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wel
3 years ago
12

The distance by road from town A to town B is 257 km. What is 50% of that distance?

Mathematics
1 answer:
Marina86 [1]3 years ago
3 0

Answer: 128.5km

Step-by-step explanation:

Since we are given the information that the distance by road from town A to town B is 257 km. To get 50% of the distance, we simply have to multiply the distance given by 50%. This will be:

= 50% × 257km

= 50/100 × 257km

= 0.5 × 257km

= 128.5km

Therefore, 50% of the distance is 128.5km.

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A car goes 400 miles on 10 gallons of
Olenka [21]

Answer:

A. 1200

Step-by-step explanation:

Since 10 times 3 is 30, then 400 times 3 is 1200. Hope I helped!

4 0
4 years ago
Find the area and the circumference of a circle with radius 6 m.
Ratling [72]

Answer:

(a) 113.04m²

(b) 37.68m

Step-by-step explanation:

(a)

Area = πr²

=π(6)²

=36π

=113.04m²

(b)

Circumference = 2πr

2π(6) = 12π

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6 0
3 years ago
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Greg is designing a clock face. Using the center of the clock face as the origin, he keeps its diameter at 10 units. Match the p
Nat2105 [25]

The center of the clock is taken as the origin.The clock is a circle with a diameter 10 units.Radius is half the diameter .Radius = 10 ÷2= 5 units.

The clock is divided in four quadrants .On x axis y=0 and on y axis x=0.

When it is  12 o'clock the hour hand is on positive of y axis.Coordinates of the point at 12 o'clock=(0,5)

When it is 3 o 'clock the hour hand is on positive of x axis .Coordinate of the point at 3o'clock is (5,0)

When it is 6 o'clock the hour hand is on negative of y axis .The coordinates of the point at 6o'clock is (0,-5)

At 9o'clock the hour hand is on negative of x axis .The coordinate of the point at 6o'clock is(-5,0)

3 0
3 years ago
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Help me out guys!!!!!!!​
Dmitrij [34]

Answer:

\frac{5 \sqrt{11} }{11}

you are correct !!

3 0
2 years ago
Given: ABC is a right triangle with right angle C. AC=15 centimeters and m∠A=40∘ . What is BC ? Enter your answer, rounded to th
konstantin123 [22]

In order to answer this question, the figure in the first picture will be helpful to understand what a right triangle is. Here, a right angle refers to 90\°.


However, if we want to solve the problem we have to know certain things before:


In the second figure is shown a general right triangle with its three sides and another given angle, we will name it \alpha:


  • The side <u>opposite to the right angle</u> is called The Hypotenuse (h)
  • The side <u>opposite to the angle \alpha</u> is called the Opposite (O)
  • The side <u>next to the angle \alpha</u> is called the Adjacent (A)

So, going back to the triangle of our question (first figure):


  • The Hypotenuse is AB
  • The Opposite is BC
  • The Adjacent is AC

Now, if we want to find the length of each side of a right triangle, we have to use the <u>Pythagorean Theorem</u> and T<u>rigonometric Functions:</u>


Pythagorean Theorem


h^{2}=A^{2} +O^{2}    (1)  


Trigonometric Functions (here are shown three of them):


Sine: sin(\alpha)=\frac{O}{h}    (2)


Cosine: cos(\alpha)=\frac{A}{h}    (3)


Tangent: tan(\alpha)=\frac{O}{A}   (4)



In this case the function that works for this problem is cosine (3), let’s apply it here:


cos(40\°)=\frac{AC}{h}    


cos(40\°)=\frac{15}{h}    (5)


And we will use the Pythagorean Theorem to find the hypotenuse, as well:



h^{2}=AC^{2}+BC^{2}    


h^{2}=15^{2}+BC^{2}    (6)


h=\sqrt{225+BC^2}   (7)



Substitute (7) in (5):


cos(40\°)=\frac{15}{\sqrt{225+BC^2}}    


Then clear BC, which is the side we want:


{\sqrt{225+BC^2}}=\frac{15}{cos(40\°)}


{{\sqrt{225+BC^2}}^2={(\frac{15}{cos(40\°)})}^2


225+BC^{2}=\frac{225}{{(cos(40\°))}^2}


BC^2=\frac{225}{{(cos(40\°))}^2}-225


BC=\sqrt{158,41}


BC=12.58


Finally BC is approximately 13 cm



7 0
4 years ago
Read 2 more answers
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