = $(x+4)$ days

On extension, his daily expense is cut by $3

New daily expense in dollars = $(\frac{360}{x}-3)$

Total expense in dollars can now be given as: $(x+4)(\frac{360}{x}-3)$

Simplifying by using distribution (FOIL).

$(x.\frac{360}{x})+(x(-3)+(4.\frac{360}{x})+(4(-3))$

$360-3x+\frac{1440}{x}-12$

$348-3x+\frac{1440}{x}$

We know total expense remains the same which is = $360.

So, we have the equation as:

$348-3x+\frac{1440}{x}=360$

Multiplying each term with $x$ to remove fractions.

$348x-3x^2+1440=360x$

Subtracting $348x$ both sides

$348x-348x-3x^2+1440=360x-348x$

$-3x^2+1440=12x$

Dividing each term with -3.

$\frac{-3x^2}{-3}+\frac{1440}{-3}=\frac{12x}{-3}$

$x^2-480=-4x$

Adding $4x$ both sides.

$x^2+4x-480=-4x+4x$

$x^2+4x-480=0$

Solving using quadratic formula.

For a quadratic equation: $ax^2+bx+c=0$

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

Plugging in values from the equation we got.

$x=\frac{-4\pm\sqrt{(4)^2-4(1)(-480)}}{2(1)}$

$x=\frac{-4\pm\sqrt{16+1920}}{2}$

$x=\frac{-4\pm\sqrt{1936}}{2}$

$x=\frac{-4\pm44}{2}$

So, we have

$x=\frac{-4+44}{2}$ and $x=\frac{-4-44}{2}$

$x=\frac{40}{2}$ and $x=\frac{-48}{2}$

∴ $x=20$ and $x=-24$

Since number of days cannot be negative, so we take $x=20$ as the solution for the equation.

Thus, the original duration of the tour = 20 days

6 0

3 years ago

Someone help plss show work

Mazyrski [523]

Answer:

Slope: $\frac{3}{5}$

Y-Intercept: $(0,-5)$

Step-by-step explanation:

1. To find the slope, we need to have the equation in slope-intercept form, which is $y=mx+b$ . In this case, our equation is already in this form. $m$ is what your slope is. In this case, our

2. To find the y-intercept we look at value $b$ that is in $y=mx+b$ . In this case, $b$ is $-5$ . So, our ordered pair is $(0,-5)$ .

4 0

3 years ago

Read 2 more answers

Help please ASAP thank you!!!

almond37 [142]

Answer:

B, coefficient vertically stretches or compress not horizontally, plus it is negative so x axis dlip

8 0

3 years ago

Read 2 more answers

Justin is constructing a line through point Q that is perpendicular to line n. He has already constructed the arcs shown. He pla

krek1111 [17]

B. It must be the same as when he constructed the arc centered at point A. This problem would be a lot easier if you had actually supplied the diagram with the "arcs shown". But thankfully, with a few assumptions, the solution can be determined. Usually when constructing a perpendicular to a line through a specified point, you first use a compass centered on the point to strike a couple of arcs on the line on both sides of the point, so that you define two points that are equal distance from the desired intersection point for the perpendicular. Then you increase the radius of the compass and using that setting, construct an arc above the line passing through the area that the perpendicular will go. And you repeat that using the same compass settings on the second arc constructed. This will define a point such that you'll create two right triangles that are reflections of each other. With that in mind, let's look closely at your problem to deduce the information that's missing. "... places his compass on point B ..." Since he's not placing the compass on point Q, that would imply that the two points on the line have already been constructed and that point B is one of those 2 points. So let's look at the available choices and see what makes sense. A .It must be wider than when he constructed the arc centered at point A. Not good. Since this implies that the arc centered on point A has been constructed, then it's a safe assumption that points A and B are the two points defined by the initial pair of arcs constructed that intersect the line and are centered around point Q. If that's the case, then the arc centered around point B must match exactly the setting used for the arc centered on point A. So this is the wrong answer. B It must be the same as when he constructed the arc centered at point A. Perfect! Look at the description of creating a perpendicular at the top of this answer. This is the correct answer. C. It must be equal to BQ. Nope. If this were the case, the newly created arc would simply pass through point Q and never intersect the arc centered on point A. So it's wrong. D.It must be equal to AB. Sorta. The setting here would work IF that's also the setting used for the arc centered on A. But that's not guaranteed in the description above and as such, this is wrong.

8 0

3 years ago

Read 2 more answers

Helene wrote a report. It took her 4 hours to write 24 pages.

sammy [17]

Answer:

C

Step-by-step explanation:

24÷6= 6 Hope this helps.

4 0

3 years ago