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3 years ago

Please help me

Mathematics

1 answer:

yKpoI14uk [10]3 years ago

6 0

9514 1404 393

Answer:

A) 5x+12 = -12x-12

D) 5x+12 = -5x-12

Step-by-step explanation:

If you subtract the right side expression from both sides, you will get an equation with something equal to zero. If the 'something' has a variable in it, there is exactly one solution.

A: (5x+12) -(-12x-12) = 17x+24 = 0 . . . one solution

B: (5x+12) -(5x-5) = 17 = 0 . . . . no solutions

C: (5x+12)-(5x+12) = 0 = 0 . . . . infinite solutions

D: (5x+12) -(-5x-12) = 10x +24 = 0 . . . one solution

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If Justin can type 408 words in 4 minutes, how many words can he type per minute?

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What is the reference angle of -139 degrees

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41º

1) Since we want to find the reference angle for a -139º angle, we need to resort to the following formula

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2 years ago

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The sum of the squares of two positive integers is 394. If one integer is 2 less than the other and the larger integer is x. Fin

IgorLugansk [536]

Larger is x
the other is called y
sum of squares is 394
x^2+y^2=394
one is 2 less that other
x>y so
y+2=x
subsitute y+2 for x

(y+2)^2+y^2=394
expand
y^2+4y+4+y^2=394
add like terms
2y^2+4y+4=394
divideboth sides by 2
y^2+2y+2=197
subtract 197 from both sides
y^2+2y-195=0
factor
(y+15)(y-13)=0
set each to zero
y+15=0
y=-15
impossible since it is stated that they are positive so get rid of this solution
y-13=0
y=13

find other
y+2=x
13+2=x
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numbers are 13 and 15

7 0

3 years ago

Jan'ai was asked to determine the minimum for a function with zeros located at –1 and 5, which also has a y-intercept of (0, –25

sergey [27]

The first error in Jan'ai's work in determining the considered function is given by: Option D: She incorrectly determined the x-coordinate of the vertex.

<h3>What are the coordinates of vertex for a quadratic function?</h3>

For a quadratic function of the form $y = ax^2 + bx + c$ , its vertex form is obtained as:

$y = ax^2 + bx + c\\y =a(x^2 + bx/a) + c\\y = a(x^2 + 2(b/2a)x + (b/2a)^2 -(b/2a)^2 )+ c\\y = a(x^2 + 2(b/2a)x + (b/2a)^2) -a \times (b/2a)^2 + c\\\\y = a(x+b/2a)^2 - a \times (b/2a)^2 + c$

For the form $y = a(x-h)^2 + k$ , the vertex has coordinates (h, k)

Thus, for the obtained equation $y = a(x+b/2a)^2 - a \times (b/2a)^2 + c$ , we get the coordinates of vertex as:

$h = -b/2a$ , $k = c - a\times(b/2a)^2$

Thus, the coordinates of vertex of $y = ax^2 + bx + c$ is:

$(h,k) = (-b/2a, c - a \times (b/2a)^2 )$

The missing steps of work of Jan'ai are:

Begin to write a function in factored form. $f(x) = a(x+1)(x-5)$
Substitute x = 0, y = f(x) = -25 to determine a. $-25 = a(0+1)(0-5)$
Simplify and solve to find a. $a=5$
Rewrite the function. $f(x) = 5(x+1)(x-5)$
Rewrite in standard form. $f(x) = 5x^2-20x-25$
Find the x-coordinate of the vertex. $x = -20/2(5) = -20/10; x = -2$
Find the y-coordinate of the vertex.

$y = 5x^2-20x-25$

$y = 5(-2)^2-20(-2)-25$

$y = 35$

so (-2,35) is the coordinate of the vertex, which denotes the minimum.

So, as we see, in the 5th step, Jan'ai had the quadratic function $f(x) = 5x^2-20x-25$ ,

Comparing this to $f(x) = ax^2 + bx + c$ , we get a = 5, b = -20, c = -25

The vertex's x-coordinate will be on -b/2a = -(-20)/ 2(5) = 20/10 = 2

But Jan'ai didn't putted that negative sign before b. in the 6th step.

Thus, the first error in Jan'ai's work in determining the considered function is given by: Option D: She incorrectly determined the x-coordinate of the vertex.

Learn more about the vertex form of a quadratic equation here:

brainly.com/question/9912128

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3 years ago

Match the sets of points representing one-to-one functions with the sets of points representing their inverse functions. h = {(1

geniusboy [140]

The one-to-one functions given as sets of points and their possible inverse functions are given as
h = { (1,1), (2,2), (3,3), (4,4), (5,5), (6,6)}
g = { (1,3), (2,6), (3,9), (4,12), (5,15), (6,18)}
f = { (1,2), (2,3), (3,4), (4,5), (5,6), (6,7)}
i = { (1,1), (2,3), (3,5), (4,7), (5,9), (6,11)}

h⁻¹ = { (1,1), (2,2), (3,3), (4,4), (5,5), (6,6)}
i⁻¹ = { (1,1), (3,2), (5,3), (7,4), (9,5), (11,6)}
g⁻¹ = {(3,1), (6,2), (9,3), (12,4), (15,5), (18,6)}
f⁻¹ = {(2,1), (3,2), (4,3), (5,4), (6,5), (7,6)}

The inverse function of a given function should have the coordinates reversed.
Therefore the matches between the given functions and their inverse functions are given in the table below.

function Inverse function
----------- ------------------------
h h⁻¹
g g⁻¹
f f⁻¹
i i⁻¹

Answer:
The given functions and their corresponding inverses are correct.

7 0

4 years ago