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3 years ago

Simplify 3/4(1/2x-12)+4/5

Mathematics

1 answer:

Nitella [24]3 years ago

8 0

Answer:

3x/8 - 41/5

Step-by-step explanation:

Simplify each term.

3x/8 − 9+4/5

To write −9 as a fraction with a common denominator, multiply by 5/5

3x/8−9 x 5/5 + 4/5

Combine −9 and 5/5

Combine the numerators over the common denominator.

3x/8 + −41/5

Move the negative in front of the fraction

3x/8 - 41/5

Hope this helps you

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If 3/5 of the girls in class have long brown hair and 1/3 of them have green eyes how many girls in class have long brown hair a

goblinko [34]

I'm guessing 1/3 of 3/5 is what you mean

1/3 of 3/5=1/3 times 3/5=3/15=1/5

1/5 of class has long brown hair and green eyes

3 0

3 years ago

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Write an equation in slope-intercept form of the line with the given characteristics:

kobusy [5.1K]

Answer:

1. y = -3x + 5

2. y = 1/4x - 2

Step-by-step explanation:

y=mx+b

1. 5 = -3(0) + b, so b = 5

y = -3x + 5

2. (-2-(-1))/(0-4) = -1/-4 = 1/4

-2 = 1/4(0) + b, so b = -2

y = 1/4x - 2

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3 years ago

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Find the equation of the quadratic function f whose graph is shown below.

Marianna [84]

Step-by-step explanation:

A quadratic function is a second-degree polynomial function with the general form

$f(x) \ = \ ax^{2} \ + \ bx \ + \ c$ ,

where $a$ , $b$ , and $c$ are real numbers, and $a \ \neq \ 0$ .

The standard form or the vertex form of a quadratic function is, however, a little different from the general form. To get the standard form from the general form, we need to use the "complete the square" method.

$f(x) \ = \ ax^{2} \ + \ bx \ + \ c \\ \\ \\ f(x) \ = \ a\left(x^{2} \ + \ \displaystyle\frac{b}{a}x \right) \ + \ c \\ \\ \\ f(x) \ = \ a\left[x^{2} \ + \ \displaystyle\frac{b}{a}x \ + \ \left(\displaystyle\frac{b}{2a}\right)^{2} \ - \ \left(\displaystyle\frac{b}{2a}\right)^{2} \right] \ + \ c \\ \\ \\ f(x) \ = \ a\left[x^{2} \ + \ \displaystyle\frac{b}{a}x \ + \ \left(\displaystyle\frac{b}{2a}\right)^{2}\right] \ - \ a\left(\displaystyle\frac{b}{2a}\right)^{2} \ + \ c$

$f(x) \ = \ a\left(x \ + \ \displaystyle\frac{b}{2a}\right)^{2} \ + \ c \ - \ a\left(\displaystyle\frac{b^{2}}{4a^{2}}\right) \\ \\ \\ f(x) \ = \ a\left(x \ + \ \displaystyle\frac{b}{2a}\right)^{2} \ + \ c \ - \ \displaystyle\frac{b^{2}}{4a}$

Let

$h \ = \ -\displaystyle\frac{b}{2a}$ and $k \ = \ c \ - \ \displaystyle\frac{b^{2}}{4a}$ ,

then the expression reduces into

$f(x) \ = \ a \left(x \ - \ h\right)^{2} \ + \ k$ ,

where the point (h, k) are the coordinates for the vertex of the quadratic function.

There are two different methods to approach this question. First, we consider the general form of the quadratic function, it is observed that has a y-intercept at the point $\left(0, \ 2\right)$ , so

$f(0) \ = \ -2 \\ \\ \\ f(0) \ = \ a(0)^{2} \ + \ b(0) + c \\ \\ \\ c = \ -2$ .

Additionally, it is pointed that two distinct points $(-1, \ -3)$ and $(-4, \ 6)$ lies on the quadratic graph, hence

$f(-1) \ = \ -3 \\ \\ \\ f(-1) \ = \ a(-1)^{2} \ + \ b(-1) \ -2 \\ \\ \\ \-\hspace{0.36cm} -3 \ = \ a \ - \ b \ -2 \\ \\ \\ \-\hspace{0.3} a \ - \ b \ = \ -1 \ \ \ \ \ \ $-----$ \ (1)$

and

$\-\hspace{0.18cm}f(-4) \ = \ 6 \\ \\ \\ \-\hspace{0.18cm} f(-4) \ = \ a(-4)^{2} \ + \ b(-4) \ -2 \\ \\ \\ \-\hspace{0.97cm} 6 \ = \ 16a \ - \ 4b \ -2 \\ \\ \\ \-\hspace{0.98cm} 8 \ = \ 16a \ - \ 4b \\ \\ \\ 4a \ - \ b \ = \ 2 \ \ \ \ \ \ $-----$ \ (2)$ .

Subtract equation (1) from equation (2) term-by-term,

$\-\hspace{0.72cm} (4a \ - \ b) \ - \ (a \ - \ b) \ = \ 2 \ - \ (-1) \\ \\ \\ (4a \ - \ a) \ + \ \left[-b \ - \ (-b)\right] \ = \ 2 \ + \ 1 \\ \\ \\ \-\hspace{3.8cm} 3a \ = \ 3 \\ \\ \\ \-\hspace{4cm} a \ = \ 1$

Substitute $a \ = \ 1$ into equation (1),

$1 \ - \ b \ = \ -1 \\ \\ \\ \-\hspace{0.86cm} b \ = \ 2$ .

Therefore, the equation of the quadratic function is

$f(x) \ = \ x^2 \ + \ 2x \ -2$ .

$\rule{12.5cm}{0.02cm}$

Alternatively, the vertex of the quadratic function is given as the point $(-1, \ -3)$ , substitute these coordinates into the vertex form of a quadratic function.

$f(x) = a\left(x \ + \ 1\right)^{2} \ - \ 3$ .

Substitute the point $(-4, \ 6)$ into the function above,

$f(-4) \ = \ 6 \\ \\ \\ f(-4) \ = \ a\left[(-4) \ + \ 1\right]^{2} \ - \ 3 \\ \\ \\ \-\hspace{0.75cm} 6 \ = \ a(-3)^{2} \ - \ 3 \\ \\ \\ \-\hspace{0.55cm} 9a \ = \ 9 \\ \\ \\ \-\hspace{0.75cm} a \ = \ 1$ .

Therefore, the general form of the quadratic function is

$f(x) \ = \ (x \ + \ 1)^{2} \ - \ 3 \\ \\ \\ f(x) \ = \ (x^2 \ + \ 2x \ + \ 1) \ - \ 3 \\ \\ \\ f(x) \ = \ x^2 \ + \ 2x \ - \ 2$ .

6 0

2 years ago

Combine like terms. (4x^2-5x+6)+(9x^2-2x)-(11x-3)

antiseptic1488 [7]

Answer:

13x^2 - 18x + 9

Step-by-step explanation:

Step 1: Distribute the plus and minus signs

(4x^2 - 5x + 6) + (9x^2 - 2x) - (11x - 3)

4x^2 - 5x + 6 + 9x^2 - 2x - 11x + 3

Step 2: Combine like terms

4x^2 - 5x + 6 + 9x^2 - 2x - 11x + 3

13x^2 - 18x + 9

Answer: 13x^2 - 18x + 9

3 0

3 years ago

Solve each equation by graphing. Round to the nearest tenth. -2x^2+2=-3x

AlekseyPX

Answer:

x = -0.5 or x = 2

Step-by-step explanation:

Finding solutions graphically is often easier if the equation can be put in the form f(x) = 0. Here, we can do that by subtracting the right-side expression to give ...

(-2x^2 +2) -(-3x) = 0

This could be put in standard form, but there is no need. A graphing calculator can deal with this directly.

The solutions are x = -0.5 and x = 2.

5 0

3 years ago