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3 years ago

CAN SOMEONE HELP EXPLAIN THIS TO ME !!!!

Mathematics

1 answer:

Jet001 [13]3 years ago

8 0

The quadratic formula is above.

You want each equation in standard form: ax^2 + bx + c = 0
I begin each problem by defining variables.
For instance 3. Is in standard form. X^2 -2x - 3 = 0. a = 1, b = -2, c = -3

Now use quadratic formula: x = [- b + or - sqrt(b^2 - 4ac)]\2a

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A line that includes the point (-3,-4) and (-4,f ) has a slope of 6. What is the value of f?

Eduardwww [97]

Answer:

f = -10

( -3, -4) ( - 4 , -10)

Step-by-step explanation:

Formula for slope

m =( y_2 - y_1)/(x _2 - x _1)

m = 6

6 = (y_2 - y_1) / (x _2 - x_1)

Substitute the points

( -3 , -4) ( -4 , f)

x_1 = -3

y_1 = -4

x_2 = -4

y_2 = f

Insert the values into

6 = (y_2 - y_1) / (x_2 - x_1)

6 = (f - (-4))/ ( -4 - (-3)

6 = ( f + 4)/ ( -4 + 3)

Cross multiply

6 ( -4 + 3) = f + 4

6 ( -1) = f + 4

-6 = f + 4

-6 - 4 = f

f = -6 - 4

f = -10

( -3 , -4) ( -3 , f)

The value of f is -10

( -3 , -4) ( -3 , -10)

8 0

4 years ago

A student has a 2% salt water solution and a 7% salt water solution. To best imitate salt water at a local beach, he needs 1 lit

FromTheMoon [43]

Let

x-------> the amount of $2\%$ solution

y--------> the amount of $7\%$ solution

we know that

$2\%=0.02\\7\%=0.07\\3.5\%=0.035$

$x+y=1$

$y=1-x$ -------> equation A

$0.02x+0.07y=0.035$ -------> equation B

substitute equation A in equation B

$0.02x+0.07(1-x)=0.035$

$0.02x+0.07-0.07x=0.035$

$0.05x=0.035$

$x=0.7\ liters$

find the value of y

$y=1-x$

$y=1-0.7=0.3\ liters$

therefore

The student need $0.7\ liters$ of $2\%$ solution and $0.3\ liters$ of $7\%$ solution

<u>the answer is</u>

A) The percent values were written incorrectly in the equation

B) The amount of 7% solution should be written as 1 – x, not x – 1.

8 0

3 years ago

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1. Write a polynomial function of minimum degree with real coefficients whose zeros include those listed. Write the polynomial i

Katena32 [7]

Polynomial with real coefficients always has even number of complex roots. We know that one of them is 2 + 5i so the second one will be 2 - 5i and:

$f(x)=\big(x-4\big)\big(x-(-8)\big)\big(x-(2+5i)\big)\big(x-(2-5i)\big)=\\\\=(x-4)(x+8)(x-2-5i)(x-2+5i)=\\\\=(x^2-4x+8x-32)(x^2-2x+5ix-2x+4-10i-5ix+10i-25i^2)\\\\=\big(x^2+4x-32\big)\big(x^2-4x+4-25\cdot(-1)\big)=\\\\=(x^2+4x-32)(x^2-4x+29)=\\\\=x^4-4x^3+29x^2+4x^3-16x^2+116x-32x^2+128x-928=\\\\=\boxed{x^4-19x^2+244x-928}$

Answer B.

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3 years ago

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Double-Angle and Half-Angle Identiies [See Attachment] Question 4

Nesterboy [21]

Please make it brainleist