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4 years ago

7

If you "ate a ton" of Quater Pounders and they weighed exactly a quarter pound each, how many would you have eaten? ( 1 ton= 2,0

00 pounds) Please show your work

1 answer:

horrorfan [7]4 years ago

8 0

You would have to do 2,000*4 Because there are 4 per pound. You would have to eat 8,000 to have eaten a actual ton of them. XD Nice question btw.

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If a(x) = 3x + 1 and b (x) = StartRoot x minus 4 EndRoot, what is the domain of (b circle a) (x)?

Vladimir79 [104]

Answer:

x ≥ 1

Step-by-step explanation:

There are no domain restrictions on a(x), so we can find the domain of the composite function by looking at that composition:

$(b \circ a)(x)=b(a(x)) = b(3x+1)=\sqrt{(3x+1)-4}\\\\(b \circ a)(x)=\sqrt{3x-3}=\sqrt{3(x-1)}$

The argument of the square root function must be non-negative, so we must have ...

3(x -1) ≥ 0

x -1 ≥ 0 . . . . . divide by 3

x ≥ 1 . . . . . . . add 1

The domain of b(a(x)) is x ≥ 1.

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A = 0.1
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3 years ago

Match the numerical expressions to their simplified forms

eduard

Answer:

$1.\ \ p^2q = (\frac{p^5}{p^{-3}q^{-4}})^{\frac{1}{4}}$

$2.\ \ pq^{\frac{3}{2}}} = (\frac{p^2q^7}{q^{4}})^{\frac{1}{2}}$

$3.\ \ pq^2 = \frac{(pq^3)^{\frac{1}{2}}}{(pq)^{\frac{-1}{2}}}$

$4.\ \ p^2q^{\frac{1}{2}} = (p^6q^{\frac{3}{2}})^{\frac{1}{3}}$

Step-by-step explanation:

Required

Match each expression to their simplified form

1.

$(\frac{p^5}{p^{-3}q^{-4}})^{\frac{1}{4}}$

Simplify the expression in bracket by using the following law of indices;

$\frac{a^m}{a^n} = a^{m-n}$

The expression becomes

$(\frac{p^{5-(-3)}}{q^{-4}})^{\frac{1}{4}}$

$(\frac{p^{5+3}}{q^{-4}})^{\frac{1}{4}}$

$(\frac{p^8}{q^{-4}})^{\frac{1}{4}}$

Split the fraction in the bracket

$(p^8*\frac{1}{q^{-4}})^{\frac{1}{4}}$

Simplify the fraction by using the following law of indices;

$\frac{1}{a^{-m}} = a^m$

The expression becomes

$(p^8*q^4)^{\frac{1}{4}}$

Further simplify the expression in bracket by using the following law of indices;

$(ab)^m = a^m * b^m$

The expression becomes

$(p^{8*\frac{1}{4}}\ *\ q^4*^{\frac{1}{4}})$

$(p^{\frac{8}{4}}\ *\ q^{\frac{4}{4}})$

$p^2q$

Hence,

$(\frac{p^5}{p^{-3}q^{-4}})^{\frac{1}{4}} = p^2q$

2.

$(\frac{p^2q^7}{q^{4}})^{\frac{1}{2}}$

Simplify the expression in bracket by using the following law of indices;

$\frac{a^m}{a^n} = a^{m-n}$

The expression becomes

$({p^2q^{7-4}}})^{\frac{1}{2}}$

$({p^2q^3}})^{\frac{1}{2}}$

Further simplify the expression in bracket by using the following law of indices;

$(ab)^m = a^m * b^m$

The expression becomes

${p^{2*\frac{1}{2}}q^{3*\frac{1}{2}}}}$

$pq^{\frac{3}{2}}}$

Hence,

$pq^{\frac{3}{2}}} = (\frac{p^2q^7}{q^{4}})^{\frac{1}{2}}$

3.

$\frac{(pq^3)^{\frac{1}{2}}}{(pq)^{\frac{-1}{2}}}$

Simplify the numerator as thus:

$\frac{p^{\frac{1}{2}} * q^3*^{\frac{1}{2}}}{(pq)^{\frac{-1}{2}}}$

$\frac{p^{\frac{1}{2}} * q^{\frac{3}{2}}}{(pq)^{\frac{-1}{2}}}$

Simplify the denominator as thus:

$\frac{p^{\frac{1}{2}} * q^{\frac{3}{2}}}{p^{\frac{-1}{2}}q^{\frac{-1}{2}}}$

Simplify the expression in bracket by using the following law of indices;

$\frac{a^m}{a^n} = a^{m-n}$

The expression becomes

$p^{\frac{1}{2} - (\frac{-1}{2} )} * q^{\frac{3}{2} - (\frac{-1}{2}) }$

$p^{\frac{1}{2} +\frac{1}{2} } * q^{\frac{3}{2} + \frac{1}{2} }$

$p^{\frac{1+1}{2}} * q^{\frac{3+1}{2}}$

$p^{\frac{2}{2}} * q^{\frac{4}{2}}$

$pq^2$

Hence,

$pq^2 = \frac{(pq^3)^{\frac{1}{2}}}{(pq)^{\frac{-1}{2}}}$

4.

$(p^6q^{\frac{3}{2}})^{\frac{1}{3}}$

Simplify the expression in bracket by using the following law of indices;

$(ab)^m = a^m * b^m$

The expression becomes

$p^6*^{\frac{1}{3}}\ *\ q^{\frac{3}{2}}*^{\frac{1}{3}}$

$p^{\frac{6}{3}}\ *\ q^{\frac{3*1}{2*3}}$

$p^2 *\ q^{\frac{3}{6}}$

$p^2 *\ q^{\frac{1}{2}$

$p^2q^{\frac{1}{2}$

Hence

$p^2q^{\frac{1}{2}} = (p^6q^{\frac{3}{2}})^{\frac{1}{3}}$

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