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2 years ago

12

Graph the function. y=-x-6 ifx<-6 y = 2x + 12 if x>-6

1 answer:

Naddik [55]2 years ago

7 0

Answer:

see attached

Step-by-step explanation:

For x < -6, the function has a slope of -1 and an x-intercept of -6.

For x > -6, the function has a slope of 2 and an x-intercept of -6.

The function given here is not defined at x=6, so there is a hole at (-6, 0).

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The length of a rectangle is 7 inches more than the width. The perimeter is 30 inches. Find the length and the width of the rect

saul85 [17]

Answer:

that's all about my answer

8 0

2 years ago

What is the answer to 4|x-1|=12

STatiana [176]

Answer:

(4,0) and (- 2,0): Answer

Step-by-step explanation:

Absolute value questions have two essential steps.

1. Solve the equation as it is written.

2. Solve it changing the sign of the right hand side. I will include a graph to confirm my answer

4*abs(x - 1) = 12 Divide by 4

abs(x - 1) = 12/4
abs(x - 1) = 3 Equate this to 3
x - 1 = 3 Add 1 to both sides
x - 1 + 1 = 3 + 1 Combine
x = 4

4*abs(x - 1) = - 12 Divide by 4

abs(x - 1) = - 12/4
abs(x - 1) = - 3
x - 1 = - 3 Add 1 to both sides
x - 1 + 1 = -3 + 1
x = - 2

So this has 2 answers

(4,0) and (- 2,0)

5 0

2 years ago

Find the components of the vertical force Bold Upper FFequals=left angle 0 comma negative 10 right angle0,−10 in the directions

quester [9]

Solution :

Let $v_0$ be the unit vector in the direction parallel to the plane and let $F_1$ be the component of F in the direction of $v_0$ and $F_2$ be the component normal to $v_0$ .

Since, $|v_0| = 1,$

$(v_0)_x=\cos 60^\circ= \frac{1}{2}$

$(v_0)_y=\sin 60^\circ= \frac{\sqrt 3}{2}$

Therefore, $v_0 = \left$

From figure,

$|F_1|= |F| \cos 30^\circ = 10 \times \frac{\sqrt 3}{2} = 5 \sqrt3$

We know that the direction of $F_1$ is opposite of the direction of $v_0$ , so we have

$F_1 = -5\sqrt3 v_0$

$=-5\sqrt3 \left$

$= \left$

The unit vector in the direction normal to the plane, $v_1$ has components :

$(v_1)_x= \cos 30^\circ = \frac{\sqrt3}{2}$

$(v_1)_y= -\sin 30^\circ =- \frac{1}{2}$

Therefore, $v_1=\left< \frac{\sqrt3}{2}, -\frac{1}{2} \right>$

From figure,

$|F_2 | = |F| \sin 30^\circ = 10 \times \frac{1}{2} = 5$

∴ $F_2 = 5v_1 = 5 \left< \frac{\sqrt3}{2}, - \frac{1}{2} \right>$

$=\left$

Therefore,

$F_1+F_2 = \left< -\frac{5\sqrt3}{2}, -\frac{15}{2} \right> + \left< \frac{5 \sqrt3}{2}, -\frac{5}{2} \right>$

$= = F$

3 0

2 years ago

Somebody please help with this problem

Mrrafil [7]

Step-by-step explanation:

We have been given that AE=BE and $\angle1\cong \angle2$ .

We can see that angle CEA is vertical angle of angle DEB, therefore, $m\angle CEA=m\angle DEB$ as vertical angles are congruent.

We can see in triangles CEA and DEF that two angles and included sides are congruent.

$\angle 1\cong \angle 2$

$AE=BE$

$\angle CEA\cong\angle DEB$ or $\angle 3\cong \angle 4$

Therefore, $\Delta CEA\cong \Delta DEB$ by ASA postulate.

Since corresponding parts of congruent triangles are congruent, therefore CE must be congruent to DE.

5 0

2 years ago

shut up bought a pen on sale. she saved 17% of the original price of $29. about how much money did she save? a $1 b $5 c $10 or

vredina [299]

Answer:

$4.93 cents so round it to $5.00

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers