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3 years ago

14

PLZ PLZ PLZ help me with this question

2 answers:

Umnica [9.8K]3 years ago

6 0

Take the 40% and turn it into a decimal and 40% in decimal form is 0.40
and then do
70x0.40=28
28 is the 40% discount from 70
and then subtract 70 from 28
70-28=42
therefore the price is now $42

aalyn [17]3 years ago

3 0

The price would be $42.

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yuradex [85]

Answer:

246913578

Step-by-step explanation:

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3 years ago

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What is 10+8+30x20 equal

Leviafan [203]

Using the Order of Operations, this simplifies to 10+8+600 because you multiply before adding. Then you would simply add up the remaining numbers, making it 618.
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3 0

3 years ago

Write an expression to represent: 3 minus the quotient of x and 4

alisha [4.7K]

The expression representing the sentence is $$3 -\frac{x}{4}$ .

We have a sentence - "3 minus the quotient of x and 4".

We have to determine the expression that represents the sentence.

<h3>Express the statement - "The difference of x and y is equal to an irrational number" in the form of expression.</h3>

The expression representing the above statement is -

x - y = π.

According to the question, we have -

3 minus the quotient of x and 4.

Using the Division Algorithm -

x = 4q + r

Substituting r = 0 by introducing decimals in the quotient, we get -

q = $$\frac{x}{4}$

Therefore -

$$3 -\frac{x}{4}$

Hence, the expression representing the sentence is $$3 -\frac{x}{4}$ .

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8 0

2 years ago

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Which word best describes the degree of overlap between the two data sets?

snow_tiger [21]

Answer:

Moderate

Step-by-step explanation:

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6 0

3 years ago

How much of a radioactive kind of calcium will be left after 10 days if the half-life is 5 days

Sophie [7]

Answer: 16 g

Step-by-step explanation:

We can solve this with the Radioactive Half Life Formula:

$A=A_{o}.2^{\frac{-t}{h}}$

Where:

$A$ is the final amount of radioactive calcium

$A_{o}=64 g$ is the initial amount of radioactive calcium

$t=10 days$ is the time elapsed

$h=5 days$ is the half life of radioactive calcium

Solving with the given data:

$A=(64 g)2^{\frac{-10 days}{5 days}}$

$A=16 g$ This is the amount of radioactive calcium left

7 0

4 years ago