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3 years ago

Which is the graph of f(x) = (x + 3)(x - 2

Mathematics

1 answer:

yanalaym [24]3 years ago

5 0

Answer:

Step-by-step explanation:

f(x) = (x + 3)(x - 2) has two zeros: One stems from (x + 3) = 0 and is (-3, 0); the other stems from (x - 2) = 0 and is (2, 0).

The axis of symmetry is a vertical line located halfway between -3 and 2:

x = -1/2.

The graph opens up because the given (x + 3)(x - 2) has a positive leading coefficient (+1).

With this information we can eliminate the last two possible answers. Note that the x-intercepts of the first graph are -3 and 2, Thus, the first graph is the correct one.

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Convert to smaller parts if possible

Triss [41]

Answer:

Row 1 -

1/3, 1/2, 1/3, 2/3, 2/3, 8/15, 1/2.

Row 2 -

3/4, 3/4, 2/7, 21/25, 5/6, 7/9, 1/3.

Row 3 -

3/20, 7/20, 3/25, 3/5, 3/5, 1, 3/2 OR 1 1/2.

Hope this helped you out.

7 0

3 years ago

BRAINLIESTTT ASAP!! PLEASE HELP ME :)

Lostsunrise [7]

Answer:

See below

Step-by-step explanation:

(a) Field lines

A negatively charged particle has an electric field associated with it.

The field lines spread out radially from the centre of the point. They are represented by arrows pointing in the direction that a positive charge would move if it were in the field.

Opposite charges attract, so the field lines point toward the centre of the particle.

For an isolated negative particle, the field lines would look like those in Figure 1 below.

If two negative charges are near each other, as in Figure 2, the field lines still point to the centre of charge.

A positive charge approaching from the left is attracted to both charges, but it moves to the closer particle on the left.

We can make a similar statement about appositive charge approaching from the left.

Thus, there are few field lines in the region between the two particles.

(b) Coulomb's Law

The formula for Coulomb's law is

F = (kq₁q₂)/r²

It shows that the force varies inversely as the square of the distance between the charges.

Thus, the force between the charges decreases rapidly as they move further apart.

5 0

3 years ago

Find the measure of CDE

Free_Kalibri [48]

Step-by-step explanation:

Here,

CDE+27+103+50+35=360(BEING THE SUM OF PIE CHART

CDE+215=360

CDE=360-215

CDE=145

(i amnt sure about answer)

5 0

3 years ago

Read 2 more answers

Can someone help me asap!<br><br>

lilavasa [31]

Answer:

B negative

Step-by-step explanation:

use real numbers?

n is +9

m is -11

m/n * (-m*n^2)

-11/9 * (11*81)

-11/1 * (11*9)

-11 * 99

-1089

6 0

2 years ago

The angle of elevation from me to the top of a hill is 51 degrees. The angle of elevation from me to the top of a tree is 57 deg

julia-pushkina [17]

Answer:

Approximately $101\; \rm ft$ (assuming that the height of the base of the hill is the same as that of the observer.)

Step-by-step explanation:

Refer to the diagram attached.

Let $\rm O$ denote the observer.
Let $\rm A$ denote the top of the tree.
Let $\rm R$ denote the base of the tree.
Let $\rm B$ denote the point where line $\rm AR$ (a vertical line) and the horizontal line going through $\rm O$ meets. $\angle \rm B\hat{A}R = 90^\circ$ .

Angles:

Angle of elevation of the base of the tree as it appears to the observer: $\angle \rm B\hat{O}R = 51^\circ$ .
Angle of elevation of the top of the tree as it appears to the observer: $\angle \rm B\hat{O}A = 57^\circ$ .

Let the length of segment $\rm BR$ (vertical distance between the base of the tree and the base of the hill) be $x\; \rm ft$ .

The question is asking for the length of segment $\rm AB$ . Notice that the length of this segment is $\mathrm{AB} = (x + 20)\; \rm ft$ .

The length of segment $\rm OB$ could be represented in two ways:

In right triangle $\rm \triangle OBR$ as the side adjacent to $\angle \rm B\hat{O}R = 51^\circ$ .
In right triangle $\rm \triangle OBA$ as the side adjacent to $\angle \rm B\hat{O}A = 57^\circ$ .

For example, in right triangle $\rm \triangle OBR$ , the length of the side opposite to $\angle \rm B\hat{O}R = 51^\circ$ is segment $\rm BR$ . The length of that segment is $x\; \rm ft$ .

$\begin{aligned}\tan{\left(\angle\mathrm{B\hat{O}R}\right)} = \frac{\,\rm {BR}\,}{\,\rm {OB}\,} \; \genfrac{}{}{0em}{}{\leftarrow \text{opposite}}{\leftarrow \text{adjacent}}\end{aligned}$ .

Rearrange to find an expression for the length of $\rm OB$ (in $\rm ft$ ) in terms of $x$ :

$\begin{aligned}\mathrm{OB} &= \frac{\mathrm{BR}}{\tan{\left(\angle\mathrm{B\hat{O}R}\right)}} \\ &= \frac{x}{\tan\left(51^\circ\right)}\approx 0.810\, x\end{aligned}$ .

Similarly, in right triangle $\rm \triangle OBA$ :

$\begin{aligned}\mathrm{OB} &= \frac{\mathrm{AB}}{\tan{\left(\angle\mathrm{B\hat{O}A}\right)}} \\ &= \frac{x + 20}{\tan\left(57^\circ\right)}\approx 0.649\, (x + 20)\end{aligned}$ .

Equate the right-hand side of these two equations:

$0.810\, x \approx 0.649\, (x + 20)$ .

Solve for $x$ :

$x \approx 81\; \rm ft$ .

Hence, the height of the top of this tree relative to the base of the hill would be $(x + 20)\; {\rm ft}\approx 101\; \rm ft$ .

6 0

3 years ago

Which is the graph of f(x) = (x + 3)(x - 2​

Which is the graph of f(x) = (x + 3)(x - 2