I NEED HELP ON A TEST HELP QUICK

To solve an equation, we can only have one variable. In this case, we are given the system 2x = 5y + 14. However, we are given the value of x in terms of y. We can substitute 10y in for 2x and solve.

10y = 5y + 14

First, we need to get all the variables on one side of the equation, so we subtract 5y.

5y = 14

Divide by 5

y = 14/5 or 2 4/5 or 2.8

Now for the x part.

2x = x + 14

Subtract 1x

x = 14

Now we check our answer.

2(14) = 5(2.8) + 14

28 = 14 + 14

28 = 28

Because 28 equals 28, our solutions are correct.

7 0

3 years ago

Suppose that you are headed toward a plataeu 70m high. If the angle of elevation to the top of the platau is 35 degree how far a

SIZIF [17.4K]

Answer:

The distance you are from the base of plateau = $99.97\ m\approx 100\ m$

Step-by-step explanation:

Given:

Height of plateau = 70 m

Angle of elevation to the top of plateau = 35°

To find the distance you are from the base of plateau.

We will construct a triangle ABC to model the given situation. The triangle would be a right triangle for which we know an angle and its opposite side. We need to find the adjacent side of the triangle.

We will apply trigonometric ratio to find the adjacent side.

$\tan\theta=\frac{Opposite\ Side}{Adjacent\ Side}$

where $\theta$ represents the angle of reference.

Plugging in the values from the triangle.

$\tan 35\° =\frac{BC}{AC}$

$\tan 35\° =\frac{70}{AC}$

Multiplying both sides by AC.

$AC\times \tan 35\° =AC\times \frac{70}{AC}$

$AC\times \tan 35\° =70$

Dividing both sides by $\tan 35\°$

$\frac{AC\times \tan 35\°}{\tan 35\°} =\frac{70}{\tan 35\°}$

∴ $AC=99.97\ m\approx 100\ m$

The distance you are from the base of plateau = $99.97\ m\approx 100\ m$

7 0

3 years ago

From home, Mary’s work is two thirds along the way to training. Training is 2.5km from work. Mary normally goes to work, then tr

polet [3.4K]

First thing to do is to illustrate the problem, Since it was mentioned that work was along the way to training, the order is shown in the picture. Mary's home and workplace are nearer compared to her training center. It is also mentioned that the distance between work and home, denoted as x, is 2/3 of the total distance from home to training. The total distance is (x + 2.5). Thus,

x = 2/3(x+2.5)
x = 2/3 x + 5/3
1/3 x = 5/3
x = 5 km

Thus, the distance from home to work is 5 km. This means that Mary has to walk this distance twice to return home to get her shoes. Then, she will travel again the total distance of 5+2.5 = 7.5 km to get to her training center. So,

Total distance = 2(5km) + 7.5 km
Total distance = 17.5 km

3 0

3 years ago