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Radda [10]
3 years ago
10

12x+7<-11 or 5x-8>40

Mathematics
1 answer:
DaniilM [7]3 years ago
6 0

Answer:

A

Step-by-step explanation:

Given

12x + 7 < - 11 or 5x - 8 > 40

Solve each inequality

12x + 7 < - 11 ( subtract 7 from both sides )

12x < - 18 ( divide both sides by 12 )

x < - \frac{3}{2}

OR

5x - 8 > 40 ( add 8 to both sides )

5x > 48 ( divide both sides by 5 )

x > \frac{48}{5}

Solution is

x < - \frac{3}{2} or x > \frac{48}{5} → A

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Stella [2.4K]
You would is this case use the perfect square rule. So your would squareroot 9r^2 and 25 to get 3r+5 then you would square this equation to get (3r+5)^2
7 0
2 years ago
Samuel is 4 1/6 feet tall. His dad is 5 3/4 feet tall.How much taller is Samuels dad then Samuel
Sloan [31]

Answer:

1 7/12

Step-by-step explanation:

Just subtract the fractions if you cant do it by head or paper use a calc so 5 3/4 - 4 1/6= 1 7/12 samuels dad is 1 7/12 taller than samuel.

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2 years ago
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Kryger [21]

Answer:

1)

Given the triangle RST with Coordinates  R(2,1), S(2, -2) and T(-1 , -2).

A dilation is a transformation which produces an image that is the same shape as original one, but is different size.  

Since, the scale factor \frac{5}{3} is greater than 1, the image is enlargement or a stretch.  

Now, draw the dilation image of the triangle RST with center (2,-2) and scale factor \frac{5}{3}

Since, the center of dilation at S(2,-2) is not at the origin, so the point S and its image S{}' are same.

Now, the distances from the center of the dilation at point S to the other points R and T.  

The dilation image will be\frac{5}{3} of each of these distances,

SR=3, so S{}'R{}'=5 ;


ST=3, so S{}'T{}'=5  

Now, draw the image of RST i.e R'S'T'

Since, RT=3\sqrt{2} [By using hypotenuse of right angle triangle] and R{}'T{}'=5\sqrt{2}.


2)

(a)

Disagree with the given statement.

Side Angle Side postulate (SAS) states that:

If two sides and the included angle of one triangle are congruent to two sides and the included angle of another triangle then these two triangles are congruent.

Given: B is the midpoint of \overline{AC} i.e \overline{AB}\cong \overline{BC}

In the triangle ABD and triangle CBD, we have

\overline{AB}\cong \overline{BC}   (SIDE)            [Given]

\overline{BD}\cong \overline{BD}   (SIDE)            [Reflexive post]

Since, there is no included angle in these triangles.

∴ \Delta ABD is not congruent to \Delta CBD .

Therefore, these triangles does not follow the SAS congruence postulates.

(b)

SSS(SIDE-SIDE-SIDE) states that if three sides of one triangle are congruent to three sides of another triangle, then the triangles are congruent.

Since it is also given that  \overline{AD}\cong \overline{CD}.

therefore, in the triangle ABD and triangle CBD, we have

\overline{AB}\cong \overline{BC}   (SIDE)            [Given]

\overline{AD}\cong \overline{CD}   (SIDE)           [Given]

\overline{BD}\cong \overline{BD}   (SIDE)            [Reflexive post]

therefore by, SSS postulates \Delta ABD\cong \Delta CBD.

3)

Given that:  \angle1=\angle 3 are vertical angles, as they are formed by intersecting lines.

Therefore

, by the definition of linear pairs

\angle 1 and \angle 2 and \angle 3  and \angle 2 are linear pair.

By linear pair theorem, \angle 1 and \angle 2   are supplementary, \angle 2 and \angle 3  are supplementary.

m\angle1+m\angle 2=180^{\circ}

m\angle2+m\angle 3=180^{\circ}

Equate the above expressions:

m\angle 1+m\angle 2=m\angle 2+m\angle 3

Subtract the angle 2 from both sides in the above expressions

∴m\angle 1=m\angle 3

By Congruent Supplement theorem: If two angles are supplements of the same angle, then the two angles are congruent.


therefore, \angle 1\cong \angle 3.















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2 years ago
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Citrus2011 [14]

Answer:

how many feet of fencing does he have?

\

Step-by-step explanation:

8 0
3 years ago
What is the length of AB?
Xelga [282]

Answer:

<em>AB = 7.35 cm</em>

Step-by-step explanation:

From the attachment,

In ΔDEF,

DF = GH-(GD+FH) = 6 - (2+3) = 1 cm

DE = 2+3 = 5 cm (sum of two radius)

Applying Pythagoras theorem,

EF=\sqrt{DE^2-DF^2}=\sqrt{5^2-1^2}=\sqrt{24}=2\sqrt6

In ΔCDI,

DI = GH-(GD+IH) = 6 - (2+1.5) = 2.5 cm

CD = 2+1.5 = 3.5 cm (sum of two radius)

Applying Pythagoras theorem,

CI=\sqrt{CD^2-DI^2}=\sqrt{3.5^2-2.5^2}=\sqrt6

AB = EF + CI = 2\sqrt6+\sqrt6=3\sqrt6=7.35\ cm

6 0
3 years ago
Read 2 more answers
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