All categories

2 years ago

HELP

1 answer:

5 0

Given the radius of 50 miles and the line joining the cities at (0, 56) and (58, 0), the transmitter signal can be picked during 59.24 miles of the drive.

<h3>How can the duration of signal reception be found?</h3>

Radius of broadcast of the transmitter = 50 miles

Location of starting point = 56 miles north of the transmitter

Location of destination city = 58 miles east of the transmitter

Therefore we have;

Slope of the line joining the two cities

= 56 ÷ (-58) = -0.966

Which gives the equation of the line as follows;

y = -0.966•x + 56

The equation of the circle is;

${x}^{2} + {y}^{2} = {50}^{2}$

${x}^{2} + { ( - 0.966x + 56)}^{2} = {50}^{2}$

1.933156•x^2 - 108.192•x + 636 = 0

Which gives;

x = 6.67 or x = 49.29

Therefore;

When x = 6.67, we have;

y = -0.966 × 6.67 + 56 = 49.56

When x = 49.29, we have;

y = -0.966 × 49.29 + 56 = 8.4

The length of the drive, during which the driver can pick the signal, l, is therefore;

l = √((49.56-8.4)^2 + (49.29-6.67)^2) = 59.24 miles

The length of the drive during which the signal is received is 59.24 miles

Learn more about the equation of a circle here:

brainly.com/question/502872

#SPJ1

You might be interested in

F(x)=x 3−9xf, left parenthesis, x, right parenthesis, equals, x, cubed, minus, 9, x What is the average rate of change of f over

mel-nik [20]

Answer:

$A(x) = 243$

Step-by-step explanation:

Given

$f(x) = x^3 - 9x$

$Interval: (12,6)$

Required

Determine the average rate of change

Average rate is calculated as thus;

$A(x) = \frac{f(b) - f(a)}{b - a}$

In this case; b = 12 and a = 6

Calculating f(12)

$f(12) = 12^3 - 9 * 12$

$f(12) = 1728 - 108$

$f(12) = 1620$

Calculating f(6)

$f(6) = 6^3 - 9 * 6$

$f(6) = 216 - 54$

$f(6) = 162$

Substitute 12 for b and 6 for a in $A(x) = \frac{f(b) - f(a)}{b - a}$

$A(x) = \frac{f(12) - f(6)}{12 - 6}$

Substitute values for f(6) and f(12)

$A(x) = \frac{1620 - 162}{12 - 6}$

$A(x) = \frac{1458}{ 6}$

$A(x) = 243$

4 0

3 years ago

Consider the statement : Congruent triangles are always similar. Which of the following statements is an example of the statemen

swat32

Answer:

Yes, the congruent triangles are always similar. The following statements are applicable

1) Angles are the same, but sides are proportional to eachother.

2) Sides are the same size.

3)Corresponding angles and corresponding sides are congruent.

Step-by-step explanation:

Yes, they are similar.

For two triangles to be similar, it is sufficient if two angles of one triangle are equal to two angles of the other triangle. Note that if two angles of one are equal to two angles of the other triangle, the third angles of the two triangles too will be equal.

If two triangles are congruent then all corresponding sides as well as corresponding angles of one triangle are equal to those of other triangles. This can happen in four cases

1)- when all sides of triangles are equal,

2) - if one side and two angles of one are equal to one side and two angles of other triangle.

3) - if two sides and included angle of one triangle are equal to two sides and included angle of other triangle

4)- if in two right angled triangles, one side and hypotenuse of one triangle are equal to one side and hypotenuse of other triangle.

Observe that for triangles to be similar, we just need all angles to be equal. But for triangles to be congruent, angles as well as sides should be equal.

Hence, while congruent triangles are similar, similar triangles may not be congruent ( but the converse is not true) .

4 0

3 years ago

2.8 = 2y What does y equal? 1.4 0.5 2.8 0.6

igor_vitrenko [27]

Answer: 1.4

Step-by-step explanation: First, swap the sides of the equation so that the one with the variable can be in front.

So, its 2y=2.8

To solve this, you simply divide both sides of the equation by 2.

2 divided by 2 is 0. That leaves you with the y by itself. Then, 2.8 divided by 2 is 1.4

So, y=1.4

7 0