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3 years ago

Compare.Write <,>,or = 25 lb____ 384 oz

2 answers:

7 0

25lb is larger than 384 oz.

Because there are sixteen ounces in one pound, and if you convert it 25 pounds is equal to 400 ounces.

Hope this helps.<span />

raketka [301]3 years ago

5 0

25 lb > 384 oz because 25 pounds is 400 ounces.

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Write the equation in slope-intercept form. 5x+3y=12

Umnica [9.8K]

5x+3y=12
Subtract 5x from both sides
3y=-5x+12
Divide by 3
Y=-5/3x +4

7 0

3 years ago

Find the limit

Lana71 [14]

Step-by-step explanation:

<h3>Appropriate Question :-</h3>

Find the limit

$\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right]$

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

Given expression is

$\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right]$

On substituting directly x = 1, we get,

$\rm \: = \: \sf \dfrac{1-2}{1 - 1}-\dfrac{1}{1 - 3 + 2}$

$\rm \: = \sf \: \: - \infty \: - \: \infty$

which is indeterminant form.

Consider again,

$\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right]$

can be rewritten as

$\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x( {x}^{2} - 3x + 2)}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x( {x}^{2} - 2x - x + 2)}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x( x(x - 2) - 1(x - 2))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x(x - 2) \: (x - 1))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ {(x - 2)}^{2} - 1}{x(x - 2) \: (x - 1))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ (x - 2 - 1)(x - 2 + 1)}{x(x - 2) \: (x - 1))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ (x - 3)(x - 1)}{x(x - 2) \: (x - 1))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ (x - 3)}{x(x - 2)}\right]$

$\rm \: = \: \sf \: \dfrac{1 - 3}{1 \times (1 - 2)}$

$\rm \: = \: \sf \: \dfrac{ - 2}{ - 1}$

$\rm \: = \: \sf \boxed{2}$

Hence,

$\rm\implies \:\boxed{ \rm{ \:\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right] = 2 \: }}$

$\rule{190pt}{2pt}$

7 0

3 years ago

Read 2 more answers

Jason baked 7 pans of brownies. He gave ¼ of the brownies to his two sisters. How many pans of brownies did he give to his siste

boyakko [2]

Step-by-step explanation:

7×1/4 = 1.75

1.75 ÷ 2 = 0.875 pans each

8 0

3 years ago

Triangle G E F is shown with its exterior angles. Line G F extends through point B. Line F E extends through point A. Line E G e

Radda [10]

The measure of angle ∠EGF is 65°. And the measure of the angle ∠CGE is 115°.

<h3>What is the triangle?</h3>

A triangle is a three-sided polygon with three angles. The angles of the triangle add up to 180 degrees.

Triangle GEF is shown with its exterior angles.

Line GF extends through point B.

Line FE extends through point A.

Line EG extends through point C.

Angles ∠FEG and ∠EGF are congruent.

∠FEG = ∠EGF = x

Sides EF and GF are congruent.

Angle ∠EFG is 50° degrees.

∠EFG + ∠FGE + ∠GEF = 180°

50° + x + x = 180°

2x = 130°

x = 65°

∠FEG = ∠EGF = 65°

Then angle ∠CGF will be

∠CGF + ∠FGE = 180°

∠CGF + 65° = 180°

∠CGF = 115°

More about the triangle link is given below.

brainly.com/question/25813512

#SPJ1

4 0

2 years ago

I need help with this ASAP plzzzzz

miss Akunina [59]

Answer:

23. $\frac{4\sqrt{3} }{3}$

24. $\frac{7\sqrt{3}}{3}$

25. $6\sqrt{3}$

26. $\frac{10\sqrt{3} }{3}$

27. $14$

28. $4\sqrt{3}$

Step-by-step explanation:

To solve these i used SOHCAHTOA

$sin=\frac{opposite}{adjacent}\\ cos=\frac{adjacent}{hypotenuse} \\tan=\frac{opposite}{adjacent}$

23.

Find the missing side using Tangent

$tan(30)=\frac{x}{4}$

$4*tan(30)=x$

$\frac{4\sqrt{3} }{3}$

24.

Find the missing side using Tangent

$tan(30)=\frac{x}{7}$

$7*tan(30)=x$

$\frac{7\sqrt{3}}{3}$

25.

Find the missing side using Tangent

$tan(30)=\frac{x}{18}$

$18*tan(30)=x$

$6\sqrt{3}$

26.

Find the missing side using Tangent

$tan(30)=\frac{x}{10}$

$10*tan(30)=x$

$\frac{10\sqrt{3} }{3}$

27.

Find the missing side using Tangent

$tan(30)=\frac{x}{14\sqrt{3} }$

$(14\sqrt{3}) *tan(30)=x$

$14$

28.

Find the missing side using Tangent

$tan(30)=\frac{x}{12}$

$12*tan(30)=x$

$4\sqrt{3}$

5 0

3 years ago