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2 years ago

Answer before 11:59 plzzz

Mathematics

1 answer:

Alex17521 [72]2 years ago

5 0

0.0392 or 3.92%
The exponential growth formula can find the interest rate.

y - tolat value (P +I) (216)
a - principal (200)
r- rate (unkown)
x - time in years (2 yrs)
Substitute known values

divide both sides by 200

take the square root of both sides

subtract 1 on both sides
r = 0.0392 or 3.92%

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Ill give brainlist if you can answer this for me

NeTakaya

Answer:

The third one

Step-by-step explanation:

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$\frac{10}{\sqrt{2} } =5\sqrt{2}$

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3 years ago

I don’t know how to do this

stich3 [128]

To check for continuity at the edges of each piece, you need to consider the limit as $x$ approaches the edges. For example,

$g(x)=\begin{cases}2x+5&\text{for }x\le-3\\x^2-10&\text{for }x>-3\end{cases}$

has two pieces, $2x+5$ and $x^2-10$ , both of which are continuous by themselves on the provided intervals. In order for $g$ to be continuous everywhere, we need to have

$\displaystyle\lim_{x\to-3^-}g(x)=\lim_{x\to-3^+}g(x)=g(-3)$

By definition of $g$ , we have $g(-3)=2(-3)+5=-1$ , and the limits are

$\displaystyle\lim_{x\to-3^-}g(x)=\lim_{x\to-3}(2x+5)=-1$

$\displaystyle\lim_{x\to-3^+}g(x)=\lim_{x\to-3}(x^2-10)=-1$

The limits match, so $g$ is continuous.

For the others: Each of the individual pieces of $f,h$ are continuous functions on their domains, so you just need to check the value of each piece at the edge of each subinterval.

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3 years ago

What is the result of substituting for y in the bottom equation y=x+3 y=x^2+2x-4

8_murik_8 [283]

The solutions are (2.1925, 5.1925) and (-3.1925, -0.1925)

Solution:

Given that,

$y = x + 3 ------- eqn 1\\\\y = x^2 + 2x - 4 ----- eqn 2$

We have to substitute eqn 1 in eqn 2

$x + 3 = x^2 + 2x - 4$

$\mathrm{Switch\:sides}\\\\x^2+2x-4=x+3\\\\\mathrm{Subtract\:}3\mathrm{\:from\:both\:sides}\\\\x^2+2x-4-3=x+3-3\\\\\mathrm{Simplify}\\\\x^2+2x-7=x\\\\\mathrm{Subtract\:}x\mathrm{\:from\:both\:sides}\\\\x^2+2x-7-x=x-x\\\\\mathrm{Simplify}\\\\x^2+x-7=0$

$\mathrm{Solve\:with\:the\:quadratic\:formula}\\\\\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}\\\\x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\\\\mathrm{For\:}\quad a=1,\:b=1,\:c=-7\\\\x =\frac{-1\pm \sqrt{1^2-4\cdot \:1\left(-7\right)}}{2\cdot \:1}$

$x = \frac{-1 \pm \sqrt{ 1 + 28}}{2}\\\\x = \frac{ -1 \pm \sqrt{29}}{2}$

$x = \frac{ -1 \pm 5.385 }{2}\\\\We\ have\ two\ solutions\\\\x = \frac{ -1 + 5.385 }{2}\\\\x = 2.1925$

$Also\\\\x = \frac{ -1 - 5.385 }{2}\\\\x = -3.1925$

Substitute x = 2.1925 in eqn 1

y = 2.1925 + 3

y = 5.1925

Substitute x = -3.1925 in eqn 1

y = -3.1925 + 3

y = -0.1925

Thus the solutions are (2.1925, 5.1925) and (-3.1925, -0.1925)

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3 years ago

2v^2-12 =-12v Would really appreciate it loves❤️

Black_prince [1.1K]

For this case we have the following equation:

$2v ^ 2-12 = -12v$

Rewriting we have:

$2v ^ 2 + 12v-12 = 0$

Dividing by 2 to both sides of the equation:

$v ^ 2 + 6v-6 = 0$

We apply the quadratic formula:

$x = \frac {-b \pm \sqrt {b ^ 2-4 (a) (c)}} {2a}$

We have to:

$a = 1\\b = 6\\c = -6$

Substituting:

$x = \frac {-6 \pm \sqrt {6 ^ 2-4 (1) (- 6)}} {2 (1)}\\x = \frac {-6 \pm \sqrt {36 + 24}} {2}\\x = \frac {-6 \pm \sqrt {60}} {2}\\x = \frac {-6 \pm \sqrt {4 * 15}} {2}\\x = \frac {-6 \pm2 \sqrt {15}} {2}$

Thus, we have two roots:

$x_ {1} = - 3+ \sqrt {15}\\x_ {2} = - 3- \sqrt {15}$

ANswer:

$x_ {1} = - 3+ \sqrt {15}\\x_ {2} = - 3- \sqrt {15}$

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2 years ago

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Inga [223]

Answer:

Just put the numbers in order then the interquartile range would be the largest number from the high quartile and the lowest number from the lower quartile.

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3 years ago