Solve for x.
3x - 4 = 2x - 10
O x = 6
OX= -6
OX= 14
O x=-14
Please help
Will give you brainly
Answers
Answer:
x = -6
Step-by-step explanation:
We have the equation 3x - 4 = 2x - 10 and are asked to find "x".
To solve , we need to get "x" alone entirely.
3x - 4 = 2x - 10
Subtract 2x from both sides :
x - 4 = -10
Add 4 to both sides to get our answer :
x = -6
Answer:
the answer is equal to-6
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Answers
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Step-by-step explanation:
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The length of a rectangular garden is 8m greater than twice the width the area of the garden is 280m^2 what is the width of the garden
Answers
Step-by-step explanation:
Given :-
The length of the garden 8m greater than 2 times the width.
Area of the garden is 280 m²
Let us consider the length as x and width as y.
Sp, we can day length as :-
x = 8 + 2y ---(1)
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We can replace the value of x now,
280 = y × ( 8 + 2y)
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Factorise it.
(y -10)(y + 14)
Cancelling -ve value, we get the width as 10 metres.
Hope it helps :)
Answer:
Step-by-step explanation:
Width = w
Length = 2w + 8
Area of rectangular garden = 280 square meter
length * width = 280
(2w + 8 ) *w = 280
2w * w + 8*w = 280
2w² + 8w = 280
2w² + 8w - 280 = 0
Divide the whole equation by 2
w² + 4w - 140 = 0
w² + 14w - 10w - 14 *10 = 0
w(w + 14) - 10(w + 14) = 0
(w + 14)(w - 10)= 0
w - 10 = 0 {Ignore w + 14, as measurements will not be -ve}
w = 10 m
l = 2*10 +8
= 20 +8
l = 28 m
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if we use the quadratic expression below to complete a diamond, what value will go at every bottom of the diamond
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Answers
Answer:
if we use the quadratic expression below to complete a diamond, what value will go at every bottom of the diamond
is -4
Answer:
if we use the quadratic expression below to complete a diamond, what value will go at every bottom of the diamond
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When you solve x^2 - 8x + 7 = 0 what is the last step of the equation
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O x = -4 13
O x = 3 + 4
O x = 4 + 3
Answers
Answer:
x = 1 and x = 7
Step-by-step explanation:
The given equation is [tex]x^2 - 8x + 7 = 0[/tex].
We need to solve the equation.
It is a quadratic equation whose solution is given by :
[tex]x=\dfrac{-b\pm \sqrt{b^2-4ac} }{2a}[/tex]
Here, a = 1, b = -8 and c = 7
Put values,
[tex]x=\dfrac{-(-8)\pm \sqrt{(-8)^2-4(1)(7)} }{2(1)}\\\\=\dfrac{8\pm \sqrt{64-28}}{2}\\\\=\dfrac{8\pm 6}{2}\\\\=\dfrac{8+6}{2},\dfrac{8-6}{2}\\\\=7,1[/tex]
So, the values of x are x = 1 and x = 7.