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3 years ago

What is -3/11 as a decimal

1 answer:

6 0

It is possible to convert a fraction to a decimal in mathematics. First, -3/11 is a fraction. To convert a fraction which is -3/11, divide 3 by 11 which you get .272727... and you get the idea. The '27' repeats. Therefore, the answer would be -.27 with a <span>Vinculum symbol which is a horizontal line above the number to indicate its repeating. Also, don't forget to put a negative sign!</span>

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2x+4y=6<br> 3x=12-6y<br> Solve by elimination

faltersainse [42]

Answer:

no solution

Step-by-step explanation:

2x + 4y = 6......reduces to x + 2y = 3

3x = 12 - 6y....reduces to x = 4 - 2y...rearranged is x + 2y = 4

so now we have :

x + 2y = 3

x + 2y = 4

ok....I dont have to go any farther to know that this has no solution because ur equations have the same slope and different y int, this means ur lines are parallel and have no solution because they never cross each others path.

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3 years ago

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PLZ HELP **URGENT** BRAINLIEST TO FIRST ANSWER **HELP PLZ**

kumpel [21]

Answer:

Step-by-step explanation:

EF - base1

HG - base2

J-midpoint of EH

K-midpoint of FG

We are solving it by using formula:

JK=(base1+base2):2

25=(3x+2+7x-2):2

25*2=10x

50=10x

x=50:10

x=5

Good luck!

5 0

3 years ago

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Please help got to get grade up

dusya [7]

Answer:

Step-by-step explanation:

The ratio of abc/def=1.5

15/1.5=10

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2 years ago

I AM SO STESSED ILL GIVE 13 POINTS FOR THE RIGHT ANSWER!!!!!!!

Blizzard [7]

The coordinates of the midpoint are (-6,3)

6 0

3 years ago

Cosθ=−2√3 , where π≤θ≤3π2 .

Alex787 [66]

Answer:

$sin(\theta + \beta) = -\frac{\sqrt{7}}{5}-4\frac{\sqrt{2}}{15}$

Step-by-step explanation:

step 1

Find the $sin(\theta)$

we know that

Applying the trigonometric identity

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

we have

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

substitute

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

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

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

$sin^2(\theta)= \frac{7}{9}$

$sin(\theta)=\pm\frac{\sqrt{7}}{3}$

Remember that

π≤θ≤3π/2

Angle θ belong to the III Quadrant

That means ----> The sin(θ) is negative

$sin(\theta)=-\frac{\sqrt{7}}{3}$

step 2

Find the sec(β)

Applying the trigonometric identity

$tan^2(\beta)+1= sec^2(\beta)$

we have

$tan(\beta)=\frac{4}{3}$

substitute

$(\frac{4}{3})^2+1= sec^2(\beta)$

$\frac{16}{9}+1= sec^2(\beta)$

$sec^2(\beta)=\frac{25}{9}$

$sec(\beta)=\pm\frac{5}{3}$

we know

0≤β≤π/2 ----> II Quadrant

sec(β), sin(β) and cos(β) are positive

$sec(\beta)=\frac{5}{3}$

Remember that

$sec(\beta)=\frac{1}{cos(\beta)}$

therefore

$cos(\beta)=\frac{3}{5}$

step 3

Find the sin(β)

we know that

$tan(\beta)=\frac{sin(\beta)}{cos(\beta)}$

we have

$tan(\beta)=\frac{4}{3}$

$cos(\beta)=\frac{3}{5}$

substitute

$(4/3)=\frac{sin(\beta)}{(3/5)}$

therefore

$sin(\beta)=\frac{4}{5}$

step 4

Find sin(θ+β)

we know that

$sin(A + B) = sin A cos B + cos A sin B$

In this problem

$sin(\theta + \beta) = sin(\theta)cos(\beta)+ cos(\theta)sin (\beta)$

we have

$sin(\theta)=-\frac{\sqrt{7}}{3}$

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

$sin(\beta)=\frac{4}{5}$

$cos(\beta)=\frac{3}{5}$

substitute the given values in the formula

$sin(\theta + \beta) = (-\frac{\sqrt{7}}{3})(\frac{3}{5})+ (-\frac{\sqrt{2}}{3})(\frac{4}{5})$

$sin(\theta + \beta) = (-3\frac{\sqrt{7}}{15})+ (-4\frac{\sqrt{2}}{15})$

$sin(\theta + \beta) = -\frac{\sqrt{7}}{5}-4\frac{\sqrt{2}}{15}$

8 0

3 years ago