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inna [77]
3 years ago
6

Please help I think I picked the wrong answer !!

Mathematics
1 answer:
insens350 [35]3 years ago
6 0

Answer:

one solution

Step-by-step explanation:

it has one solution because the slopes of the two equations(1x & 2x) are different, which means the lines of the two equations are not parallel, they have an intersection point(also means one solution).

You might be interested in
What is the solution to the inequality 2n+5| > 1?
hodyreva [135]

Answer:

n < - 3 or n > - 2

Step-by-step explanation:

Inequalities of the type | x | > a , have solutions of the form

x < - a or x > a

Then

2n + 5 < - 1 or 2n + 5 > 1

Solve both inequalities

2n + 5 < - 1 ( subtract 5 from both sides )

2n < - 6 ( divide both sides by 2 )

n < - 3

OR

2n + 5 > 1 ( subtract 5 from both sides )

2n > - 4 ( divide both sides by 2 )

n > - 2

Solution is n < - 3 or n > - 2

3 0
3 years ago
105-30-45-3028+10000000000000038492=?
Volgvan

Answer:

105-30-45-3028+10000000000000038492= 1.000000000000003e19

Step-by-step explanation:

I did the math

have a good day!!

8 0
3 years ago
The highest point in Louisiana is Driskill Mountain at 585 feet. The lowest points -8 feet in New Orleans. What is the differenc
Eva8 [605]

The answer is a 577 foot difference

7 0
3 years ago
Find a third number so that three numbers form a right triangle.<br><br>9,41​
expeople1 [14]

Answer:

40

Step-by-step explanation:

a^2+b^2=c^2

therefore c^2-b^2=a^2

41x41=1681

9x9=81

1681-81=1600

Sqrt1600=40

5 0
2 years ago
Read 2 more answers
If tan a = 3/4 find the value of sin 3a
Effectus [21]
T=-1

sinA=sin(π/2-3A), A=2nπ+π/2-3A, 4A=2nπ+π/2, A=nπ/2+π/8 where n is an integer.

Also, π-A=2nπ+π/2-3A, 2A=2nπ-π/2, A=nπ-π/4.

The hard way:

cos3A=cos(2A+A)=cos(2A)cosA-sin(2A)sinA.

Let s=sinA and c=cosA, then s²+c²=1.

cos3A=(2c²-1)c-2c(1-c²)=c(4c²-3).

s=c(4c²-3) is the original equation.

Let t=tanA=s/c, then c²=1/(1+t²).

t=4c²-3=4/(1+t²)-3=(4-3-3t²)/(1+t²)=(1-3t²)/(1+t²).

So t+t³=1-3t², t³+3t²+t-1=0=(t+1)(t²+2t-1).

So t=-1 is a solution.

t²+2t-1=0 is a solution, t²+2t+1-1-1=0=(t+1)²-2, so t=-1+√2 and t=-1-√2 are solutions.

Therefore tanA=-1, -1+√2, -1-√2 are the three solutions from which:

A=-π/4, π/8, -3π/8 radians and these values +2πn where n is an integer.

Replacing π by 180° converts the solutions to degrees.

3 0
2 years ago
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