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2 years ago

A cabinet is 1.5 meters tall. If 1 meter is approximately 1.09 yards, how tall is the cabinet, in yards?

Mathematics

1 answer:

Firlakuza [10]2 years ago

3 0

Answer:

1.64041995 yards

Step-by-step explanation:

If one meter is 1.09 yards, just add half that amount.

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A bowl contains 25 balls numbered 1 to 25. A ball is drawn and its number is noted. Without replacing the first ball, another ba

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Well i think that if there is 25 ball there is more likely to be odd cause there is not 26 balls to make even and odds have the same amount of odd and even balls so there is one more odd than there is even

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3 years ago

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Explain how two amounts of change can be the same but the percent of change can be different

densk [106]

If I understand the question correctly, you can have the same amount of change be two different percentages by having 4 quarters and 100 pennies.
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3 years ago

Which inequality describes this graph?

Alborosie

Answer:

x≠5

Step-by-step explanation:

An open circle means "less than" or "greater than", this means that anything is greater than or less than where the circle is placed. So, when an open circle is placed on a number line without any arrow, this means that that number is not equal to x.

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2 years ago

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Consider the logarithms base 5. For each logarithmic expression below, either calculate the value of the expression or explain w

Tom [10]

Answer:

log₅(3125) = 5

Step-by-step explanation:

Given:

log₅(3125)

Now,

using the property of log function that

logₐ(b) = $\frac{\log(b)}{\log(a)}$

thus,

Therefore, applying the above property, we get

⇒ $\frac{\log(3125)}{\log(5)}$ (here log = log base 10)

now,

3125 = 5⁵

thus,

⇒ $\frac{\log(5^5)}{\log(5)}$

Now,

we know from the properties of log function that

log(aᵇ) = b × log(a)

therefore applying the above property we get

⇒ $\frac{5\log(5)}{\log(5)}$

⇒ 5

Hence,

log₅(3125) = 5

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3 years ago

The figure shows △ABC . BD is the angle bisector of ∠ABC .

Misha Larkins [42]

From angle bisector theorem, we know that:

$\dfrac{|AD|}{|DC|}=\dfrac{|BA|}{|BC|}\\\\\\\dfrac{|AD|}{|DC|}=\dfrac{8}{10}=\dfrac{4}{5}$

Moreover:

$|AD|+|DC|=6$

so:

$|DC|=6-|AD|$

and

$\dfrac{|AD|}{|DC|}=\dfrac{4}{5}\\\\\\\dfrac{|AD|}{6-|AD|}=\dfrac{4}{5}\qquad\qquad\text{cross multiplying}\\\\\\ 5\cdot|AD|=4\cdot\big(6-|AD|\big)\\\\5\cdot|AD|=24-4\cdot|AD|\\\\5\cdot|AD|+4\cdot|AD|=24\\\\9\cdot|AD|=24\qquad|:9\\\\\\|AD|=\dfrac{24}{9}=\boxed{\dfrac{8}{3}}$

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3 years ago

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