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2 years ago

Can someone help me with this please

Mathematics

1 answer:

MrRa [10]2 years ago

7 0

Answer:

Andre's Method is the only one that works

Step-by-step explanation:

Noah's method won't work because he doesn't follow the order of operations. You can pull the the 9 out first you have to distribute the 2 first.

Elena's method doesn't work because to eliminate the 18 you must use the reverse operation it must be added not subtracted.

Andre's Method is the only one that works

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0.0025 cm << in scientific notation

Leno4ka [110]

2.5 X 10^-3 is the scientific notation for 0.0025

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2 years ago

Find the coordinates if the midpoint of a line segment with endpoints (-3,4) and (7,9).

svetoff [14.1K]

Using the midpoint formula
$\frac{ x_{1}+ x_{2} }{2}, 
 \frac{ y_{1}+ y_{2} }{2}$
$\frac{-3+7}{2},
 \frac{4+9}{2}$
$\frac{4}{2},
 \frac{13}{2}$
$(2,6.5)$
So, your answer would be (2,6.5).

5 0

2 years ago

Read 2 more answers

Describe the slope of the line<br> Find the slope.

frosja888 [35]

Use equation (y2-y1)/(x2-x1)
(3-(-1))/(0-5)
4/-5 is the slope

3 0

1 year ago

I need some help with this. If you could help me, that would be great.

Vesnalui [34]

Answer:

.31 g / mL

Step-by-step explanation:

Density = mass / volume

Sample 3

mass = 7 1/4 = 7.25

Volume 23.1

Density = 7.25/ 23.1 = .313852814

To the nearest hundredth

.31 g / mL

8 0

2 years ago

WHAT IS THE REMAINDER WHEN <img src="https://tex.z-dn.net/?f=32%5E%7B37%5E%7B32%7D%20%7D" id="TexFormula1" title="32^{37^{32} }"

Feliz [49]

Recall Euler's theorem: if $\gcd(a,n) = 1$ , then

$a^{\phi(n)} \equiv 1 \pmod n$

where $\phi$ is Euler's totient function.

We have $\gcd(9,32) = 1$ - in fact, $\gcd(9,32^k)=1$ for any $k\in\Bbb N$ since $9=3^2$ and $32=2^5$ share no common divisors - as well as $\phi(9) = 6$ .

Now,

$37^{32} = (1 + 36)^{32} \\\\ ~~~~~~~~ = 1 + 36c_1 + 36^2c_2 + 36^3c_3+\cdots+36^{32}c_{32} \\\\ ~~~~~~~~ = 1 + 6 \left(6c_1 + 6^3c_2 + 6^5c_3 + \cdots + 6^{63}c_{32}\right) \\\\ \implies 32^{37^{32}} = 32^{1 + 6(\cdots)} = 32\cdot\left(32^{(\cdots)}\right)^6$

where the $c_i$ are positive integer coefficients from the binomial expansion. By Euler's theorem,

$\left(32^{(\cdots)\right)^6 \equiv 1 \pmod9$

so that

$32^{37^{32}} \equiv 32\cdot1 \equiv \boxed{5} \pmod9$

7 0

1 year ago