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3 years ago

HELLO Everyone ....I hope your safe and doing welll :))) anywayyyys have a good day:))

Mathematics

2 answers:

Alborosie3 years ago

7 0

Answer:

hope you have a good day to,☺

dalvyx [7]3 years ago

7 0

Answer:

you to keep safe and to your family...the student will appreciate of u because you will served the word positive

Step-by-step explanation:

thank you for cheking...

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Solve for X. * (1 Point) x + 70 76° 410

Yuki888 [10]

Answer:

x = -7

Step-by-step explanation:

41 + 76 + x + 70 = 180

x + 187 = 180

x = 180 - 187

x = -7

7 0

2 years ago

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Consider a graph for the equation y= -3x+4. What is the y intercept?

Alexeev081 [22]

Answer:

Step-by-step explanation:

The equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

y = - 3x + 4 is in this form

with y- intercept c = 4 → A

3 0

3 years ago

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What is a coefficient

Oksi-84 [34.3K]

Answer:

a numerical or constant quantity placed before and multiplying the variable in an algebraic expression

Step-by-step explanation:

7 0

2 years ago

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Given the Arithmetic sequence A1,A2,A3,A4 53, 62, 71, 80 What is the value of A38?

Murrr4er [49]

Answer:

$A_{38} = 350$

Step-by-step explanation:

The 5th term of the arithmetic sequence is 53. We can write the equation:

$a + 4d = 53...(1)$

The 6th term of the arithmetic sequence is 62. We can write the equation:

$a + 5d = 62...(2)$

Subtract the first equation from the second one to get:

$5d - 4d = 62 - 53$

$d = 9$

The first term is

$a + 4(9) = 53$

$a + 36 = 53$

$a = 53 - 36$

$a = 17$

The 38th term of the sequence is given by:

$A_{38} = a + 37d$

$A_{38} = 17+ 37(9)$

$A_{38} = 350$

3 0

3 years ago

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Solve the equation in the complex number system. x^2-12x+40=0

mylen [45]

solving the equation $x^2-12x+40=0$ we get $x=6+2i\,\,and\,\,x=6-2i$

Step-by-step explanation:

Solve the equation:

$x^2-12x+40=0$

We can solve using quadratic Formula:

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

a= 1, b = -12, c = 40

Putting values:

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\x=\frac{-(-12)\pm\sqrt{(-12)^2-4(1)(40)}}{2(1)}\\x=\frac{12\pm\sqrt{144-160}}{2}\\x=\frac{12\pm\sqrt{-16}}{2}\\We\,\,know\,\,\sqrt{-1}=i\\ x=\frac{12\pm4i}{2}\\ x=\frac{12+4i}{2}\,\,and\,\, x=\frac{12-4i}{2}\\ x=\frac{4(3+i)}{2}\,\,and\,\, x=\frac{4(3-i)}{2}\\x=2(3+i)\,\,and\,\,x=2(3-i)\\x=6+2i\,\,and\,\,x=6-2i$

So, solving the equation $x^2-12x+40=0$ we get $x=6+2i\,\,and\,\,x=6-2i$

Keywords: Solving quadratic equation

Learn more about Solving quadratic equation at:

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5 0

3 years ago