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asambeis [7]
2 years ago
10

Write an equation of the line in slope-intercept form through the given points: (0, 2) and (3, 3).

Mathematics
1 answer:
dybincka [34]2 years ago
4 0
The answer is y=1/3x+2
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If T (n) =6-1 what is the 1st term.
Mariulka [41]
T(n)=6-1=5
that means all terms are 5 because n does not appear on the right hand side.

However, if T(n)=6n-1, then
first term = T(1)=6(1)-1=5,
but second term T(2)=6(2)-1=11...
8 0
3 years ago
the time t required to drive a certain distance varies inversely with the speed, r. if it takes 4 hours to drive the distance at
valina [46]
Hello,

Answer A

d :the distance in miles
t: the time in hours
v: the speed in miles/hours (vistesse)


d=v*t
==>d=40*4=160(mi)

d=55*x=160==>x=160/55=2.909090... (mi/(mi/h)=h)≈2.91(hours)
8 0
3 years ago
When listing the domain (and range), we<br> list the repeated numbers
svetoff [14.1K]

Answer:hope help

Step-by-step explanation:

Find the range by listing all the y values from the ordered pairs. Repeated values within the domain or range don't have to be listed more than once. In order for a relation to be a function, each x must correspond with only one y value.

8 0
2 years ago
Read 2 more answers
Geometry Proof - Thanks in advance for the help!
RideAnS [48]

Answer:

Using reflexive property (for side), and the transversals of the parallel lines, we can prove the two triangles are congruent.

Step-by-step explanation:

  • Since AB and DC are parallel and AC is intersecting in the middle, you can make out two pairs of alternate interior angles<em>.</em> These angle pairs are congruent because of the alternate interior angles theorem. The two pairs of congruent angles are: ∠DAC ≅ ∠BCA, and ∠BAC ≅ ∠DCA.
  • With the reflexive property, we know side AC ≅ AC.
  • Using Angle-Side-Angle theorem, we can prove ΔABC ≅ ΔCDA.
8 0
3 years ago
Pleasee help ASAP!! No links pleaseee!!!
Ilya [14]

Given:

The vertices of the rectangle ABCD are A(0,1), B(2,4), C(6,0), D(4,-3).

To find:

The area of the rectangle.

Solution:

Distance formula:

D=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

Using the distance formula, we get

AB=\sqrt{(2-0)^2+(4-1)^2}

AB=\sqrt{(2)^2+(3)^2}

AB=\sqrt{4+9}

AB=\sqrt{13}

Similarly,

BC=\sqrt{(6-2)^2+(0-4)^2}

BC=\sqrt{(4)^2+(-4)^2}

BC=\sqrt{16+16}

BC=\sqrt{32}

BC=4\sqrt{2}

Now, the length of the rectangle is AB=\sqrt{13} and the width of the rectangle is BC=4\sqrt{2}. So, the area of the rectangle is:

A=length \times width

A=\sqrt{13}\times 4\sqrt{2}

A=4\sqrt{26}

A\approx 20

Therefore, the area of the rectangle is 20 square units.

3 0
3 years ago
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