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2 years ago

Please answer my question

Mathematics

1 answer:

Nataly [62]2 years ago

4 0

Answer:

66000cm/660m

Step-by-step explanation:

perimeter of a circle=pi x d

perimeter=132cm

132 cm = 1 revolution

x = 500 revolution

132 x 500=66000cm

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Does anyone know the answer to this scale and unit conversion GED math problem? Thank you very much!

aliya0001 [1]

Answer:

Flour $5\dfrac{1}{2}\ c$

Sugar $2\dfrac{2}{3}\ c$

Butter $1\dfrac{1}{2}\ c$

Eggs $2$

Baking powder $7\ t$

Vanilla extract $3\ t$

Step-by-step explanation:

There will be 16 people at the party. Each person will eat 3 cookies, so all people will eat $16\cdot 3=48$ cookies.

Given cookie recipe makes 24 cookies, so to make 48 cookies you need

$48:24=2$ portions.

Thus,

1 portion 2 portions

Flour $2 \dfrac{3}{4}\ c$ $2\dfrac{3}{4}\cdot 2=2\cdot 2+\dfrac{3}{4}\cdot 2=4+\dfrac{3}{2}=4+1 \dfrac{1}{2}=5\dfrac{1}{2}\ c$

Sugar $1\dfrac{1}{3}\ c$ $1\dfrac{1}{3}\cdot 2=1\cdot 2+\dfrac{1}{3}\cdot 2=2+\dfrac{2}{3}=2\dfrac{2}{3}\ c$

Butter $\dfrac{3}{4}\ c$ $\dfrac{3}{4}\cdot 2=\dfrac{3}{2}=1\dfrac{1}{2}\ c$

Eggs $1$ $1\cdot 2=2$

Powder $3\dfrac{1}{2}\ t$ $3\dfrac{1}{2}\cdot 2=3\cdot 2+\dfrac{1}{2}\cdot 2=6+1=7\ t$

Vanilla $1\dfrac{1}{2}\ t$ $1\dfrac{1}{2}\cdot 2=1\cdot 2+\dfrac{1}{2}\cdot 2=2+1=3\ t$

8 0

2 years ago

Data represented by quantities is called data.

AnnyKZ [126]

Four types of data (nominal, ordinal, interval, and ratio) that represent values or observations that can be sorted into a category

3 0

2 years ago

<img src="https://tex.z-dn.net/?f=%5Csqrt%7B%5Cfrac%7B-140%7D%7B%5Csqrt%7B28%7D%20%7D" id="TexFormula1" title="\sqrt{\frac{-140}

Zanzabum

This is the answer

8 0

1 year ago

The cost of a CD at Ernesto's Music Emporium is $12, including tax. Brian has a $100 gift card from Ernesto's. What is his balan

Allushta [10]

$52 because 4 times 12 equals 48 so 100 minus 48 equals 52

6 0

2 years ago

(2pm^-1q^0)^-4 • 2m ^-1 p^3 / 2pq^2

Montano1993 [528]

Answer:

$\dfrac{m^3}{16p^2q^2}$

Step-by-step explanation:

Given:

$(2pm^{-1}q^0)^{-4}\cdot \dfrac{ 2m^{-1} p^3}{2pq^2}$

$m^{-1}=\dfrac{1}{m}$

$q^0=1$

$2pm^{-1}q^0=2p\cdot \dfrac{1}{m}\cdot 1=\dfrac{2p}{m}$

$(2pm^{-1}q^0)^{-4}=\left(\dfrac{2p}{m}\right)^{-4}=\left(\dfrac{m}{2p}\right)^4=\dfrac{m^4}{(2p)^4}=\dfrac{m^4}{16p^4}$

$m^{-1}=\dfrac{1}{m}$

$2m^{-1} p^3=2\cdot \dfrac{1}{m}\cdot p^3=\dfrac{2p^3}{m}$

$\dfrac{ 2m^{-1} p^3}{2pq^2}=\dfrac{\frac{2p^3}{m}}{2pq^2}=\dfrac{2p^3}{m}\cdot \dfrac{1}{2pq^2}=\dfrac{p^2}{mq^2}$

$(2pm^{-1}q^0)^{-4}\cdot \dfrac{ 2m^{-1} p^3}{2pq^2}=\dfrac{m^4}{16p^4}\cdot \dfrac{p^2}{mq^2}=\dfrac{m^3}{16p^2q^2}$

8 0

2 years ago