4 oz is how much of a cup

PLEASE HELP TIMER!!!!!!!!!!!!!

xz_007 [3.2K]

Answer:

Answer: -3/2

Step-by-step explanation:

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5 0

2 years ago

Prove that (81/16)^-3/4 ×[(25/9)^-3/2 ÷ (5/2)^-3]=1

andreev551 [17]

Answer:

First write them in positive exponent form

(16/81)¾ × [ (9/25)^3/2 ÷ (2/5)³ ]

(2⁴×¾)/ (3⁴×¾) × [ (3² × ^3/2) / (5² ×^3/2) ÷ 2³/5³)

Simplify the terms

2³/3³ × ( 3³ / 5³ ÷ 2³/5³)

Solve the terms in the bracket

2³/3³ × (3³/5³×5³/2³)

You will get

2³/3³ × 3³/2³ = 1

They will cancel each other so the answer will be 1

Hope this helps.

8 0

4 years ago

How do you know when a term is defined or undefined

lord [1]

In geometry, definitions are formed using known words or terms to describe a new word. There are three words in geometry that are not formally defined. These three undefined terms are point, line and plane.

POINT (an undefined term) 
In geometry, a point has no dimension (actual size). Even though we represent a point with a dot, the point has no length, width, or thickness. A point is usually named with a capital letter. In the coordinate plane, a point is named by an ordered pair, (x,y). 

LINE (an undefined term) 
In geometry, a line has no thickness but its length extends in one dimension and goes on forever in both directions. A line is depicted to be a straight line with two arrowheads indicating that the line extends without end in two directions. A line is named by a single lowercase written letter or by two points on the line with an arrow drawn above them. 

PLANE (an undefined term) 
In geometry, a plane has no thickness but extends indefinitely in all directions. Planes are usually represented by a shape that looks like a tabletop or wall. Even though the diagram of a plane has edges, you must remember that the plane has no boundaries. A plane is named by a single letter (plane m) or by three non-collinear points (plane ABC). 

Undefined terms can be combined to define other terms. Noncollinear points, for example, are points that do not lie on the same line. A line segment is the portion of a line that includes two particular points and all points that lie between them, while a ray is the portion of a line that includes a particular point, called the end point, and all points extending infinitely to one side of the end point. 

Defined terms can be combined with each other and with undefined terms to define still more terms. An angle, for example, is a combination of two different rays or line segments that share a single end point. Similarly, a triangle is composed of three noncollinear points and the line segments that lie between them. 

Everything else builds on these and adds more information to this base. Those added things include all the theorems and other "defined" terms like parallelogram or acute angle.

5 0

3 years ago

Which point is a solution of x + 2y ≤ 4?

VARVARA [1.3K]

x+2*y-(4)≤0

here yhu go

3 0

3 years ago

CALC- limits please show your method

gladu [14]

A. Factor the numerator as a difference of squares:

$\displaystyle\lim_{x\to9}\frac{x-9}{\sqrt x-3}=\lim_{x\to9}\frac{(\sqrt x-3)(\sqrt x+3)}{\sqrt x-3}=\lim_{x\to9}(\sqrt x+3)=6$

c. As $x\to\infty$ , the contribution of the terms of degree less than 2 becomes negligible, which means we can write

$\displaystyle\lim_{x\to\infty}\frac{4x^2-4x-8}{x^2-9}=\lim_{x\to\infty}\frac{4x^2}{x^2}=\lim_{x\to\infty}4=4$

e. Let's first rewrite the root terms with rational exponents:

$\displaystyle\lim_{x\to1}\frac{\sqrt[3]x-x}{\sqrt x-x}=\lim_{x\to1}\frac{x^{1/3}-x}{x^{1/2}-x}$

Next we rationalize the numerator and denominator. We do so by recalling

$(a-b)(a+b)=a^2-b^2$
$(a-b)(a^2+ab+b^2)=a^3-b^3$

In particular,

$(x^{1/3}-x)(x^{2/3}+x^{4/3}+x^2)=x-x^3$
$(x^{1/2}-x)(x^{1/2}+x)=x-x^2$

so we have

$\displaystyle\lim_{x\to1}\frac{x^{1/3}-x}{x^{1/2}-x}\cdot\frac{x^{2/3}+x^{4/3}+x^2}{x^{2/3}+x^{4/3}+x^2}\cdot\frac{x^{1/2}+x}{x^{1/2}+x}=\lim_{x\to1}\frac{x-x^3}{x-x^2}\cdot\frac{x^{1/2}+x}{x^{2/3}+x^{4/3}+x^2}$

For $x\neq0$ and $x\neq1$ , we can simplify the first term:

$\dfrac{x-x^3}{x-x^2}=\dfrac{x(1-x^2)}{x(1-x)}=\dfrac{x(1-x)(1+x)}{x(1-x)}=1+x$

So our limit becomes

$\displaystyle\lim_{x\to1}\frac{(1+x)(x^{1/2}+x)}{x^{2/3}+x^{4/3}+x^2}=\frac{(1+1)(1+1)}{1+1+1}=\frac43$

3 0

3 years ago