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Alik [6]
3 years ago
11

Write the equation in standard form 3x + 1 = 2y

Mathematics
1 answer:
yulyashka [42]3 years ago
8 0
The standard form of an equation is
ax + by = c

In the question, we have the following formula
3x + 1 = 2y

To get the formula in standard form, we first have to subtract 2y from both sides (since we need to get 2y on the left side)
3x - 2y + 1 = 0

Finally we subtract 1 from both sides (since we need the number on the right side)
3x - 2y = - 1

Hence, 4. is the correct answer.
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-4(7.5x-8)=-5(6.3x+8)
grin007 [14]

Answer:

x = -48

Step-by-step explanation:

-4(7.5x - 8) = -5(6.3x + 8)

-30x + 32 = -31.5x - 40

-30x + 31.5x = -40 - 32

1.5x = -72

x = -72/(1.5)

x = -48

8 0
3 years ago
14.88 + x = 17.05 What is the value of x?
julia-pushkina [17]

Answer:

Step-by-step explanation:

14.88 + x = 17.05

x = 17.05 - 14.88

x = 2.17

hope it helps

7 0
3 years ago
Read 2 more answers
Can someone help me with this and not get any wrong please!
Luden [163]

Number 1 is B b/c is u count backwards 5 u get b


8 0
3 years ago
A triangle has one leg measuring 10 inches and the hypotenuse measuring 20 inches. the other leg measures 10\sqrt[]{3} 10 3 inch
stiv31 [10]
From the description given for the triangle above, I think the type of triangle that is represented would be a right triangle. This type of triangle contains a right angle and two acute angles. In order to say or prove that it is a right triangle, it should be able to satisfy the Pythagorean Theorem which relates the sides of the triangle. It is expressed as follows:

c^2 = a^2 + b^2

 where c is the hypotenuse or the longest side and a, b are the two shorter sides.

To prove that the triangle is indeed a right triangle, we use the equation above.

c^2 = a^2 + b^2
c^2 = 20^2 = 10^2 + (10sqrt(3))^2
400 = 100 + (100(3))
400 = 400
5 0
4 years ago
Solve: 2x+7/5 - x-3/10 = x+1/15<br>find the value of x and verify the result will RHS ​
choli [55]

Step-by-step explanation:

<h3><u>Given Question :- </u></h3>

Solve for x :-

\dfrac{2x + 7}{5} -  \dfrac{x - 3}{10}  =  \dfrac{x + 1}{15}

\red{\large\underline{\sf{Solution-}}}

Given linear equation is

\rm :\longmapsto\: \dfrac{2x + 7}{5} -  \dfrac{x - 3}{10}  =  \dfrac{x + 1}{15}

\rm :\longmapsto\: \dfrac{2(2x + 7) - (x - 3)}{10}  =  \dfrac{x + 1}{15}

\rm :\longmapsto\: \dfrac{4x + 14 - x  +  3}{10}  =  \dfrac{x + 1}{15}

\rm :\longmapsto\: \dfrac{(4x - x)  + (14 + 3)}{10}  =  \dfrac{x + 1}{15}

\rm :\longmapsto\: \dfrac{3x  + 17}{10}  =  \dfrac{x + 1}{15}

On multiply by 5 on both sides,

\rm :\longmapsto\: \dfrac{3x  + 17}{2}  =  \dfrac{x + 1}{3}

On cross multiplication, we get

\rm :\longmapsto\:3(3x + 17) = 2(x + 1)

\rm :\longmapsto\:9x +51 = 2x + 2

\rm :\longmapsto\:9x  - 2x = 2 - 51

\rm :\longmapsto\:7x = - 49

\bf\implies \:x =  - 7

<h3><u>VERIFICATION</u></h3>

Consider, LHS

\red{\rm :\longmapsto\: \dfrac{2x + 7}{5} -  \dfrac{x - 3}{10}}

On substituting the value of x, we get

\red{\rm \:  =  \:  \dfrac{2( - 7) + 7}{5} -  \dfrac{ - 7 - 3}{10}}

\red{\rm \:  =  \:  \dfrac{ - 14 + 7}{5} -  \dfrac{ - 10}{10}}

\red{\rm \:  =  \:  \dfrac{ - 7}{5}  + 1}

\red{\rm \:  =  \:  \dfrac{ - 7 + 5}{5}}

\red{\rm \:  =  \:  \dfrac{ - 2}{5}}

Consider RHS

\green{\rm :\longmapsto\:\dfrac{x + 1}{15}}

On substituting the value of x, we get

\green{\rm \:  =  \: \dfrac{ - 7 + 1}{15}}

\green{\rm \:  =  \: \dfrac{ - 6}{15}}

\green{\rm \:  =  \: \dfrac{ - 2}{5}}

\rm \implies\:LHS=RHS

HENCE, VERIFIED

5 0
3 years ago
Read 2 more answers
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