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Nostrana [21]
3 years ago
6

Please helpppppp? It's geometry

Mathematics
1 answer:
stepan [7]3 years ago
7 0
B is the best answer.
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Write the equation of the line that passes through the point (3,5) and is perpendicular to the line y=3.
lorasvet [3.4K]

Answer:

x=3 and y=5 passes through(3,5)

x=2 and x=3 is perpendicular to y=3

so anwer is x=3

4 0
4 years ago
HURRRRRRRRRRRYYY PLEASE!Brainlest
earnstyle [38]

The scale factor is some number between 0 and 1 (not including either endpoint) because the image is smaller than the preimage.

Triangle ABC is the preimage, or the "before"; while triangle A'B'C' is the image, which is the "after".

preimage ----> image

before ----> after

8 0
3 years ago
Sebuah balok berukuran panjang 3 cm, lebar 4 cm, dan panjang diagonal ruang 13 cm. Volume balok adalah
kramer

Answer:

V=144\ cm^3

Step-by-step explanation:

#We use the base diagonal and the height diagonal to calculate the beam's height:

h=\sqrt{13^2-(4^2+3^2)}\\\\h=12\ cm

#The volume of the beam can then be calculated as:

V=lwh\\\\=4\times 3\times 12\\\\=144\ cm^3

Hence, the beam's volume is 144\ cm^3

5 0
3 years ago
If the base is 4, what is the value if the exponent is 2? What if the exponent is -2?
Vikentia [17]

Answer:

In the world of exponents, 4 is the number being raised by the exponent, which is 2.

Let's answer the first question:

If the base is 4, what is the value if the exponent is 2?

  • The base is 4 and the exponent is 2.  We would multiply 4 two times.

4^2= 4*4 = 16

So, it would be 16.

Let's answer the second question:

What if the exponent is -2?

<u>This is the rule for negative exponents:</u>

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

Using this rule, we can solve 4^{-2}.

4^{-2}= \frac{1}{4^2}=\frac{1}{16}

So, our answer for the 2nd question is 1/16.

3 0
4 years ago
Using fermat's little theorem, find the least positive residue of $2^{1000000}$ modulo 17.
torisob [31]
Fermat's little theorem states that
a^p≡a mod p

If we divide both sides by a, then
a^{p-1}≡1 mod p
=>
a^{17-1}≡1 mod 17
a^{16}≡1 mod 17

Rewrite
a^{1000000} mod 17  as
=(a^{16})^{62500} mod 17
and apply Fermat's little theorem
=(1)^{62500} mod 17
=>
=(1) mod 17

So we conclude that
a^{1000000}≡1 mod 17

6 0
4 years ago
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