All categories

3 years ago

Solve for d.

Mathematics

1 answer:

Hoochie [10]3 years ago

8 0

Answer:

-8 = d

Step-by-step explanation:

-10 = -2+d

Add 2 to each side

-10 +2 = -2+2 +d

-8 = d

You might be interested in

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.

3 0

3 years ago

Find the Least Common Multiple (LCM) of 50 and 90

Effectus [21]

Answer:

450

(I can't post the answer if it's less than 20 letters so just ignore this I'm trying to fill up)

7 0

3 years ago

How would i write the converse of the statement : if two angles are bothe obtuse, the two angles are equal

Pepsi [2]

Answer:

The converse statement is:If two angles are not equal, then the two angles are not obtuse.

Step-by-step explanation:

we know that if in any statement A implies B then the converse of this statement is given by Not B implies Not A.

Here A=two angles are both obtuse

B=two angles are equal.

here it was given that : if two angles are both obtuse, the two angles are equal.

So it's converse statement is given by: If two angles are not equal, the two angles are not both obtuse.

4 0

3 years ago

Write a rule that you can use to solve addition equations without using models or drawing

marta [7]

The simple rules in adding equation is

- simplify like terms first
- then get one variable on one side of the equation
- then solve the rest

3 0

4 years ago

Read 2 more answers

If the circle has the same diameter as the edge length of the square, then the area of this circle is ___________the area of the

WINSTONCH [101]

Answer:

The area of this circle is $(\frac{\pi}{2} )$ the area of the square.

For the uniform electric field normal to the surface, the flux through the surface is electric field multiplied by the area of this surface.

Therefore, Φsquare is $(\frac{2}{\pi} )$ ϕcircle

Step-by-step explanation:

Area of the circle is given by;

$A_c = \frac{\pi d^2}{4}$

Area of the square is given by;

$A_s = L^2$

relationship between the edge length of the square, d, and length of its side, L,

$d = \sqrt{L^2 + L^2} \\\\d = \sqrt{2L^2}$

But area of the square , $A_s = L^2$

$d = \sqrt{2A_s}$

Then, the area of the square in terms of the edge length is given by;

$A_s = \frac{d^2}{2}$

Area of the circle in terms of area of the square is given by;

$A_c = \frac{\pi d^2}{4} = \frac{\pi}{2}(\frac{d^2}{2} )\\\\But \ A_s = \frac{d^2}{2} \\\\A_c = \frac{\pi}{2}(\frac{d^2}{2} )\\\\A_c = \frac{\pi}{2}(A_s )$

For the uniform electric field normal to the surface, the flux through the surface is electric field multiplied by the area of this surface.

Ф = E.A

Flux through the surface of the circle is given by;

$\phi _{circle} = E.(\frac{\pi d^2}{4})$

Flux through the surface of the square is given by;

$\phi _{square} = E.(\frac{d^2}{2} )\\\\\phi _{square} =E.(\frac{d^2}{2} ).(\frac{\pi}{2} ).(\frac{2}{\pi} )\\\\\phi _{square} =E.(\frac{\pi d^2}{4} ).(\frac{2}{\pi} )\\\\\phi _{square} =(\phi _{circle}).(\frac{2}{\pi} )$

Therefore, Φsquare is $(\frac{2}{\pi} )$ ϕcircle

5 0

4 years ago