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3 years ago

What is (5^-2)^3 in standard form? Please answer!

Mathematics

2 answers:

Marina86 [1]3 years ago

8 0

(5^-2)= (1/5^2)... (1/25) now cube that so it will be 1/15625. Hope that helps

neonofarm [45]3 years ago

6 0

Answer:

6.4x10^-5

Step-by-step explanation:

hello, there.

I hope this helps:

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!PLEASE HELP! Solve for MN. A) 21 B)9 C)6 D)4

Oksanka [162]

The measure of MN from the diagram is 6

<h3>Similarity theorem of triangles</h3>

From the given triangle, the expression below is true;

ML/LK = MN/NO

Given the following parameters

ML = 4

LK = 10

NO= 15

Substitute the given parameters into the formula

4/10 = MN/15

Cross multiply

10MN = 4 * 15

10MN = 60

MN = 6

Hence the measure of MN from the diagram is 6

Learn more on similar triangles here: brainly.com/question/14285697

7 0

2 years ago

Which of the following numbers is a multiple of 6?

LiRa [457]

The multiples of 6. Answer : 6,12,18,24,30,36,42,48,54,60,66,72,78,84,90,96,102,108,114,120,126,132,138,144,150,156,1

4 0

3 years ago

Read 2 more answers

Please help me find the area of this triangle

Paladinen [302]

Answer:

120m^2

Step-by-step explanation:

24*10=240

240/2=120

7 0

3 years ago

Read 2 more answers

Karl is making a pot of chili. The recipe calls for StartFraction 3 Over 8 EndFraction cup of chili powder, but Karl only wants

mina [271]

Given:

The recipe calls for $\dfrac{3}{8}$ cup of chili powder.

Karl only wants to use half as much so it won’t be so spicy.

To find:

How much chili powder should Karl use?

Solution:

We have,

Chill power for recipe = $\dfrac{3}{8}$ cup

Chill power used by Karl is half of chill power for recipe.

$\text{Chill power used by Karl}=\dfrac{1}{2}\times \dfrac{3}{8}$ cup $\text{Chill power used by Karl}=\dfrac{3}{16}$ cup

Therefore, $\dfrac{3}{16}$ cup of chili powder should Karl use.

7 0

3 years ago

Cot^2x/cscx-1=1+sinx/sinx

KATRIN_1 [288]

$\bf \textit{difference of squares}
\\\\
(a-b)(a+b) = a^2-b^2\qquad \qquad 
a^2-b^2 = (a-b)(a+b)
\\\\\\
sin^2(\theta)+cos^2(\theta)=1\implies cos^2(\theta)=1-sin^2(\theta)\\\\
-------------------------------\\\\
\cfrac{cot^2(x)}{csc(x)-1}=\cfrac{1+sin(x)}{sin(x)}\impliedby \textit{let's do the left-hand-side}$

$\bf \cfrac{\quad \frac{cos^2(x)}{sin^2(x)}\quad }{\frac{1}{sin(x)}-1}\implies \cfrac{\quad \frac{cos^2(x)}{sin^2(x)}\quad }{\frac{1-sin(x)}{sin(x)}}\implies \cfrac{cos^2(x)}{sin^2(x)}\cdot \cfrac{sin(x)}{1-sin(x)}
\\\\\\
\cfrac{cos^2(x)}{sin(x)}\cdot \cfrac{1}{1-sin(x)}\implies \cfrac{cos^2(x)}{sin(x)[1-sin(x)]}$

$\bf \cfrac{1-sin^2(x)}{sin(x)[1-sin(x)]}\implies \cfrac{1^2-sin^2(x)}{sin(x)[1-sin(x)]}
\\\\\\
\cfrac{\underline{[1-sin(x)]}~[1+sin(x)]}{sin(x)\underline{[1-sin(x)]}}\implies \cfrac{1+sin(x)}{sin(x)}$

5 0

3 years ago