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ryzh [129]
2 years ago
11

35/[(13+7)/(5-1)]+15=

Mathematics
1 answer:
Ket [755]2 years ago
3 0

Answer:

nose lo lamento mucho de no poder ayudar

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2. Determine the value of x. Assume the segments that appear tangent
dimulka [17.4K]

Answer:

x = 16.8

Step-by-step explanation:

To get the value of x, we use Pythagoras’ theorem since what we have is a right angled triangle

Mathematically, we have it that;

the square of the hypotenuse (21) equals the sum of the squares of the two other sides

21^2 = 12.6^2 + x^2

x^2 = 21^2 - 12.6^2

x^2 = 282.24

x = √282.24)

x = 16.8

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3 years ago
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7 0
2 years ago
How do I solve this equation? 3|13-2t|=15
Nikolay [14]
3 |13 - 2t | =15 ⇒ |13-2t| = 15\div3 ⇒ \boxed {|13-2t| = 5}

\hbox {We have two equations : 1. } |13-2t|=5 
                                               2.|13-2t|=-5

\hbox{The first: } ⇒ 13-2t=5 ⇒ 2t=13-5 ⇒ 2t=8 ⇒ \boxed {t=4}

\hbox { The second: } 13-2t=-5 ⇒ 2t=13-(-5)=18 ⇒ \boxed {t=9}
5 0
3 years ago
Read 2 more answers
P and q are integers that are multiple of 5. Which of this
zhuklara [117]

Answer:

we conclude that the only option (a) is true.

Step-by-step explanation:

As we know that the multiples of 5 are the numbers which we get when we multiply by 5.

i.e.

5×1=5

5×2=10

Here, 5 and 10 are multiples of 5.

Let p and q are integers that are multiples of 5.

Let us consider

p=5

q=10

so

p+q=5+10

     = 15

A number is divisible by 5 if it ends in 5 or 0.

i.e. 15/5 = 3

so p+q is divisible by 3 as there is no remainder left.

Therefore, option (a) is true.

Checking the other options:

(b) P –q is divisible by 10

As

p=5

q=10

so

p-q=5-10

     = -5

Numbers that are divisible by 10 need to be even or divisible by 2 and divisible by 5.

As -5 is not divisible by 2.

So, option b is NOT true.

(c) P +q is divisible by 20

As

p=5

q=10

so

p-q=5+10

     = 15

Divisibility rule of  20  implies that the last two digits of the number are either  00  or divisible by  20 .

Therefore, P + q= 20 is not divisible by 20 as we don't get the whole number.

(d) P + q is divisible by 25​

As

p=5

q=10

so

p-q=5+10

     = 15

p+q=15 is not divisible by 25 as it does not end with 00, 25, 50, or 75.

so, option d is NOT correct.

Therefore, we conclude that the only option (a) is true.

5 0
2 years ago
How many solutions does the system of equations below have?
soldier1979 [14.2K]

Answer:

One solution                    

Step-by-step explanation:

5x + y = 8

15x + 15y = 14

Lets solve using substitution, first we need to turn "5x + = 8" into "y = mx + b" or slope - intercept form

So we solve for "y" in the equation "5x + y = 8"

5x + y = 8

Step 1: Subtract 5x from both sides.

5x + y − 5x = 8 − 5x

Step 2: 5x subtracted by 5x cancel out and "8 - 5x" are flipped

y = −5x + 8

Now we can solve using substitution:

We substitute "-5x + 8" into the equation "15x + 15y = 14" for y

So it would look like this:

15x + 15(-5x + 8) = 14

Now we just solve for x

15x + (15)(−5x) + (15)(8) = 14(Distribute)

15x − 75x + 120 = 14

(15x − 75x) + (120) = 14(Combine Like Terms)

−60x + 120 = 14

Step 2: Subtract 120 from both sides.

−60x + 120 − 120 = 14 − 120

−60x = −106

Divide both sides by -60

\dfrac{ -60x  }{ -60  }   =   \dfrac{ -106  }{ -60  }

Simplify

x =   \dfrac{ 53  }{ 30  }

Now that we know the value of x, we can solve for y in any of the equations, but let's use the equation "y = −5x + 8"

\mathrm{So\:it\:would\:look\:like\:this:\ y =  -5 \left(  \dfrac{ 53  }{ 30  }    \right)  +8}

\mathrm{Now\:lets\:solve\:for\:"y"\:then}

y =  -5 \left(  \dfrac{ 53  }{ 30  }    \right)  +8}

\mathrm{Express\: -5 \times   \dfrac{ 53  }{ 30  }\:as\:a\:single\:fraction}

y =   \dfrac{ -5 \times  53  }{ 30  }  +8

\mathrm{Multiply\:-5 \:and\:53\:to\:get\:-265 }

y =   \dfrac{ -265  }{ 30  }  +8

\mathrm{Simplify\:  \dfrac{ -265  }{ 30  }    \:,by\:dividing\:both\:-265\:and\:30\:by\:5} }

y =   \dfrac{ -265 \div  5  }{ 30 \div  5  }  +8

\mathrm{Simplify}

y =  - \dfrac{ 53  }{ 6  }  +8

\mathrm{Turn\:8\:into\:a\:fraction\:that\:has\:the\:same\:denominator\:as\: - \dfrac{ 53  }{ 6  }}

\mathrm{Multiples\:of\:1: \:1,2,3,4,5,6}

\mathrm{Multiples\:of\:6: \:6,12,18,24,30,36,42,48}

\mathrm{Convert\:8\:to\:fraction\:\dfrac{ 48  }{ 6  }}

y =  - \dfrac{ 53  }{ 6  }  + \dfrac{ 48  }{ 6  }

\mathrm{Since\: - \dfrac{ 53  }{ 6  }\:have\:the\:same\:denominator\:,\:add\:them\:by\:adding\:their\:numerators}

y =   \dfrac{ -53+48  }{ 6  }

\mathrm{Add\: -53 \: and\: 48\: to\: get\:  -5}

y =  - \dfrac{ 5  }{ 6  }

\mathrm{The\:solution\:is\:the\:ordered\:pair\:(\dfrac{ 53  }{ 30  }, - \dfrac{ 5  }{ 6  })}

So there is only one solution to the equation.

5 0
2 years ago
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