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3 years ago

8y 2 + 17y + 2 What’s the answer for this question

1 answer:

3 0

So you first need to calculate 8y 2 which is basically 8x2 which equals 16 then add 16y + 17y which is 33 so your answer is 33y+2

You might be interested in

How to do a k over 8 equals 7

11Alexandr11 [23.1K]

Let's deconstruct the equation.

K x (1/8) = 7

You want to isolate K, so you multiply by 8 for both sides (since 8 x 1/8 = 1)
So, K = 7 x 8

K=56

6 0

3 years ago

Which sequence has a common ratio of 2? a{20, 40, 80, 160, 320, 640, …} b{20, 10, 5, 2.5, 1.25, 0.625, …} c{20, 15, 10, 5, 0, -5

Elis [28]

Answer:

Step-by-step explanation:

40/20=2

80/40=2

Therefore the common ration is 2

4 0

3 years ago

Help me ASAP I’ll mark you as a BL Question attached below

mojhsa [17]

Answer:

I think it might be A if not then its be

Step-by-step explanation:

sorry if its not right

8 0

3 years ago

In a class of 500 students a hundred and sixty have 320 have brown eyes and 20 green in a class of 25 students how many would ha

saveliy_v [14]

Is your question right because you said 500 students a hundred and sixty have 320 have brown eyes?

4 0

3 years ago

Show that the line integral is independent of path by finding a function f such that ?f = f. c 2xe?ydx (2y ? x2e?ydy, c is any p

Juli2301 [7.4K]

I'm reading this as

$\displaystyle\int_C2xe^{-y}\,\mathrm dx+(2y-x^2e^{-y})\,\mathrm dy$

with $\nabla f=(2xe^{-y},2y-x^2e^{-y})$ .

The value of the integral will be independent of the path if we can find a function $f(x,y)$ that satisfies the gradient equation above.

You have

$\begin{cases}\dfrac{\partial f}{\partial x}=2xe^{-y}\\\\\dfrac{\partial f}{\partial y}=2y-x^2e^{-y}\end{cases}$

Integrate $\dfrac{\partial f}{\partial x}$ with respect to $x$ . You get

$\displaystyle\int\dfrac{\partial f}{\partial x}\,\mathrm dx=\int2xe^{-y}\,\mathrm dx$
$f=x^2e^{-y}+g(y)$

Differentiate with respect to $y$ . You get

$\dfrac{\partial f}{\partial y}=\dfrac{\partial}{\partial y}[x^2e^{-y}+g(y)]$
$2y-x^2e^{-y}=-x^2e^{-y}+g'(y)$
$2y=g'(y)$

Integrate both sides with respect to $y$ to arrive at

$\displaystyle\int2y\,\mathrm dy=\int g'(y)\,\mathrm dy$
$y^2=g(y)+C$
$g(y)=y^2+C$

So you have

$f(x,y)=x^2e^{-y}+y^2+C$

The gradient is continuous for all $x,y$ , so the fundamental theorem of calculus applies, and so the value of the integral, regardless of the path taken, is

$\displaystyle\int_C2xe^{-y}\,\mathrm dx+(2y-x^2e^{-y})\,\mathrm dy=f(4,1)-f(1,0)=\frac9e$

8 0

3 years ago