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3 years ago

5

Which is a reasonable estimate of the amount of water in a small, inflatable pool used by children? A) 5 pints B) 5 quarts C) 50

cups D) 50 gallons

2 answers:

zzz [600]3 years ago

6 0

Answer:

50 gallons

Step-by-step explanation: i had it on my test

Ede4ka [16]3 years ago

3 0

I would say the amount used would be at least 50 gallons because it makes most sense. Hope I helped :))

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Find the Two Values that x can have if X^2 -10=26

Zolol [24]

Answer:

x1= -6 x2=6

Step-by-step explanation:

x^2-10 = 26

x^2 = 26 +10

x^2 = 36

x = sqrt(36)

x = -6 or x = 6 because both 6 and -6 squared are equal to 36

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How much interest would be earned on a $3,200 deposit at 3.5% for 1 year?

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Cual es el complementario de un angulo

jonny [76]

Answer:

La suma de los ángulos complementarios es 90 °

Step-by-step explanation:

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3 years ago

Need the a answer for this

FinnZ [79.3K]

Answer:

Given expression:

$\dfrac{14a^4b^6c^{-10}}{8a^{-2}b^3c^{-5}}$

Separate the variables:

$\implies \dfrac{14}{8} \cdot \dfrac{a^4}{a^{-2}} \cdot \dfrac{b^6}{b^3} \cdot \dfrac{c^{-10}}{c^{-5}}$

Reduce the first fraction:

$\implies \dfrac{7}{4} \cdot \dfrac{a^4}{a^{-2}} \cdot \dfrac{b^6}{b^3} \cdot \dfrac{c^{-10}}{c^{-5}}$

$\textsf{Apply Division Property of Exponents rule} \quad \dfrac{a^b}{a^c}=a^{b-c}:$

$\implies \dfrac{7}{4} \cdot a^{4-(-2)} \cdot b^{6-3} \cdot c^{-10-(-5)}$

$\implies \dfrac{7}{4} \cdot a^{6} \cdot b^{3} \cdot c^{-5}$

$\textsf{Apply Negative Property of Exponents rule} \quad a^{-n}=\dfrac{1}{a^n}$

$\implies \dfrac{7}{4} \cdot a^{6} \cdot b^{3} \cdot \dfrac{1}{c^5}$

Therefore:

$\implies \dfrac{7a^6b^3}{4c^5}$

4 0

1 year ago

The sum of the first two terms of an infinite GP is 6 and each term is 5 times the sum of the succeeding termsthen the second te

kozerog [31]

Answer:

$\dfrac{6}{7}$ .

Step-by-step explanation:

It is given that the sum of first two terms of an infinite GP is 6.

nth term of a GP is

$a_n=ar^{n-1}$

$a+ar=6$ ...(1)

Each term is 5 times the sum of the succeeding terms.

$a_n=5(a_{n+1}+a_{n+2}+...+\infty)$

$ar^{n-1}=5(ar^n+ar^{n+1}+...+\infty)$

$ar^{n-1}=5ar^n(1+r+r^2+...+\infty)$

Divide both sides by a.

$r^{n-1}=5r^n(\dfrac{1}{1-r})$ $[\text{Sum of infinite GP}=\dfrac{a}{1-r}]$

$\dfrac{r^n}{r}=\dfrac{5r^n}{1-r}$

$\dfrac{1}{r}=\dfrac{5}{1-r}$

$1-r=5r$

$1=6r$

$\dfrac{1}{6}=r$

The common ratio is 1/6.

Put r=1/6 in (1).

$a+a(\dfrac{1}{6})=6$

$6a+a=36$

$7a=36$

$a=\dfrac{36}{7}$

Second term $ar=\dfrac{36}{7}\times \dfrac{1}{6}=\dfrac{6}{7}$

Therefore, the second term is $\dfrac{6}{7}$ .

4 0

2 years ago