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yaroslaw [1]
3 years ago
10

What is 5 less than the product of 10 and c?

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

Hi, the equation you typed is 10c-5

less than is subtraction and the product of two numbers is multiplying them together.

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What is the 15th term in the sequence using the given formula?<br><br> yn = 12n-1
Veseljchak [2.6K]

Answer:

Step-by-step explanation:

hello :

yn = 12n-1

the 15th term is :y15 when x= 15

y15 = 12(15)-1

y15 = 179

7 0
3 years ago
What methods do you know that can be used to solve a quadratic equation?
zhuklara [117]

Answer: If the quadratic equation has real, rational solutions, the quickest way to solve it is often to factorise into the form (px + q)(mx + n), where m, n, p and q are integers. This is especially true where the coefficient of x2 is 1.

Example 1 - Solve x2+7x+12=0

Step-by-step explanation:

that's the only one I remember

8 0
3 years ago
Read 2 more answers
Currently, the sum of the ages of Yumi, Rana and Victoria is 42 years. Four years ago, the sum of the ages of Rana and Victoria
Setler [38]
Y + R + V = 42
but 4 YEARS AGO : (R-4) +(V-4) = Y

Replace Y in the 1st equation by:  (R-4) +(V-4)

(R-4) +(V-4) + R + V = 42

2R - 2V - 8 = 42;  2R -2V  = 50, or R+V=25
IF R+V =25 AND Y + R +V = 42, then Y + 25 = 42 and Y =17

7 0
3 years ago
PLZ HELP, AND PLZ EXPLAIN
shutvik [7]

Answer:

C

Step-by-step explanation:

To make it easy let's start by organizing our information :

  • AC=12 AND BD=8
  • ABCD is a rhombus
  • K and L are the midpoints of sides AD and CD
  • we notice that the rhombus ABCD is divided into four right triangles

What do you think of when you hear a right triangle ?

  • The pythagorian theorem !

AC and BD  are khown so let's focus on them .

If we concentrated we can notice that AB and BD are cossing each other in the midpoints . why ?

Simply because they are the diagonals of a rhombus .

ow let's apply the pythagorian theorem :

  • (AC/2)² + (BD/2)² = BC²
  • 6²+4²=52
  • BC²= 52⇒\sqrt{52}=BC

Now we khow that : AB=BC=CD=AD=\sqrt{52}

This isn't enough . Let's try to figure out a way to calculate the length of KL  wich is the base of the triangle

  • KL is parallel to AC
  • k is the midpoint of AD and L of DC

I smell something . yes! Thales theorem

  • KL/AC=DL/DC=DK/AD WE4LL TAKE OLY ONE
  • KL/12=\sqrt{52}/2*\sqrt{52}  
  • KL/12=1/2⇒ KL=6

Now we have the length of the base kl

Now the big boss the height :

  • notice that you khow the length of KL
  • BD crosses kl from its midpoint and DL = \sqrt{52} /2

What I want to do is to apply the pythgorian thaorem to khow the lenght of that small part that is not a part of the height of the triangle . I will call it D

  • DL²=(KL/2)²+D²
  • 52/4= 9+ D²
  • D² = 52/4-9 +4 SO D=2

now the height of the trigle is H= BD-D= 8-2=6

NOw the area of the triangle is :

  • A=(KL*H)/2 ⇒ A= (6*6)/2=18

THE ANSWER IS 18 SQ.UN

5 0
3 years ago
Help plz I really need help could you show your work if you do it plz.
Vika [28.1K]
What do you need help with?
7 0
3 years ago
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