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2 years ago

Solve for x. -6x+6=-4x-2

Mathematics

1 answer:

Tcecarenko [31]2 years ago

4 0

The answer would be x = 4. Hope this helps!

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Each side of a square is increasing at a rate of 6cm/s. At what rate is the area of the square increasing when the area of the s

diamong [38]

Area of a square = length²
A = x²
16 = x²; x = 4
dA/dx = 2x cm²/s
dx/dt = 6 cm/s
Using chain rule:
dA/dt = dA/dx * dx/dt
dA/dt = 2x * 6
dA/dt = 12x
At x = 4,
dA/dt = 12(4) = 48 cm²/s

7 0

4 years ago

How do I solve number 7?

borishaifa [10]

Answer:

16ᵗʰ Term of the sequence is 1010

Step-by-step explanation:

7.)

Here,

First Term = a₁ = 5

Common Difference = d = 67

Now, For 16ᵗʰ term, n = 16

a₁₆ = 5 + (16 - 1) 67

a₁₆ = 5 + 15 × 67

a₁₆ = 5 + 1005

a₁₆ = 1010

Thus, 16ᵗʰ Term of the sequence is 1010

<u>-TheUnknownScientist</u>

8 0

3 years ago

How do you write a linear equation in slope-intercept form

inessss [21]

Identify the slope find the y-intercept it can be done by substituting the slope and coordinates

7 0

4 years ago

I need help with an area math question

Ede4ka [16]

Answer:

Step-by-step explanation:

Area of outer triangle = $\dfrac{4x*(4x+3)}{2}= 2x*(4x +3)$

= 2x*4x + 2x*3

= 8x² + 6x

Area of inner triangle= $\dfrac{2x*(2x+5)}{2}=x*(2x+5)=x*2x+x*5=2x^{2} + 5x$

Area of aluminium = area of outer triangle - area of inner triangle

= 8x² + 6x + 2x² + 5x

= 10x² + 11x

Area of 20 aluminum borders = 20*(10x² + 11x) = 200x² + 220x

b) x = 25 cm

Area of 20 aluminum borders = 200*25² + 220*25

= 200 * 625 + 220*25

= 125000 + 5500

= 130500 cm²

c) Area of inner triangle (Glass) = 5500 cm²

Cost of aluminium = 130500 * 0.06 = $ 7830

Cost of glass = 5500 * 0.03 = $ 165

Cost of making 20 windows = 7830 + 165 = $ 7995

7 0

3 years ago

Boyle's Law says that the volume of a gas varies inversely with the pressure. When the volume of a certain gas is 7L, the pressu

klasskru [66]

$\bf \qquad \qquad \textit{inverse proportional variation}\\\\
\textit{\underline{y} varies inversely with \underline{x}}\qquad \qquad y=\cfrac{k}{x}\impliedby 
\begin{array}{llll}
k=constant\ of\\
\qquad variation
\end{array}\\\\
-------------------------------$

$\bf \textit{"volume of a gas varies inversely with the pressure"}\quad v=\cfrac{k}{p}
\\\\\\
\textit{we also know that }
\begin{cases}
v=7\\
p=410
\end{cases}\implies 7=\cfrac{k}{210}\implies 1470=k
\\\\\\
thus\qquad \qquad \boxed{v=\cfrac{1470}{p}}
\\\\\\
\textit{when p=287, what is \underline{v}?}\qquad v=\cfrac{410}{287}$

7 0

4 years ago