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docker41 [41]
3 years ago
7

PLEASE HELP

Mathematics
2 answers:
Gala2k [10]3 years ago
8 0
First we have to determine a linear equation. We will take two points (0,900) and (1,850):
y=mx+b
900=m*0+b,   b=900
850=m*1+900
m=-50
Linear equation is:
y=-50x+900
Part A: y-interception is y=900 or(0,900).
It means that a car has to travel a distance of 900 miles to reach a final destination.
Part B:Average rate of change:\frac{y2-y1}{x2-x1} = \frac{750-850}{3-1}= \frac{-100}{2}=-50. It represents how many miles per one hour changes the distance from a destination.
Part C: Domain of this function if the car traveled the same rate until it reached its destination:
0=-50x+900
50x=900
x=900/50=18
Domain: x∈ [0, 18].
 
Lorico [155]3 years ago
8 0

Answer:

Step-by-step explanation:

First we have to determine a linear equation. We will take two points (0,900) and (1,850):

y=mx+b

900=m*0+b, b=900

850=m*1+900

m=-50

Linear equation is:

y=-50x+900

Part A: y-interception is y=900 or(0,900).

It means that a car has to travel a distance of 900 miles to reach a final destination.

Part B:Average rate of change:=-50. It represents how many miles per one hour changes the distance from a destination.

Part C: Domain of this function if the car traveled the same rate until it reached its destination:

0=-50x+900

50x=900

x=900/50=18

Domain: x∈ [0, 18

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A company manufactures aluminum mailboxes in the shape of a box with a half-cylinder top. The company will make 1960 mailboxes t
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Answer:

  7088 m²

Step-by-step explanation:

The area of the half-cylinder is half the area of a cylinder with the same dimensions. This half-cylinder has a radius of 0.3 m and a length of 0.8 m, so its area is ...

  A = πr(r +l) = π(0.3 m)(0.3 m+0.8 m) = 0.33π m² ≈ 1.0362 m²

The area of the bottom part of the box is the sum of its lateral area and its base area.

  lateral area = Ph = (2(0.6m +0.8m))(0.75 m) = 2.1 m²

  base area = lw = (0.8 m)(0.6 m) = 0.48 m²

The total area is then ...

  total area = half-cylinder area + lateral area + base area

  = 1.0362 m² +2.1 m² +0.48 m²

  = 3.6162 m²

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The area for 1960 mailboxes will be 1960 times this area, or ...

  area for week's production = 1960 × 3.6162 m² = 7087.752 m²

It will take about 7088 square meters of aluminum to make these mailboxes.

5 0
3 years ago
What is 100,000 written as a power with a base of 10? ANSWER QUICK
DochEvi [55]

Answer:

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Step-by-step explanation:

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What is 2/3 divided by 4/5?
m_a_m_a [10]

Answer:

5/6

Step-by-step explanation:

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If f(x)=3x+2and g(x)=x^2-9 find (f-g)(x)
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Step-by-step explanation:

If f(x)=3x+2and g(x)=x^2-9

find (f-g)(x)

f(x)- g(x) = [3x + 2] - [x^2 - 9]

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5 0
3 years ago
HELP ME HELP ME HELP ME HELP ME HELP ME HELP ME HELP ME HELP ME HELP ME HELP ME HELP ME HELP ME HELP ME HELP ME HELP ME HELP ME
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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3 years ago
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