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3 years ago

20 POINTS HELP ASAP

Mathematics

1 answer:

Anon25 [30]3 years ago

6 0

Answer:

Step-by-step explanation:

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Graph this please due today

vaieri [72.5K]

Answer: see last picture

Step-by-step explanation:

We see that the y-intercept is 10 and the slope is -3

so when x = 0, y = 10

Graph this first point (picture 1)

Since the slope is -3, every time you go one unit to the left, you go down 3 units, so graph this second point (picture 2)

Continue this until you have no more room on the graph (picture 3)

now draw a line through the dots (picture 4)

4 0

3 years ago

How is this solved using trig identities (sum/difference)?

GenaCL600 [577]

FIRST PART
We need to find sin α, cos α, and cos β, tan β
α and β is located on third quadrant, sin α, cos α, and sin β, cos β are negative

Determine ratio of ∠α
Use the help of right triangle figure to find the ratio
tan α = 5/12
side in front of the angle/ side adjacent to the angle = 5/12
Draw the figure, see image attached

Using pythagorean theorem, we find the length of the hypotenuse is 13
sin α = side in front of the angle / hypotenuse
sin α = -12/13

cos α = side adjacent to the angle / hypotenuse
cos α = -5/13

Determine ratio of ∠β
sin β = -1/2
sin β = sin 210° (third quadrant)
β = 210°

$cos \beta = -\frac{1}{2} \sqrt{3}$

$tan \beta= \frac{1}{3} \sqrt{3}$

SECOND PART
Solve the questions
Find sin (α + β)
sin (α + β) = sin α cos β + cos α sin β
$sin( \alpha + \beta )=(- \frac{12}{13} )( -\frac{1}{2} \sqrt{3})+( -\frac{5}{13} )( -\frac{1}{2} )$
$sin( \alpha + \beta )=(\frac{12}{26}\sqrt{3})+( \frac{5}{26} )$
$sin( \alpha + \beta )=(\frac{5+12\sqrt{3}}{26})$

Find cos (α - β)
cos (α - β) = cos α cos β + sin α sin β
$cos( \alpha + \beta )=(- \frac{5}{13} )( -\frac{1}{2} \sqrt{3})+( -\frac{12}{13} )( -\frac{1}{2} )$
$cos( \alpha + \beta )=(\frac{5}{26} \sqrt{3})+( \frac{12}{26} )$
$cos( \alpha + \beta )=(\frac{5\sqrt{3}+12}{26} )$

Find tan (α - β)
$tan( \alpha - \beta )= \frac{ tan \alpha-tan \beta }{1+tan \alpha tan \beta }$
$tan( \alpha - \beta )= \frac{ \frac{5}{12} - \frac{1}{2} \sqrt{3} }{1+(\frac{5}{12}) ( \frac{1}{2} \sqrt{3})}$

Simplify the denominator
$tan( \alpha - \beta )= \frac{ \frac{5}{12} - \frac{1}{2} \sqrt{3} }{1+(\frac{5\sqrt{3}}{24})}$
$tan( \alpha - \beta )= \frac{ \frac{5}{12} - \frac{1}{2} \sqrt{3} }{ \frac{24+5\sqrt{3}}{24} }$

Simplify the numerator
$tan( \alpha - \beta )= \frac{ \frac{5}{12} - \frac{6}{12} \sqrt{3} }{ \frac{24+5\sqrt{3}}{24} }$
$tan( \alpha - \beta )= \frac{ \frac{5-6\sqrt{3}}{12} }{ \frac{24+5\sqrt{3}}{24} }$

Simplify the fraction
$tan( \alpha - \beta )= (\frac{5-6\sqrt{3}}{12} })({ \frac{24}{24+5\sqrt{3}})$
$tan( \alpha - \beta )= \frac{10-12\sqrt{3} }{ 24+5\sqrt{3}}$

7 0

2 years ago

A middle school took all of its 6th grade students on a field trip to see a symphony. The students filled 300 seats, which was 1

prisoha [69]

Answer:

3,000

Step-by-step explanation:

Basically, there are 300 seats filled which is 10%

100% is 10x10

300=10%

multiply both sides by 10

300x10=3000 and 10x10=100

therefore, the total number of seats is 3000 since 100% represents the whole thing.

Hope this was helpful!

5 0

2 years ago

Given the original number n. Multiply the number by 9. Add 99. Divide this sum by 9. Subtract the original number, n, from the q

tangare [24]

Answer:

Step-by-step explanation:

Hello,

Given the original number n.

$\large \boxed{n}$

Multiply the number by 9.

$\large \boxed{9n}$

Add 99.

$\large \boxed{9n+99}$

Divide this sum by 9.

$\large \boxed{\dfrac{9n+99}{9}=n+11}$

Subtract the original number, n, from the quotient.

$\large \boxed{n+11-n=11}$

Thank you.

4 0

2 years ago

An elementary ntimesn scaling matrix with k on the diagonal is the same as the ntimesn identity matrix with _______ exactly one

Dafna11 [192]

Answer:

exactly one, 0's, triangular matrix, product and 1.

Step-by-step explanation:

So, let us first fill in the gap in the question below. Note that the capitalized words are the words to be filled in the gap and the ones in brackets too.

"An elementary ntimesn scaling matrix with k on the diagonal is the same as the ntimesn identity matrix with EXACTLY ONE of the (0's) replaced with some number k. This means it is TRIANGULAR MATRIX, and so its determinant is the PRODUCT of its diagonal entries. Thus, the determinant of an elementary scaling matrix with k on the diagonal is (1).

Here, one of the zeros in the identity matrix will surely be replaced by one. That is to say, the determinants = 1 × 1 × 1 => 1. Thus, it is a a triangular matrix.

3 0

2 years ago