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I am Lyosha [343]

3 years ago

5

An isosceles triangle has a perimeter of 33 cm. The unequal side has a length 3 cm more than the equal side. Find the length of

each side.

1 answer:

grigory [225]3 years ago

6 0

Answer:

let equal side=x

unequal side=x+3

perimeter =33

x+x+x+3=33

3x=30

x=10

x+3=10+3

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Mrac [35]

To justify the yearly membership, you want to pay at least the same amount as a no-membership purchase, otherwise you would be losing money by purchasing a yearly membership. So set the no-membership cost equal to the yearly membership cost and solve.

no-membership costs $2 per day for swimming and $5 per day for aerobic, in other words, $7 per day. So if we let d = number of days, our cost can be calculated by "7d"

a yearly membership costs $200 plus $3 per day, or in other words, "200 + 3d"

Set them equal to each other and solve:

7d = 200 + 3d

4d = 200

d = 50

So you would need to attend the classes for at least 50 days to justify a yearly membership. I hope that helps!

5 0

2 years ago

Help me plsss its pretty easy lol

DaniilM [7]

Answer:

D, or the last one.

Step-by-step explanation:

Brainliest?

4 0

2 years ago

Read 2 more answers

Solve y=f(x) for x. Then find the input when the output is -3.

DochEvi [55]

Answer:

Please check the explanation

Step-by-step explanation:

Given the function

$f\left(x\right)\:=\:\left(x-5\right)^3-1$

Given that the output = -3

i.e. y = -3

now substituting the value y=-3 and solve for x to determine the input 'x'

$\:\:y=\:\left(x-5\right)^3-1$

$-3\:=\:\left(x-5\right)^3-1\:\:\:$

switch sides

$\left(x-5\right)^3-1=-3$

Add 1 to both sides

$\left(x-5\right)^3-1+1=-3+1$

$\left(x-5\right)^3=-2$

$\mathrm{For\:}g^3\left(x\right)=f\left(a\right)\mathrm{\:the\:solutions\:are\:}g\left(x\right)=\sqrt[3]{f\left(a\right)},\:\sqrt[3]{f\left(a\right)}\frac{-1-\sqrt{3}i}{2},\:\sqrt[3]{f\left(a\right)}\frac{-1+\sqrt{3}i}{2}$

Thus, the input values are:

$x=-\sqrt[3]{2}+5,\:x=\frac{\sqrt[3]{2}\left(1+5\cdot \:2^{\frac{2}{3}}\right)}{2}-i\frac{\sqrt[3]{2}\sqrt{3}}{2},\:x=\frac{\sqrt[3]{2}\left(1+5\cdot \:2^{\frac{2}{3}}\right)}{2}+i\frac{\sqrt[3]{2}\sqrt{3}}{2}$

And the real input is:

$x=-\sqrt[3]{2}+5$

$x=3.74$

4 0

2 years ago

Help!!!!¡¡¡¡ i need help solving this it will increase my letter grade!!!!

Shtirlitz [24]

Answer:

The answer is below⬇️⬇️

Step-by-step explanation:

f(x) = 3x+4

g(x) = 2x

h(x) = x²+x-2

g(hx) = 2(x²+x-2)

= 2x²+2x-4

f(g(hx))=3(2x²+2x-4)+4

=6x²+6x-12+4

=6x²+6x-8

g(f(g(hx)))=2(6x²+6x-8)

=12x²+12x-16

f(g(f(g(hx))))=3(12x²+12x-16)+4

=36x²+36x-48+4

=36x²+36x-44

h(f(g(f(g(hx)))))=(36x²+36x-44)²+36x²+36x-44-2

=1296x⁴+2592x³-1872x²-3168x+1936+36x²+36x-46

=1296x⁴+2592x³-1836x²-3132x+1890

f(h(f(g(f(g(hx))))))=3(1296x⁴+2592x³-1836x²-3132x+1890)+4

=3888x⁴+7776x³-5508x²-9396x+5674

h(f(h(f(g(f(g(hx)))))))=(3888x⁴+7776x³-5508x²-9396x+5674)²+3888x⁴+7776x³-5508x²-9396x+5674-2

=15116544x⁸+60466176x⁷+17635968x⁶-158723712x⁵-71663616x⁴+657591048x³-255531048x²-106635204x+32194276

6 0

3 years ago

Sam entered three functions into a graphing calculator, y1 = x + 2, y2 = x2 + 2, and y3 = 2x. A portion of the table created by

V125BC [204]

For this case we have the following functions:
$y1 = x + 2

y2 = x ^ 2 + 2

y3 = 2 ^ x$

When x = 0 we have:
For y1:
$y1 = 0 + 2

y1 = 2$
For y2:
$y2 = 0 ^ 2 + 2

y2 = 0 + 2

y2 = 2$
Therefore, we have to:
$y1 = y2
$

When x = 5 we have:
For y2:
$y2 = 5 ^ 2 + 2

y2 = 25 + 2

y2 = 27$
For y3:
$y3 = 2 ^ 5

y3 = 32$
Therefore, we have to:
$y2 \ \textless \ y3
$

When x = -1 we have:
For y1:
$y1 = -1 + 2

y1 = 1$
For y2:
$y2 = (-1) ^ 2 + 2

y2 = 1 + 2

y2 = 3
$
For y3:
$y3 = 2 ^ {-1}

y3 = 1/2

y3 = 0.5$
Therefore, we have to:
$y3 \ \textless \ y1 \ \textless \ y2
$

Answer:
When x = 0, y1 = y2
When x = 5, y2 <y3
When x = -1, y3 <y1 <y2

4 0

3 years ago