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3 years ago

12⋅(4^−2/4^−4) 1/4 3/4 16 192

Mathematics

2 answers:

mars1129 [50]3 years ago

7 0

My caculation it would be 16 i hope that this help you

Bond [772]3 years ago

6 0

It would be 16. I got this bc well 4^-2 is 0.0625 and 4^-4 is 0.00390625. so you divide those and that would give u 16.

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Justine and her cousin Natalie are picking out snacks at the corner store for movie night. Natalie spends $5.30. Justine spends

Kitty [74]

Answer:

34.45

Step-by-step explanation:

11÷2 = 5.5

so 5.5x5.30 = 29.15 + 5.30 = 34.45

8 0

3 years ago

Why do you think we need a horizontal axis and a vertical axis?

Dimas [21]

A vertical line on a sheet of lined graph paper is a vertical axis (normally called the X axis) that denotes everything to the left of the line is negative and everything to the right is positive. A horizontal line on that graph paper (normally called the Y axis) denotes everything below the line is negative and every thing above is positive. Each axis is graduated in scale. The equation X=Y has no single answer, but we can plot a locus of points on your graph by assigning values to X ( or Y) and create a diagonal line which passes thru the origin (where the lines cross) that denotes all possible answers to the equation that exist within the limits of the graph paper. Every equation will plot a locus of points.

5 0

3 years ago

Element X decays radioactively with a half life of 11 minutes. If there are 870 grams of Element X, how long, to the nearest ten

serg [7]

Answer:

It would take 27.5 minutes the element to decay to 154 grams.

Step-by-step explanation:

The decay equation:

$\frac {dN}{dt}\propto -N$

$\Rightarrow \R\frac {dN}{dt}=-\lambda N$

$\Rightarrow \frac {dN}N=-\lambda dt$

Integrating both sides

$\Rightarrow \int \frac {dN}N=\int-\lambda dt$

$\Rightarrow ln|N|=-\lambda t+c$

When t=0, N= $N_0$ = initial amount

$ln|N_0|=-\lambda .0+c$

$\Rightarrow c=ln|N_0|$

$ln|N|=-\lambda t+ln|N_0|$

$\Rightarrow ln|N|-ln|N_0|=-\lambda t$

$\Rightarrow ln|\frac{N}{N_0}|=-\lambda t$

Decay equation:

$ln|\frac{N}{N_0}|=-\lambda t$

Given that, the half life of of element X is 11 minutes.

For half life, $N=\frac12 N_0$ , t= 11 min.

$ln|\frac{N}{N_0}|=-\lambda t$

$\Rightarrow ln|\frac{\frac12N_0}{N_0}|=-\lambda . 11$

$\Rightarrow ln|\frac12}|=-\lambda . 11$

$\Rightarrow -\lambda . 11=ln|\frac12}|$

$\Rightarrow \lambda =\frac{ln|\frac12|}{-11}$

$\Rightarrow \lambda =\frac{ln|2|}{11}$ [ $ln|\frac12|=ln|1|-ln|2|=-ln|2|$ , since ln|1|=0]

N=154 grams, $N_0$ = 870 grams, t=?

$ln|\frac{N}{N_0}|=-\lambda t$

$\Rightarrow ln|\frac{154}{870}|=-\frac{ln|2|}{11}.t$

$\Rightarrow t= \frac{ln|\frac{154}{870}|\times 11}{-ln|2|}$

=27.5 minutes

It would take 27.5 minutes the element to decay to 154 grams.

5 0

3 years ago

Need help with these two

Brilliant_brown [7]

Before we start to answer, let's get our formulas put in.
4 qt = 1 gal.
1 c = 0.5pt (or 2 cups to 1 pt)

Now that we've gotten the formulas out of the way, let's plug in our numbers.

We have 5 qt.
Since we have 4 qt per gal, let's subtract 4 from 5.
5 - 4 = 1. We have 1 gal (due to having a whole 4) and 1 qt left over.
5 qt = 1 gal 1 qt.

We have 7 c.
Since we have 2 cups per pint, let's divide 7 by 2.
7 / 2 = 3.5

We have 3.5 Pt.

7c = 3.5pt

Your answers are:
1 gal 1 qt
3.5pt

I hope this helps!

4 0

4 years ago

What is the value of x?

Tanya [424]

Step-by-step explanation:

19x - 4 = 110 ( alternate Exterior angle )

→ 19x = 110 + 4 = 114

→ x = 114/19

→ x = 6

therefore, value of x is 6°.

HOPE THIS ANSWER HELPS YOU DEAR! TAKE CARE

3 0

3 years ago