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Phoenix [80]
3 years ago
7

The equation y = ax describes the graph of a line. If the value of a is positive, the line:

Mathematics
2 answers:
Gala2k [10]3 years ago
8 0

Goes up and to the right. - APEX

GuDViN [60]3 years ago
3 0
Has a positive slope.  (It goes from lower left to upper right).


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Need answer asap please!!​
ELEN [110]

Answer:

a

Step-by-step explanation:

-5(x + 2) = -4x - 12 - x + 2

= -5x - 10

= -5(x + 2)

both sides are equivalent so there are infinite solutions

4 0
3 years ago
the coordinates of one point of a line segment are (-2,-7). The line segment is 12 units long. Give three possible coordinates o
valentinak56 [21]

(-2,-19) , (-2, 5), ( 2,10)

4 0
3 years ago
In the expression (44 ÷ 4 + 10) - 20 , with which operation should we start?
Luden [163]
You have to do whats in the parentheses from left to right then subtract the answer you get by 20



(9x10) - (30+30)
90 - 60 = 30



8 x (12+5) -7^2
17
17-49
18 x -32

= -576.



You have to use pemdas for your check off
P=parentheses
E=exponet
M=multiplication
D=division
A=add
S=subtrabt

HOPE IM CORRECT
4 0
3 years ago
Read 2 more answers
6th grade help please
miv72 [106K]
12 + 6x + 6x + 6y = 12 + 12x + 6y
8 0
3 years ago
Read 2 more answers
Find the exact length of the curve. x=et+e−t, y=5−2t, 0≤t≤2 For a curve given by parametric equations x=f(t) and y=g(t), arc len
Rama09 [41]

The length of a curve <em>C</em> parameterized by a vector function <em>r</em><em>(t)</em> = <em>x(t)</em> i + <em>y(t)</em> j over an interval <em>a</em> ≤ <em>t</em> ≤ <em>b</em> is

\displaystyle\int_C\mathrm ds = \int_a^b \sqrt{\left(\frac{\mathrm dx}{\mathrm dt}\right)^2+\left(\frac{\mathrm dy}{\mathrm dt}\right)^2} \,\mathrm dt

In this case, we have

<em>x(t)</em> = exp(<em>t</em> ) + exp(-<em>t</em> )   ==>   d<em>x</em>/d<em>t</em> = exp(<em>t</em> ) - exp(-<em>t</em> )

<em>y(t)</em> = 5 - 2<em>t</em>   ==>   d<em>y</em>/d<em>t</em> = -2

and [<em>a</em>, <em>b</em>] = [0, 2]. The length of the curve is then

\displaystyle\int_0^2 \sqrt{\left(e^t-e^{-t}\right)^2+(-2)^2} \,\mathrm dt = \int_0^2 \sqrt{e^{2t}-2+e^{-2t}+4}\,\mathrm dt

=\displaystyle\int_0^2 \sqrt{e^{2t}+2+e^{-2t}} \,\mathrm dt

=\displaystyle\int_0^2\sqrt{\left(e^t+e^{-t}\right)^2} \,\mathrm dt

=\displaystyle\int_0^2\left(e^t+e^{-t}\right)\,\mathrm dt

=\left(e^t-e^{-t}\right)\bigg|_0^2 = \left(e^2-e^{-2}\right)-\left(e^0-e^{-0}\right) = \boxed{e^2-\frac1{e^2}}

5 0
3 years ago
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