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alexdok [17]
3 years ago
11

b) How many white-winged cross bills will there be at the end of June 2020? (Remember that t represent

Mathematics
1 answer:
Mandarinka [93]3 years ago
4 0

Answer:

A dozen = 12

Step-by-step explanation:

I think I’m right

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PLS HELP ME ASAP WITH 42!! (MUST SHOW WORK!!) + LOTS OF POINTS!! *best if show picture with work!!*
ddd [48]
I thought this would be simple, as I'm familiar with algebra and not really "The constant of proportionality," but I will do my best.

So this said "Constant of proportionality," is referring to basically the answers for the equation when X equals certain numbers.

Make a table of different answers when you plug in X and you get the 'Constant of proportionality.'

y = 2.5x + 3
y = 2.5(1) + 3
y = 2.5 + 3
y = 5.5

Since we plugged in 1 for X and got 5.5 for Y, our input and output is (1, 5.5)

Replace X for a different value, and you will get a bunch of different numbers that will in essence be your function inputs and outputs.  Make a table of these and you have your answer.

EXAMPLE - 
-= x =-   -= y =-
-= 1 =-   -= 5.5 =-
-= 2 =-   -= 8 =-
-= 3 =-   -= 11.5 =-
-= 4 =-   -= 13 =-


So there you have it.  I hope this helps!  If you have any further questions, don't hesitate to ask.
3 0
2 years ago
The input is 8 and the output is 26 what's the rule?
LuckyWell [14K]

Answer:

It could either be x * 3.25 or x + 18

Step-by-step explanation:

8 * 3.25 = 26

8 + 18 = 26

3 0
3 years ago
PLEASE HELP ME!!!!!!!!!!!
dusya [7]

Answer:

he was being cheated

Step-by-step explanation:

3 0
3 years ago
Find center and radius<br> (x-3)^2+y^2=8
Flauer [41]

Answer:

<h2>Center: (3, 0)</h2><h2>Radius: 2√2</h2>

Step-by-step explanation:

The equation of a circle in standard form:

(x-h)^2+(y-k)^2=r^2

(h, k) - center

r - radius

We have the equation:

(x-3)^2+y^2=8\to(x-3)^2+(y-0)^2=(\sqrt8)^2

Therefore

the center: (3,\ 0)

the radius: r=\sqrt8\to r=\sqrt{4\cdot2}\to r=\sqrt4\cdot\sqrt2\to r=2\sqrt2

4 0
3 years ago
Please help me with this question
san4es73 [151]

Step-by-step explanation:

Given: f'(x) = x^2e^{2x^3} and f(0) = 0

We can solve for f(x) by writing

\displaystyle f(x) = \int f'(x)dx=\int x^2e^{2x^3}dx

Let u = 2x^3

\:\:\:\:du=6x^2dx

Then

\displaystyle f(x) = \int x^2e^{2x^3}dx = \dfrac{1}{6}\int e^u du

\displaystyle \:\:\:\:\:\:\:=\frac{1}{6}e^{2x^3} + k

We know that f(0) = 0 so we can find the value for k:

f(0) = \frac{1}{6}(1) + k \Rightarrow k = -\frac{1}{6}

Therefore,

\displaystyle f(x) = \frac{1}{6} \left(e^{2x^3} - 1 \right)

5 0
2 years ago
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