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igomit [66]
3 years ago
14

ME NEEDS SOME HELP WITH MATH PLS

Mathematics
2 answers:
SIZIF [17.4K]3 years ago
7 0
C. |-3| < |9| and -3 < 9
Angelina_Jolie [31]3 years ago
5 0

♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️

The  \:  \: correct  \:  \: answer \:  \:  is \:  \:  (( C )) . \\

♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️♥️

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Simora [160]

Answer:

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8 0
3 years ago
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Please help guys!!! due soon
MaRussiya [10]

Answer:

6

Step-by-step explanation:

37/37+5=1+5=6

3 0
3 years ago
14. Classify Each Angle Pair then find the value of X
monitta
Set the two equation equal to each other to solve for x. Angles are acute because they are less than 90 degrees.

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3 years ago
Find the lengths of nails with the following penny sizes: 3, 6, and 10. The 3-penny nail is inches. the 6 penny is__ the 3 penny
stealth61 [152]

the lengths of nails with the following penny sizes: 3, 6, and 10

  • A 3-penny nail is_5/4_inches.
  • A 6-penny nail is_2_inches.
  • A 10-penny nail is_3_inches.

This is further explained below.

<h3>What is the lengths of nails with the following penny sizes: 3, 6, and 10?</h3>

That is the equation solved for n.

The penny size is d, and n is the length of the nail, then:

if d = 3, we have:

n = (3 + 2)/4 = 5/4

if d = 6, then

n = (6 + 2)/4 = 2

if d = 10, then:

n = (10 + 2)/4 = 3

Thus:

A 3-penny nail is_5/4_inches.

A 6-penny nail is_2_inches.

A 10-penny nail is_3_inches.

if you want to learn more about linear equations:

brainly.com/question/1884491

#SPJ1

6 0
2 years ago
An equation of a hyperbola is given.
siniylev [52]

Answer:

a)

The vertices are \left(3,\:0\right),\:\left(-3,\:0\right).

The foci are \left(3\sqrt{5},\:0\right),\:\left(-3\sqrt{5},\:0\right).

The asymptotes are y=2x,\:y=-2x.

b) The length of the transverse axis is 6.

c) See below.

Step-by-step explanation:

\frac{\left(x-h\right)^2}{a^2}-\frac{\left(y-k\right)^2}{b^2}=1 is the standard equation for a right-left facing hyperbola with center \left(h,\:k\right).

a)

The vertices\:\left(h+a,\:k\right),\:\left(h-a,\:k\right) are the two bending points of the hyperbola with center \:\left(h,\:k\right) and semi-axis a, b.

Therefore,

\frac{x^2}{9}-\frac{y^2}{36}=1, is a right-left Hyperbola with \:\left(h,\:k\right)=\left(0,\:0\right),\:a=3,\:b=6 and vertices \left(3,\:0\right),\:\left(-3,\:0\right).

For a right-left facing hyperbola, the Foci (focus points) are defined as \left(h+c,\:k\right),\:\left(h-c,\:k\right) where c=\sqrt{a^2+b^2} is the distance from the center \left(h,\:k\right) to a focus.

Therefore,

\frac{x^2}{9}-\frac{y^2}{36}=1, is a right-left Hyperbola with \:\left(h,\:k\right)=\left(0,\:0\right),\:a=3,\:b=6 c=\sqrt{3^2+6^2}= 3\sqrt{5} and foci \left(3\sqrt{5},\:0\right),\:\left(-3\sqrt{5},\:0\right)

The asymptotes are the lines the hyperbola tends to at \pm \infty. For right-left hyperbola the asymptotes are: y=\pm \frac{b}{a}\left(x-h\right)+k

Therefore,

\frac{x^2}{9}-\frac{y^2}{36}=1, is a right-left Hyperbola with \:\left(h,\:k\right)=\left(0,\:0\right),\:a=3,\:b=6 and asymptotes

y=\frac{6}{3}\left(x-0\right)+0,\:\quad \:y=-\frac{6}{3}\left(x-0\right)+0\\y=2x,\:\quad \:y=-2x

b) The length of the transverse axis is given by 2a. Therefore, the lenght is 6.

c) See below.

4 0
4 years ago
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