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vovangra [49]
2 years ago
11

Consider this power.(–4) –3

Mathematics
2 answers:
DENIUS [597]2 years ago
5 0
Answer:
(-4)-3= -7
Explanation:
aksik [14]2 years ago
4 0

Answer:

12? I'm not sure. That's the answer I think?

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Write the equation of the line that passes through the point (-5,5) and has a slope of -8/5
Free_Kalibri [48]

Answer:

The equation of the line that passes through the point (-5,5) and has a slope of -8/5 will be:

                                           y=-\frac{8}{5}x-3

Step-by-step explanation:

Given

  • \left(x_1,\:y_1\right)=\left(-5,5\right)
  • m=\frac{-8}{5}

Point-Slope Form equation is

\left(y-y_1\right)=m\left(x-x_1\right)

\left(y-5\right)=\frac{-8}{5}\left(x-\left(-5\right)\right)

\mathrm{Apply\:rule}\:-\left(-a\right)=a

y-5=\frac{-8}{5}\left(x+5\right)

\mathrm{Apply\:the\:fraction\:rule}:\quad \frac{-a}{b}=-\frac{a}{b}

y-5=-\frac{8}{5}\left(x+5\right)

y-5+5=-\frac{8}{5}\left(x+5\right)+5

y=-\frac{8}{5}x-3

Therefore, the equation of the line that passes through the point (-5,5) and has a slope of -8/5 will be:

                                           y=-\frac{8}{5}x-3

5 0
3 years ago
Does 2+2 REALLY equal 5? There has been proof. But is it really correct?
kari74 [83]
I actually posted this a while ago and it got answerd by a smart person so here's what I understood


the error is that it assumes that √(x²)=x
because if x is negative, it becomes positive
that's how the 2nd line is wrong
the 2nd line is the first wrong step

it goes from 4-9/2+9/2
so it goes from 4-4.5+4.5
notice the part under the parenthsees
√((4-4.5)²)
if we evaluate 4-4.5, we get -0.5
but by squaring it we get 0.5

-0.5+4.5=4, yes
but when we square it we get
0.5+4.5=5, that is how it works
5 0
2 years ago
Find an equation of the tangent to the curve x =5+lnt, y=t2+5 at the point (5,6) by both eliminating the parameter and without e
svet-max [94.6K]

ANSWER

y = 2x -4

EXPLANATION

Part a)

Eliminating the parameter:

The parametric equation is

x = 5 +  ln(t)

y =  {t}^{2}  + 5

From the first equation we make t the subject to get;

x - 5 =  ln(t)

t =  {e}^{x - 5}

We put it into the second equation.

y =  { ({e}^{x - 5}) }^{2}  + 5

y =  { ({e}^{2(x - 5)}) }  + 5

We differentiate to get;

\frac{dy}{dx}  = 2 {e}^{2(x - 5)}

At x=5,

\frac{dy}{dx}  = 2 {e}^{2(5 - 5)}

\frac{dy}{dx}  = 2 {e}^{0}  = 2

The slope of the tangent is 2.

The equation of the tangent through

(5,6) is given by

y-y_1=m(x-x_1)

y - 6 = 2(x - 5)

y = 2x - 10 + 6

y = 2x -4

Without eliminating the parameter,

\frac{dy}{dx}  =  \frac{ \frac{dy}{dt} }{ \frac{dx}{dt} }

\frac{dy}{dx}  =  \frac{ 2t}{  \frac{1}{t} }

\frac{dy}{dx}  =  2 {t}^{2}

At x=5,

5 = 5 +  ln(t)

ln(t)  = 0

t =  {e}^{0}  = 1

This implies that,

\frac{dy}{dx}  =  2 {(1)}^{2}  = 2

The slope of the tangent is 2.

The equation of the tangent through

(5,6) is given by

y-y_1=m(x-x_1)

y - 6 = 2(x - 5) =

y = 2x -4

5 0
3 years ago
Given that the measure of ∠x is 149°, and the measure of ∠y is 110°, find the measure of ∠z.
Rom4ik [11]

Answer:

∠z is 79°

Step-by-step explanation:

8 0
2 years ago
Solve 2(x + 1) = 2x + 2.
Tom [10]

2(x + 1) = 2x + 2\\\\
2x+2=2x+2\\\\
x\in\mathbb{R}

3 0
2 years ago
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