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3 years ago

Evaluate. 21 ÷ 7 + 20 • 102 • 2 – 3

1 answer:

7 0

To avoid confusion, remember PEMDAS. This is the order of operation.
Parenthesis, Exponents, Multiply, Divide, Add, Subtract

21 ÷ 7 + 20 * 102 * 2 - 3

No parenthesis and no exponents, proceed to multiplication.

20 * 102 * 2 = 4,080

Then, perform division

21÷7 = 3

Then, do addition

3 + 4080 = 4083

Lastly, do subtraction

4,083 - 3 = 4,080

So, 21 ÷ 7 + 20 * 102 * 2 - 3 = 4,080

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12 1/2 gal= qt can you help me plsssssss

Deffense [45]

Answer:

50 qrts

Step-by-step explanation:

12.5 gal = 4x qrts

12.5 * 4 = 50

7 0

3 years ago

Read 2 more answers

The scatterplot contains data points, including the

ser-zykov [4K]

Step-by-step explanation:

what is the question? please provide a question.

5 0

3 years ago

Solve the equation after making an appropriate substitution. X^6-98x^3=3375

Leni [432]

Rearrange $x^6-98x^3=3375$ to give

$x^6-98x^3-3375=0$

Let $x^3$ be a variable 'p' and so we can write $x^6$ as $x^6=(x^3)(x^3)=(p)(p)= p^2$

Rewrite the equation in terms of 'p'

$p^2-98p-3375 = 0$

where $a=1, b=-98, c=-3375$

using the quadratic formula $\frac{-b+ \sqrt{b^2-4ac} }{2a}$ and subsitute the value of $a, b, c$

$p_1= \frac{-(-98)+ \sqrt{(-98)^2-4(1)(-3375)} }{2} =125$

$p_2= \frac{-(-98)- \sqrt{(-98)^2-4(1)(-3375)} }{2} =-27$

There are two value of p; 125 and -27

Now we find the value of x

Earlier we substitute $x^3$ for $p$ and $x^6$ for $p^2$

When $p=125$ , $x= \sqrt[3]{125}=5$
When $p = -27, x= \sqrt[3]{-27}=-3$

So the final answer is the two values of x;

x = 5 OR x = -3

3 0

3 years ago

Multiple Choice Worth 5 points)

Snezhnost [94]

Answer:

110 is the correct answer...

4 0

2 years ago

Read 2 more answers

The radioactive substance cesium-137 has a half-life of 30 years. The amount A (t)(in grams) of a sample of cesium-137 remaining

lakkis [162]

Rather than use that formula, look at the attachment and use the 4th equation.

Ending Amount = Beginning Amount / 2^n
where "n" is the number of half-lives. "n" = elapsed time / half-life
(Is 934 the beginning amount? Let's say it is.)

for 40 years, n=40/30 = 1.3333333...
Ending Amount = 934 / 2^1.3333333...
Ending Amount = 934 / 2.5198420998 
Ending Amount = 370.66 grams

for 100 years, n=100/30 = 3.333333 ...
Ending Amount = 934 / 2^ 3.333333 ...
Ending Amount = 934 / 10.0793683992 
Ending Amount = 92.66 grams

Source:
http://www.1728.org/halflife.htm

4 0

3 years ago