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3 years ago

The scale on the map is 1.:10 mi. How many miles will 5 represent

Mathematics

1 answer:

vovangra [49]3 years ago

5 0

Answer:

50 miles

Step-by-step explanation:

Given the scale :

1 : 10 miles

5 will represent x miles

1 = 10

5 = x

Cross multiply :

x = 5 * 10

x = 50 miles

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How many solutions are there for the system x^2+4y^2=100 and 4y-x^2=-20

never [62]

There are 5 solutions for this system.

x^2 + 4y^2 = 100 ____1
4y - x^2 = -20 ____2
Add both 1 & 2 together. x^2 gets cancelled
4y^2 + 4y = 80 (send 80 to the other side and divide by 4)
Then equation the becomes : y^2 + y -20 =0
Now factorise the equation: (y+5) (y-4) = 0
Solve for y : y = -5 and y = 4
Using the values of y to find the values of x. From equation 1:
x^2 = 100 - 4y^2 x = /100 - 4y^2 (/ means square root) Replace values of y
y = -5, x = /100 - 4(-5)^2 = /100 - 100 = 0
y = 4, x = /100 - 4(4)^2 = / 100 - 64 = /36 = -6 or 6
Thus we have 6 solutions y = -5, 4 and x = -6, 0, 6

6 0

3 years ago

Can someone can help me plsssssss

Darina [25.2K]

The answer is letter A

8 0

3 years ago

Read 2 more answers

Flying against the wind, a jet travels 2190 miles in 3 hours. Flying with the wind, the same jet travels 9520 miles in 8 hours.

dezoksy [38]

Answer:

Velocity of jet in still air is 970 miles per hour and velocity of wind is 210 miles per hour.

Step-by-step explanation:

Jet's velocity against wind is

3040/4 = 760 miles per hour and flying with wind it is

8260/7 = 1180 miles per hour.

Let the velocity of jet in still air be x miles per hour and velocity of wind be y miles per hour.

As such its velocity against wind is x − y and with wind is x + y and therefore

x − y = 760 and x + y = 1180

Adding the two 2 x = 1940 and x = 970 and y = 1180 − 970 = 210

Hence velocity of jet in still air is 970 miles per hour and velocity of wind is

210 miles per hour.

Actual Answer:

https://socratic.org/questions/flying-against-the-wind-a-jet-travels-3040-miles-in-4-hours-flying-with-the-wind

7 0

3 years ago

Find the particular solution of the differential equation that satisfies the initial condition(s). f ''(x) = x−3/2, f '(4) = 1,

sweet [91]

Answer:

Hence, the particular solution of the differential equation is $y = \frac{1}{6} \cdot x^{3} - \frac{3}{4}\cdot x^{2} - x$ .

Step-by-step explanation:

This differential equation has separable variable and can be solved by integration. First derivative is now obtained:

$f'' = x - \frac{3}{2}$

$f' = \int {\left(x-\frac{3}{2}\right) } \, dx$

$f' = \int {x} \, dx -\frac{3}{2}\int \, dx$

$f' = \frac{1}{2}\cdot x^{2} - \frac{3}{2}\cdot x + C$ , where C is the integration constant.

The integration constant can be found by using the initial condition for the first derivative ( $f'(4) = 1$ ):

$1 = \frac{1}{2}\cdot 4^{2} - \frac{3}{2}\cdot (4) + C$

$C = 1 - \frac{1}{2}\cdot 4^{2} + \frac{3}{2}\cdot (4)$

$C = -1$

The first derivative is $y' = \frac{1}{2}\cdot x^{2}- \frac{3}{2}\cdot x - 1$ , and the particular solution is found by integrating one more time and using the initial condition ( $f(0) = 0$ ):