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svp [43]
3 years ago
6

"x + 6(x + 2)(x - 2)" align="absmiddle" class="latex-formula">
​
Mathematics
2 answers:
frutty [35]3 years ago
7 0
X+6(x^2-4)
x+6x^2-24
6x^2+x-24
Anvisha [2.4K]3 years ago
6 0

Step-by-step explanation:

JOlN me for $€X

if any girI can $How her bøõßs JOlN me

dgj-gwud-xhm

and first show me your hairs for confirmation

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The distance around Mia's vegetable garden is 100 feet. Mia has 20 yards of
arsen [322]

Add 40 plus 60 80 160

3 0
3 years ago
In the formula for the volume of a cylinder, how is the area of the base determined?
Len [333]

Answer:

area of base =  \pi r^2

where 'r' is the radius

Step-by-step explanation:

Volume of the cylinder formula = \pi r^2h

volume of cylinder = area of base * height

The base of a cylinder is a circle. so we use the area of circle for the base of cylinder while finding volume

We know area of the circle is \pi r^2

So area of the base of cylinder is  \pi r^2

To find volume of the cylinder , we multiply the base of the cylinder with height

To determine the base of the cylinder we need radius of the cylinder

area of base =  \pi r^2

where 'r' is the radius

8 0
3 years ago
Find the area of the lateral faces of the right triangular prism with altitude of 5 cm and base edges of length 3 cm, 4 cm, and
Bingel [31]
The area of all 3 lateral triangular faces will be found by using the formula for area of a triangle.

A = 1/2bh

Face 1:
1/2 x 3 x 5
A = 7.5 square cm

Face 2:
1/2 x 4 x 5
A = 10 square cm

Face 1:
1/2 x 5 x 5
A = 12.5 square cm

12.5 + 7.5 + 10 = 30 square cm

The area of the lateral faces is 30 square cm.
3 0
4 years ago
Read 2 more answers
Suppose you rent a limousine for a formal reception. The bill for the evening is $53.00. A tax of 6% will be added, and you want
kramer
A=tip 1
b=tip 2
t=total tip
c-total cost

a+b=t
53+t=c

53*.06=a
53*.20=b
a=3.18
b=10.60

t=3.18+10.60
t=13.78

c=53+13.78
total cost=$66.78
6 0
3 years ago
If cos() = − 2 3 and is in Quadrant III, find tan() cot() + csc(). Incorrect: Your answer is incorrect.
nydimaria [60]

Answer:

\tan(\theta) \cdot \cot(\theta) + \csc(\theta) = \frac{5 - 3\sqrt 5}{5}

Step-by-step explanation:

Given

\cos(\theta) = -\frac{2}{3}

\theta \to Quadrant III

Required

Determine \tan(\theta) \cdot \cot(\theta) + \csc(\theta)

We have:

\cos(\theta) = -\frac{2}{3}

We know that:

\sin^2(\theta) + \cos^2(\theta) = 1

This gives:

\sin^2(\theta) + (-\frac{2}{3})^2 = 1

\sin^2(\theta) + (\frac{4}{9}) = 1

Collect like terms

\sin^2(\theta)  = 1 - \frac{4}{9}

Take LCM and solve

\sin^2(\theta)  = \frac{9 -4}{9}

\sin^2(\theta)  = \frac{5}{9}

Take the square roots of both sides

\sin(\theta)  = \±\frac{\sqrt 5}{3}

Sin is negative in quadrant III. So:

\sin(\theta)  = -\frac{\sqrt 5}{3}

Calculate \csc(\theta)

\csc(\theta) = \frac{1}{\sin(\theta)}

We have: \sin(\theta)  = -\frac{\sqrt 5}{3}

So:

\csc(\theta) = \frac{1}{-\frac{\sqrt 5}{3}}

\csc(\theta) = \frac{-3}{\sqrt 5}

Rationalize

\csc(\theta) = \frac{-3}{\sqrt 5}*\frac{\sqrt 5}{\sqrt 5}

\csc(\theta) = \frac{-3\sqrt 5}{5}

So, we have:

\tan(\theta) \cdot \cot(\theta) + \csc(\theta)

\tan(\theta) \cdot \cot(\theta) + \csc(\theta) = \tan(\theta) \cdot \frac{1}{\tan(\theta)} + \csc(\theta)

\tan(\theta) \cdot \cot(\theta) + \csc(\theta) = 1 + \csc(\theta)

Substitute: \csc(\theta) = \frac{-3\sqrt 5}{5}

\tan(\theta) \cdot \cot(\theta) + \csc(\theta) = 1 -\frac{3\sqrt 5}{5}

Take LCM

\tan(\theta) \cdot \cot(\theta) + \csc(\theta) = \frac{5 - 3\sqrt 5}{5}

6 0
3 years ago
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