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Zigmanuir [339]
3 years ago
12

SOMEONE PLEASE HELP ME ASAP PLEASEEEE!!!!​

Mathematics
2 answers:
Andru [333]3 years ago
8 0

Answer:

mx4. I am not completely sure but they are doing -x4 on each of them

Step-by-step explanation:

MariettaO [177]3 years ago
4 0
M times 4 is the answer
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Need help to do this step by step please <br> 3x2 - 2x=-1
sdas [7]

Answer:

Solving the expression 3x^2-2x=-1 we get: \mathbf{x=\frac{1+\sqrt{2}i }{3}\:or\:x=\frac{1-\sqrt{2}i }{3}}

Step-by-step explanation:

We need to solve the expression: 3x^2-2x=-1

This is a quadratic expression and it can be solved using quadratic formula

Solving:

3x^2-2x=-1\\

we can write it as:

3x^2-2x+1=0

The quadratic formula is: x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

where a = 3, b = -2 and c= 1

Putting values and solving:

x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\x=\frac{-(-2)\pm\sqrt{(-2)^2-4(3)(1)}}{2(3)}\\x=\frac{2\pm\sqrt{4-12}}{2(3)}\\x=\frac{2\pm\sqrt{-8}}{6}\\We\:know\:that\:\sqrt{-1}=i\\x=\frac{2\pm\sqrt{8}\sqrt{-1} }{6} \\We\:know\:\sqrt{8}=\sqrt{2\times 2 \times 2}=\sqrt{2^2 \times 2}=2\sqrt{2}   \\x=\frac{2\pm2\sqrt{2}i }{6}\\Now,\\x=\frac{2+2\sqrt{2}i }{6}\:or\:x=\frac{2-2\sqrt{2}i }{6}\\x=\frac{2(1+\sqrt{2}i) }{6}\:or\:x=\frac{2(1-\sqrt{2}i) }{6}\\x=\frac{1+\sqrt{2}i }{3}\:or\:x=\frac{1-\sqrt{2}i }{3}

So, solving the expression 3x^2-2x=-1 we get: \mathbf{x=\frac{1+\sqrt{2}i }{3}\:or\:x=\frac{1-\sqrt{2}i }{3}}

8 0
3 years ago
Based only on the information given in the diagram it is guaranteed that RST UVW
lapo4ka [179]

Answer:

The triangles are similar

Step-by-step explanation:

We are given and two angles  so by default we know the third angle.  We know the three angles are the same, so the triangles are similar by AAA.  We do not about conqruency without a side length.

6 0
3 years ago
Read 3 more answers
HELP PLEASE 30 POINTS! WILL MARK BRAINLIEST IF CORRECT!!!!!
KATRIN_1 [288]

Answer:

See Below.

Step-by-step explanation:

We want to verify:

\cos(x-y)-\cos(x+y)=2\sin(x)\sin(y)

We will utilize the following identities:

\cos(x-y)=\cos(x)\cos(y)+\sin(x)\sin(y)

And:

\cos(x+y)=\cos(x)\cos(y)-\sin(x)\sin(y)

So, by substitution, we acquire:

(\cos(x)\cos(y)+\sin(x)\sin(y))-(\cos(x)\cos(y)-\sin(x)\sin(y))=2\sin(x)\sin(y)

Distribute:

\cos(x)\cos(y)+\sin(x)\sin(y)-\cos(x)\cos(y)+\sin(x)\sin(y)=2\sin(x)\sin(y)

The first and third term will cancel:

\sin(x)\sin(y)+\sin(x)\sin(y)=2\sin(x)\sin(y)

Combine like terms:

2\sin(x)\sin(y)\stackrel{\checkmark}{=}2\sin(x)\sin(y)

5 0
3 years ago
Read 2 more answers
Find the minimum or maximum value of the function g(x)=−3x2−6x+5
polet [3.4K]

Step-by-step explanation:

g(x) = ax²+bx+c

g(x)=−3x²−6x+5

a = -3, b= -6, c = 5

since a <0 , the function has only maximum value.

=> g'(x) = 0

-6x -6 = 0

-6x = 6

x = -1

the maximum value => g(-1) =

-3(-1)²-6(-1)+5 = -3+6+5 = 9

the domain : {x | x € Real numbers}

the range : {y| y ≤ 9, y € Real numbers}

the function is increasing for x < -1

the function is decreasing for x > -1

4 0
3 years ago
"select all the numbers that are solutions to the equation x^3=27"
Ymorist [56]
B and C

3^3 = 27
the cubed root of 27 is 3
4 0
3 years ago
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