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3 years ago

The two plots below show the heights of some sixth graders and some seventh graders:

Mathematics

2 answers:

stiv31 [10]3 years ago

8 0

Answer:

i think its A

Step-by-step explanation:

DiKsa [7]3 years ago

6 0

A I think, someone correct me if I’m wrong.

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These liness pass through the same point

anyanavicka [17]

Answer:

All of them except B

Step-by-step explanation:

Bisecting is splitting a line in half

Parallel lines never touch each other

Intersecting lines intersect

Corresponding lines are 2 parallel lines with a line that is going through them

8 0

3 years ago

Two times the sum of a number and two is the same as five less than five times the number. What is the number? In number form wi

9966 [12]

2(2+x)=5x-5

x=3

hope this helps

8 0

3 years ago

The parking garage charges $3 for the first hour of parking and $1.45 for each additional hour.

Alika [10]

I don't know what you are asking but my best guess is it will be
$4.46 in total

8 0

3 years ago

Read 2 more answers

Which equation represents the line through (0,0) and (3,4)? a. y=3/4x b.y=4/3x c.y=4x d.y=3x

love history [14]

Answer:

Step-by-step explanation:

The equation of a line passing through the origin is

y = mx ( m is the slope )

Calculate m using the slope formula

m = $\frac{y_{2}-y_{1} }{x_{2}-x_{1} }$

with (x₁, y₁ ) = (0, 0 ) and (x₂, y₂ ) = (3, 4 )

m = $\frac{4-0}{3-0}$ = $\frac{4}{3}$

y = $\frac{4}{3}$ x → b

3 0

3 years ago

Find the exact value of cot 330° in simplest form with a rational denominator.

Hoochie [10]

Answer:

$\cot 330^{\circ} = -\sqrt{3}$

Step-by-step explanation:

The cotangent function can be rewritten by trigonometric relations, that is:

$\cot 330^{\circ} = \frac{1}{\tan 330^{\circ}} = \frac{\cos 330^{\circ}}{\sin 330^{\circ}}$ (1)

By taking approach the periodicity properties of the cosine and sine function (both functions have a period of 360°), we use the following equivalencies:

$\sin 330^{\circ} = \sin (-30^{\circ}) = -\sin 30^{\circ}$ (2)

$\cos 330^{\circ} = \cos (-30^{\circ}) = \cos 30^{\circ}$ (3)

By (2) and (3) in (1), we have following expression:

$\cot 330^{\circ} = -\frac{\cos 30^{\circ}}{\sin 30^{\circ}}$

If we know that $\sin 30^{\circ} = \frac{1}{2}$ and $\cos 30^{\circ} = \frac{\sqrt{3}}{2}$ , then the result of the trigonometric expression is:

$\cot 330^{\circ} = -\frac{\frac{\sqrt{3}}{2} }{\frac{1}{2} }$

$\cot 330^{\circ} = -\sqrt{3}$

6 0

3 years ago