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ladessa [460]
3 years ago
11

HELP PLEASE!!!

Mathematics
2 answers:
Akimi4 [234]3 years ago
8 0

Answer:-3

Step-by-step explanation:

lbvjy [14]3 years ago
7 0

Answer:

Create equivalent expressions in the equation that all have equal bases, then solve for  

x

.

Exact Form:

5/3

Decimal Form:

x

=

1.6

Mixed Number Form:

x

=

1  2/3

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Help with setting up these two problems there 2 different problems
ELEN [110]
I'm not good at math i cant help
7 0
3 years ago
Read 2 more answers
3
Y_Kistochka [10]

Answer:

D) y = -1/2x + 4

Step-by-step explanation:

Points on the graph: (0, 4) and (8, 0)

Slope:

m=(y2-y1)/(x2-x1)

m=(0 - 4)/(8-0)

m= -4/8

m = -1/2

Slope-intercept:

y - y1 = m(x - x1)

y - 4 = -1/2(x - 0)

y - 4 = -1/2x

y = -1/2x + 4

8 0
2 years ago
A triangular sail has a base length of 2.5 meters.The area of the sail is 3.75 square meters.How tall is the sail?
love history [14]
Area of a triangle = 1/2 x base x height

--------------------------------------
Find Height :
--------------------------------------
3.75  = 1/2 x 2.5 x height

3.75 = 1.25 x height 

height = 3.75 ÷ 1.25

height = 3 meters

----------------------------------------------------------------------------
Answer: The height is 3 meters.
----------------------------------------------------------------------------
7 0
2 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
Lines of latitude range from ___ to ___, lines of longitude range from ___ to ___
dimaraw [331]

Answer:

Step-by-step explanation: Lines of latitude range from north to south, lines of longitude range from east to west

3 0
3 years ago
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