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3 years ago

Question below. Please answer it, its math.

Mathematics

1 answer:

AleksAgata [21]3 years ago

4 0

Answer:

- 0.8

Step-by-step explanation:

The first thing we want to do here is simplify the expression -

$\frac{3}{5}$ ( 2x + 5 ) - 2x, Distribute the " $\frac{3}{5}$ " to elements within the parenthesis

= $\frac{3}{5}$ $*$ 2x + $\frac{3}{5}$ $*$ 5 - 2x, Focus on simplifying the expression " $\frac{3}{5}$ $*$ 2x + $\frac{3}{5}$ $*$ 5 "

= $2\cdot \frac{3}{5}x+5\cdot \frac{3}{5}$ - 2x

= $\frac{6x}{5}+3$ - 2x, Combine fractions

= $-\frac{4x}{5}$ + 3

= $-\frac{4}{5}$ x + 3

So we have our simplified expression " $-\frac{4}{5}$ x + 3, " with $-\frac{4}{5}$ being the coefficient of x. Our requirements are that this fraction should be expressed as a decimal, so we can simply divide the numerator by the denominator to figure that out,

- 4 / 5 = - 0.8,

Solution = - 0.8

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Hi can some please help?

Oduvanchick [21]

So I'm assuming that you're taking Calculus.
The first thing you want to do is take the integral of f(x)...
Use the power rule to get:
4X^2-13X+3.
Now solve for X when f(x)=0. This is because when the slope is 0, it is either a minimum or a maximum(I'm assuming you know this)
Now you get X=0.25 and X=3. Since we are working in the interval of (1,4), we can ignore 0.25
Thus our potential X values for max and min are X=1,X=4,X=3(You don't want to forget the ends of the bounds!)
Plugging these value in for f(x), we get
f(1)=2.833
f(3)=-8.5
f(4)=1.667
Thus X=1 is the max and X=3 is the min.
So max:(1,2.833)
min:(3,-8.5)

Hope this helps!

3 0

3 years ago

(09.02)

dmitriy555 [2]

X^2 - 7x + 12 = 0
x^2 - 4x - 3x + 12 = 0
x (x - 4) -3 ( x - 4) = 0
(x - 4) (x - 3) = 0
x = 3 , 4

Check the answer by plugging in 3 and 4 for x, if the equation equals zero then you have your answer.

4 0

2 years ago

IS THE ANSWER A??? please explain

Ne4ueva [31]

Step-by-step explanation:

D. must be the richt answer.

(x - Px)^2 + (y - Py)^2 = r^2

Where P(Px | Py) is the center of the circle.

If you insert one point of the circle, the equation must be true.

3 0

3 years ago

Which of the following graphs shows the solution set for the inequality below? 3|x + 1| < 9

Bas_tet [7]

Step-by-step explanation:

The absolute value function is a well known piecewise function (a function defined by multiple subfunctions) that is described mathematically as

$f(x) \ = \ |x| \ = \ \left\{\left\begin{array}{ccc}x, \ \text{if} \ x \ \geq \ 0 \\ \\ -x, \ \text{if} \ x \ < \ 0\end{array}\right\}$ .

This definition of the absolute function can be explained geometrically to be similar to the straight line $\textbf{\textit{y}} \ = \ \textbf{\textit{x}}$ , however, when the value of x is negative, the range of the function remains positive. In other words, the segment of the line $\textbf{\textit{y}} \ = \ \textbf{\textit{x}}$ where $\textbf{\textit{x}} \ < \ 0$ (shown as the orange dotted line), the segment of the line is reflected across the x-axis.

First, we simplify the expression.

$3\left|x \ + \ 1 \right| \ < \ 9 \\ \\ \\\-\hspace{0.2cm} \left|x \ + \ 1 \right| \ < \ 3$ .

We, now, can simply visualise the straight line, $y \ = \ x \ + \ 1$ , as a line having its y-intercept at the point $(0, \ 1)$ and its x-intercept at the point $(-1, \ 0)$ . Then, imagine that the segment of the line where $x \ < \ 0$ to be reflected along the x-axis, and you get the graph of the absolute function $y \ = \ \left|x \ + \ 1 \right|$ .

Consider the inequality

$\left|x \ + \ 1 \right| \ < \ 3$ ,

this statement can actually be conceptualise as the question

$``\text{For what \textbf{values of \textit{x}} will the absolute function \textbf{be less than 3}}"$ .

Algebraically, we can solve this inequality by breaking the function into two different subfunctions (according to the definition above).

Case 1 (when $x \ \geq \ 0$ )

$x \ + \ 1 \ < \ 3 \\ \\ \\ \-\hspace{0.9cm} x \ < \ 3 \ - \ 1 \\ \\ \\ \-\hspace{0.9cm} x \ < \ 2$

Case 2 (when $x \ < \ 0$ )

$-(x \ + \ 1) \ < \ 3 \\ \\ \\ \-\hspace{0.15cm} -x \ - \ 1 \ < \ 3 \\ \\ \\ \-\hspace{1cm} -x \ < \ 3 \ + \ 1 \\ \\ \\ \-\hspace{1cm} -x \ < \ 4 \\ \\ \\ \-\hspace{1.5cm} x \ > \ -4$

*remember to flip the inequality sign when multiplying or dividing by

negative numbers on both sides of the statement.

Therefore, the values of x that satisfy this inequality lie within the interval

$-4 \ < \ x \ < \ 2$ .

Similarly, on the real number line, the interval is shown below.

The use of open circles (as in the graph) indicates that the interval highlighted on the number line does not include its boundary value (-4 and 2) since the inequality is expressed as "less than", but not "less than or equal to". Contrastingly, close circles (circles that are coloured) show the inclusivity of the boundary values of the inequality.

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2 years ago

What is the qoutient of -48÷6? A.42 B.8 C.-8 D.-42

maks197457 [2]

Answer:

C. -8

This is the correct answer. Hope this helps :)

4 0

2 years ago

Read 2 more answers