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VikaD [51]
3 years ago
6

What is the value of the expression 30+[(6/3)+(3+4)]

Mathematics
2 answers:
kolbaska11 [484]3 years ago
4 0

30+[(6/3)+(3+4)]

30+[2+7]

30+9

39

Lady_Fox [76]3 years ago
3 0

Do you want me to show the work? If not then the answer is 39.


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Find all values of x such that sin 2x = sin x and 0 ≤ x ≤ 2π.
jarptica [38.1K]
Hello : 
<span> sin 2x = sin x and 0 ≤ x ≤ 2π.
all solutions : 
2x= x +2k</span>π    or x= π -x +2kπ ..... k in : Z
x = 2kπ  or : x = π/2 + kπ 
but :
 <span>0 ≤ x ≤ 2π
</span><span>all values of x such that sin 2x = sin x and 0 ≤ x ≤ 2π are : 
</span>k = 0 : x=0    , x= π/2
k=1 : x=2 π    , x=3π/2
5 0
3 years ago
What is the equation of the line that passes through the points (-3,5) (6,8)
Ratling [72]

(\stackrel{x_1}{-3}~,~\stackrel{y_1}{5})\qquad (\stackrel{x_2}{6}~,~\stackrel{y_2}{8}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{8}-\stackrel{y1}{5}}}{\underset{run} {\underset{x_2}{6}-\underset{x_1}{(-3)}}}\implies \cfrac{3}{6+3}\implies \cfrac{3}{9}\implies \cfrac{1}{3}

\begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{5}=\stackrel{m}{\cfrac{1}{3}}(x-\stackrel{x_1}{(-3)}) \\\\\\ y-5=\cfrac{1}{3}(x+3)\implies y-5=\cfrac{1}{3}x+1\implies y=\cfrac{1}{3}x+6

6 0
2 years ago
HELP ASAP
slega [8]

Answer:

No

Step-by-step explanation:

The sample size wasn't big enough. It could only be a trend in that grade.

7 0
3 years ago
Read 2 more answers
Determine whether the given lengths can be sides of a right triangle. Which of the following are true statements. A.The lengths
Stels [109]

Answer:

Given sides 12, 16 and 20 can be the sides of right triangle.

Step-by-step explanation:

Sides of right triangle always follow the Pythagoras theorem.

i.e (base)^2 + (Height)^2 = (Hypotenuse)^2

For the given Lengths 7, 40 and 41

We need to check if

7^2 + 40^2 =41^2 \\or\\ 7^2 + 40^2 \neq41^2

Since \\7^2 + 40^2 = 1649\\and \\41^2 = 1681

That means, \\ 7^2 + 40^2 \neq 41^2

hence 7,40 and 41 can not be the sides of right triangle.

Next,

Given sides 12,16 and 20.

Again follow the similar process used in the above problem.

12^2 + 16^2 =400\\And \\20^2 = 400\\Since 12^2 + 16^2 = 20^2

Therefore given sides 12,16 and 20 can be the sides of right triangle.


7 0
3 years ago
Help me pls i don't know how to math thank you
Dimas [21]
Number 2 is 6
i dont know number 3
<span>  number 4 is _x-intercept = -12/11 = -1.09091  y-intercept = -12/4 = -3
</span>number 5 is x=12 ,<span>y=<span>−8</span></span><span>y=<span>-<span>8</span></span></span>

6 0
3 years ago
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