All categories

2 years ago

I’m really confused can anyone solve this?

Mathematics

2 answers:

Makovka662 [10]2 years ago

8 0

Answer:

x = 13

Step-by-step explanation:

Use pythagorean theorem since you have a right angle. So $c^{2} = a^{2} + b^{2}$

so in this case $c^{2} = 5^{2} + 12^{2}$ so c = $\sqrt{25+144}$ and c = $\sqrt{169}$ so c = 13

Inessa [10]2 years ago

4 0

Answer:

17 i think

Step-by-step explanation:

You might be interested in

Can someone Please help me on this easy math problem

allsm [11]

Step-by-step explanation:

Equation of a straight line

y = mx + c, m is the slope, and c the intercept

points (3,7) x = 3, y = 7, c = 1

7 = m × 3 + 1

7 = 3m + 1

3m = 7 - 1

3m = 6

m = 6/3

m = 2

6 0

2 years ago

What is the greatest common factor of 14 and 22

Alecsey [184]

Answer:

Step-by-step explanation:

greatest common factor means the greatest number we can divide by for each of the number and still get a whole number

14 and 22. they are both even so one factor is 2

if we check, 14 & 22 aren't both divisible by 3,4,5,6,7,8,9,10,11,12,13,14 so the gcf is 2

4 0

3 years ago

Read 2 more answers

Your task is to build a road joining a ranch to a highway that enables drivers to reach the city in the shortest time. The perpe

lubasha [3.4K]

Answer:

(a)In the attachment

(b)The road of length 35.79 km should be built such that it joins the highway at 19.52km from the perpendicular point P.

Step-by-step explanation:

(a)In the attachment

(b)The distance that enables the driver to reach the city in the shortest time is denoted by the Straight Line RM (from the Ranch to Point M)

First, let us determine length of line RM.

Using Pythagoras theorem

$|RM|^{2}=30^2+x^2\\|RM|=\sqrt{30^2+x^2}$

The Speed limit on the Road is 60 km/h and 110 km/h on the highway.

Time Taken = Distance/Time

Time taken on the road $=\frac{\sqrt{30^2+x^2}}{60}$

Time taken on the highway $=\frac{50-x}{110}$

Total time taken to travel, T $=\frac{\sqrt{30^2+x^2}}{60}+\frac{50-x}{110}$

Minimum time taken occurs when the derivative of T equals 0.

$T^{'}=\frac{x}{60\sqrt{30^2+x^2}}-\frac{1}{110}\\\frac{x}{60\sqrt{30^2+x^2}}-\frac{1}{110}=0\\\frac{x}{60\sqrt{30^2+x^2}}=\frac{1}{110}\\110x=60\sqrt{30^2+x^2}\\$

Square both sides

$12100x^2=3600(30^2+x^2)\\12100x^2=3240000+3600x^2\\12100x^2-3600x^2=3240000\\8500x^2=3240000\\x^2=\frac{3240000}{8500} =381.18\\x=\sqrt{381.18} =19.52$