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3 years ago

What is the sum of 3 3/4 + 1 1/2=?

1 answer:

5 0

You need to find the least common denominator first which is 4. 3 3/4 can stay the same but you must have like denominators to add so change 1 1/2 to 1 2/4 by multiplying by 2/2. You know have like denominators and you can add whole numbers first 3+1=4 and then the fractions 3/4+2/4= 5/4 which is also 1 1/4. Now you add 4 so your answer is 5 1/4

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Match the phenomenal expression on the left with the simplified version on the right

nalin [4]

Given the polynomial expression:

(y + 5)²

(y - 5)(y + 5)

Let's simplify each of the given expression:

a.) (y + 5)²

The given equation is a factor of a perfect square trinomial. For this type of expression, the following is the formula for expanding it.

$\text{ (a+b)}^2\text{ = }a^2\text{ + 2ab + }b^2$

We get,

$\mleft(y+5\mright)^2=(y)^2+2(y)(5)+(5)^2$ $(y+5)^2=y^2+10y+25$

b.) (y - 5)(y + 5)

To be able to simplify the following expression. We will be using the formula for the difference of two squares.

$(a+b)(a-b)=a^2-b^2$

We get,

$\mleft(y-5\mright)\mleft(y+5\mright)=(y)^2-(5)^2$ $(y-5)(y+5)=y^2-25^{}$

8 0

1 year ago

Help!!! EASY!!! HELP DUE SOON!!!

Alinara [238K]

4 + 3 + 2 = 9
18/9 = 2

3 x 2 = 6

Answer = 6

3 0

3 years ago

The Bowman family has $215,322 in assets and $182,009 in liabilities. What

drek231 [11]

397,331 is how much they are worth

4 0

3 years ago

I need to know what the answer to that question is and I need to know how to do it

mel-nik [20]

Answer:

$\frac{59}{8} or 7\frac{3}{8}$

Step-by-step explanation:

$\frac{80}{4} - 12\frac{5}{8}$

First you need to make the 12 5/8 into an improper fraction by multiplying 12 by 8 and then adding the total by 5.

$12\frac{5}{8}$ -> $\frac{101}{8}$

Now you have $\frac{80}{4} - \frac{101}{8}$ so you need a common denominator by multiplying the first fraction by 2

$\frac{80}{4} -> \frac{160}{8}$

Then you subtract 160 - 101 which equals 59.

$\frac{59}{8} or 7\frac{3}{8}$

5 0

3 years ago

Read 2 more answers

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 (<em>h</em>, <em>k</em>) 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