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Margaret [11]
3 years ago
11

The work of a student to find the dimensions of a rectangle of area 8 + 16x and width 4 is shown below:

Mathematics
1 answer:
enyata [817]3 years ago
4 0
Step 3
I think ?
Not sure
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How many days in the month
iren2701 [21]

January : 31 days.

February : February is the second and shortest month of the year in the Julian and Gregorian calendar with 28 days in common years and 29 days in leap years, with the quadrennial 29th day being called the leap day. It is the first of five months to have a length of less than 31 days, and the only month to have less than 30 days

March : 31 days.

April : 30 days.

May : 31 days.

June : 30 days.

July : 31 days.

August : 31 days.

Septembers : 30 days.

October : 31 days.

November : 30 days.

December : 31 days.



Hope this helps!

7 0
4 years ago
Helpppppp fastttttttttt
crimeas [40]

Answer:

d

Step-by-step explanation:

8 0
3 years ago
Find the domain of fg
vazorg [7]
We have the following functions:
 f (x) = x ^ 2 + 1
 g (x) = 1 / x
 Multiplying we have:
 (f * g) (x) = (x ^ 2 + 1) * (1 / x)
 Rewriting:
 (f * g) (x) = ((x ^ 2 + 1) / x)
 Therefore, the domain of the function is given by all the values of x that do not make zero the denominator.
 We have then:
 All reals except number 0
 Answer:
 
b. all real numbers, except 0
7 0
4 years ago
Read 2 more answers
Directions: Find the indicated trigonometric ratio as a fraction in simplest form. 1. Sin L =
Rainbow [258]

Answer:

\sin L = 0.60

tan\ N = 1.33

\cos L = 0.80

\sin N = 0.80

Step-by-step explanation:

Given

See attachment

From the attachment, we have:

MN = 6

LN = 10

First, we need to calculate length LM,

Using Pythagoras theorem:

LN^2 = MN^2 + LM^2

10^2 = 6^2 + LM^2

100 = 36 + LM^2

Collect Like Terms

LM^2 = 100 - 36

LM^2 = 64

LM = 8

Solving (a): \sin L

\sin L = \frac{Opposite}{Hypotenuse}

\sin L = \frac{MN}{LN}

Substitute values for MN and LN

\sin L = \frac{6}{10}

\sin L = 0.60

Solving (b): tan\ N

tan\ N = \frac{Opposite}{Adjacent}

tan\ N = \frac{LM}{MN}

Substitute values for LM and MN

tan\ N = \frac{8}{6}

tan\ N = 1.33

Solving (c): \cos L

\cos L = \frac{Adjacent}{Hypotenuse}

\cos L = \frac{LM}{LN}

Substitute values for LN and LM

\cos L = \frac{8}{10}

\cos L = 0.80

Solving (d): \sin N

\sin N = \frac{Opposite}{Hypotenuse}

\sin N = \frac{LM}{LN}

Substitute values for LM and LN

\sin N = \frac{8}{10}

\sin N = 0.80

7 0
3 years ago
What equals 7 in multiplication
Zanzabum
Will there's 
5.5 x 2 = 11

hope that helped
8 0
3 years ago
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