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kati45 [8]
3 years ago
5

What is 7500×5643 ÷3​

Mathematics
2 answers:
just olya [345]3 years ago
8 0

Answer:

<h2>14107500</h2>

Step-by-step explanation:

7500 \times 5643 \\  =  \frac{42322500}{3}  \\  = 14107500

tatyana61 [14]3 years ago
7 0

Answer:

\Large \boxed{14107500}

Step-by-step explanation:

There are different ways to do this;

1. ⇒ (7500×5643)÷3

⇒ 42322500÷3

⇒ = 14107500

2. ⇒ 7500*5643/3

⇒ 42322500/3

⇒ =14107500

^ Two ways you could have solved the Q.

Answer; 14107500

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Answer:

Step-by-step explanation:

(1+r)^5=\frac{535.29}{400} \\1+r=(\frac{535.29}{400}) ^{\frac{1}{5} } \\1+r \approx 1.06\\r \approx .06

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Michael Jordan is 6'6'' tall and could jump 4 feet of vertical distance off the ground.A basketball net's rim is 3.9 meters off
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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
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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



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