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3 years ago

HELP PLEASE IM STRUGGLING

Mathematics

1 answer:

ser-zykov [4K]3 years ago

4 0

Step-by-step explanation:

Both are very correct,

$5 { }^{ - 2} = \frac{1}{25} \: or \: \frac{1}{5 {}^{2} }$

because any number raised to the power of a negative

is given as

$a {}^{ - x} = \frac{1}{a {}^{x} }$

if this helped pls give brainliest

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Explain what is happening at this step in the solving equation process.

aleksklad [387]

Answer:

1. After solving we get value of x: $\mathbf{x=\frac{6}{11} }$

2. After solving we get value of x: $\mathbf{x=\frac{6}{11} }$

Step-by-step explanation:

We need to Solve the equations.

1. $10 - 6 x = 4 + 5 x$

Step 1: Subtract 5x on both sides

$10 - 6 x-5x = 4 + 5 x-5x\\10-11x=4$

Step 2: Subtract 10 on both sides

$10-11x-10=4-10\\-11x=-6$

Step 3: Divide both sides by -11

$\frac{-11x}{-11}=\frac{-6}{-11}\\x=\frac{6}{11}$

So, after solving we get value of x: $\mathbf{x=\frac{6}{11} }$

2. $7-6x=1+5x$

Step 1: Subtract 5x on both sides

$7 - 6 x-5x = 1 + 5 x-5x\\7-11x=1$

Step 2: Subtract 7 on both sides

$7-11x-7=1-7\\-11x=-6$

Step 3: Divide both sides by -11

$\frac{-11x}{-11}=\frac{-6}{-11}\\x=\frac{6}{11}$

So, after solving we get value of x: $\mathbf{x=\frac{6}{11} }$

5 0

3 years ago

4. FGH - MNP. Solve for x.

Dahasolnce [82]

Answer:

x = 6

Step-by-step explanation:

Since the triangles are similar then the ratios of corresponding sides are in proportion, that is

$\frac{MN}{FG}$ = $\frac{NP}{GH}$ , substitute values

$\frac{x}{12}$ = $\frac{10}{20}$ = $\frac{1}{2}$ ( cross- multiply )

2x = 12 ( divide both sides by 2 )

x = 6

6 0

3 years ago

Solve for t <br><br> 2t^2-14t+3=3

Dvinal [7]

Answer:

t=7,t=0

Step-by-step explanation:

2t^2-14t+3=3

Subtract 3 from both sides

2t^2 - 14t +3 - 3 = 3-3

Simpilfy

2t^2 -14t = 0

7 0

3 years ago

WILL REWARD!!!! need help with geometry

Vikentia [17]

1,1. Same Spot. I'm 97% Sure.

6 0

3 years ago

Read 2 more answers

Use the definition of taylor series to find the taylor series (centered at

Alexxandr [17]

Cos(x) is infinitely differentiable, can be expanded using Taylor's series.
The series about c can be expressed as
$f(x)=\sum_{n=0}^{\infty}\frac{f^{n}(c)}{n!}(x-c)^n$
Substituting cos(x), c= π /4,we have
$cos(x)=\sum_{n=0}^{\infty}\frac{f^{n}(\frac{\pi}{4})}{n!}(x-\frac{\pi}{4})^n$
and since
$f'(\frac{\pi}{4})=-sin(\frac{\pi}{4})=-\sqrt{2}/2$
$f"(\frac{\pi}{4})=-cos(\frac{\pi}{4})=-\sqrt{2}/2$
$f^{(iii)}(\frac{\pi}{4})=sin(\frac{\pi}{4})=\sqrt{2}/2$
$f^{(iv)}(\frac{\pi}{4})=cos(\frac{\pi}{4})=\sqrt{2}/2$

we can simplify the expansion to
$cos(x)=\frac{\sqrt{2}}{2}(1-(x-\frac{\pi}{4})/1!-(x-\frac{\pi}{4})^2/2!+(x-\frac{\pi}{4})^3/3!+(x-\frac{\pi}{4})^4/4!-(x-\frac{\pi}{4})^5/5!-...)$
Note that the sign pattern is - - + + - - + + - - ..... following the sign pattern of the derivatives of cos(x).

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3 years ago