What is the domain of this table X 1 2 3 4 Y 2 4 3 2

Raphael graphed the system of equations shown. y = – 3 y = x – 0.8 What is the best approximation for the solution to this syste

Ber [7]

Given ---›

y = -3

&

y = x - 0.8

So,
=> x = y + 0.8

=> x = -3 + 0.8

=> x = -2.2

Therefore, the best approximation for the solution to this system of equations is = (–2.2, –3)

3 0

3 years ago

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Factor completely <br> 4(x+1)^2/3 + 12(x+1)^-1/3

wolverine [178]

$\bf a^{\frac{{ n}}{{ m}}} \implies \sqrt[{ m}]{a^{ n}} \qquad \qquad
\sqrt[{ m}]{a^{ n}}\implies a^{\frac{{ n}}{{ m}}}\\\\
\left.\qquad \qquad \right.\textit{negative exponents}\\\\
a^{-{ n}} \implies \cfrac{1}{a^{ n}}
\qquad \qquad
\cfrac{1}{a^{ n}}\implies a^{-{ n}}
\qquad \qquad 
a^{{{ n}}}\implies \cfrac{1}{a^{-{{ n}}}}
\\\\
-------------------------------\\\\
%4(x+1)^2/3 + 12(x+1)^-1/3
4(x+1)^{\frac{2}{3}}+12(x+1)^{-\frac{1}{3}}\implies 4(x+1)^{\frac{2}{3}}+12\cfrac{1}{(x+1)^{\frac{1}{3}}}
\\\\\\$

$\bf 4(x+1)^{\frac{2}{3}}+\cfrac{12}{(x+1)^{\frac{1}{3}}}\impliedby \textit{so, our LCD is }(x+1)^{\frac{1}{3}}
\\\\\\
\cfrac{4(x+1)^{\frac{2}{3}}\cdot (x+1)^{\frac{1}{3}}+12}{(x+1)^{\frac{1}{3}}}\implies \cfrac{4(x+1)^{\frac{2}{3}+\frac{1}{3}}+12}{(x+1)^{\frac{1}{3}}}
\\\\\\
\cfrac{4(x+1)^{\frac{3}{3}}+12}{(x+1)^{\frac{1}{3}}}\implies \cfrac{4(x+1)+12}{(x+1)^{\frac{1}{3}}}\implies \cfrac{4x+4+12}{(x+1)^{\frac{1}{3}}}
\\\\\\
\cfrac{4x+16}{(x+1)^{\frac{1}{3}}}\implies \cfrac{4(x+4)}{\sqrt[3]{x+1}}$

3 0

3 years ago

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I really need help with this question. Can I get a explanation too?

aniked [119]

Answer:

C

Step-by-step explanation:

Find the surface area:

Bottom base: 57 × 27 = 1,539

Top base: 21 × 27 = 567

Rectangular face: 23 × 27 = 621

Trapezoidal face: (21 + 57) ÷ 2 × 15 = 585

1,539 + 567 + 621 + 621 + 585 + 585 = 4,518

The surface area is 4,518 square inches

3 0

3 years ago

Altitudes AA1 and BB1 are drawn in acute △ABC. Prove that A1C·BC=B1C·AC

Sophie [7]

Answer:

See the attached figure which represents the problem.

As shown, AA₁ and BB₁ are the altitudes in acute △ABC.

△AA₁C is a right triangle at A₁

So, Cos x = adjacent/hypotenuse = A₁C/AC ⇒(1)

△BB₁C is a right triangle at B₁

So, Cos x = adjacent/hypotenuse = B₁C/BC ⇒(2)

From (1) and (2)

∴ A₁C/AC = B₁C/BC

using scissors method

∴ A₁C · BC = B₁C · AC

7 0

3 years ago

Determine the gradient and the y-intersect point of:

maks197457 [2]

Answer:

gradient:

3x+2y=5

3(1)+2dy/dx=0

3+2dy/dx=0

2dy/dx=-3

dy/dx=-3/2.

y-intersect:

3x+2y=5

2y=5-3x

y=5-3x/2

5 0

3 years ago