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Salsk061 [2.6K]
3 years ago
9

If Sine (x) = two-fifths and tan(x) > 0, what is sin(2x)?

Mathematics
2 answers:
julsineya [31]3 years ago
8 0

Answer:

B

Step-by-step explanation:

edg 2021

ElenaW [278]3 years ago
4 0

B) \frac{4\sqrt{21}}{25}

Correct on Edge

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12 pots cost $27 . what is the cost of 8 pots?​
Alborosie

Answer:

$18

Step-by-step explanation:

you divide 27/12= 2.25 this gives you the price for one.

then you multiply 8 × 2.25 = $18

$18 dollars is the price for 8 pots

6 0
2 years ago
(04.01) Which point could be removed in order to make the relation a function? (4 points) {(0, 2), (3, 8), (−4, −2), (3, −6), (−
umka21 [38]
A function will not have any repeating x values...it can have repeating y values, just not the x ones

{(0,2),(3,8),(-4,-2),(3,-6),(-1,8),(8,3)}
u would have to remove one of the sets of points that has 3 as its x value....so either remove (3,8) or (3,-6)....because with both of them in there, u have repeating x values
7 0
3 years ago
Read 2 more answers
Nasir and Seif began with an equal number of Skittles. Nasir ate 10 Skittles, and Seif ate 6. Now the ratio of Nasir's Skittles
Finger [1]
X=Amount of skittles at the start.
Nasir ate 10 skittles and Seif ate 6 skittles so the number of skittles they had(x) - how many they both ate equals : 2/3;

(x-10)/(x-6)=2/3
3(x-10)=2(x-6)
3x-30=2x-12
x-30=-12
x=18

They each started with 18 skilttles.
7 0
3 years ago
Does anyone know the answer
alukav5142 [94]

676 out of 650 is 104%

7 0
2 years ago
A is a 3 times 3 matrix with three pivot positions.
saw5 [17]

Answer:

The following are the answer:

In option a  "No".

In option b  "Yes".

Step-by-step explanation:

In choice a:

Ax = 0 has no nontrivial solution. A would be the three-pivot matrix, it may assume, that the function has no free variable, and only if the function has had at least one free factor are their nontrivial formulas for the equations of the form Ax=0.  

It implies that since A is a 3x3 matrix, has no free variables so that it has no non-trivial choices, and Ax = 0.

In choice b:

we assume that every potential has at least one solution that is Ax=b . If A does have a three-pivot matrix, It will be a pivot element for each row and column, and for each possible b∈ R³, Ax = b has at least one solution.

7 0
3 years ago
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