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4 years ago

If I'm wrong plz correct me so I can review

Mathematics

1 answer:

creativ13 [48]4 years ago

4 0

check the picture below.

so the area of the rectangle is x*y.

the area of the triangle is (1/2)(w)(z).

if we subtract the area of the triangle, from the containing rectangle, we're in effect making a whole in the rectangle, as you see there in lower-right corner.

$\bf \stackrel{rectangle}{x\cdot y}~~-~~\stackrel{triangle}{\cfrac{1}{2}wz}$

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Over what interval is the quadratic function decreasing?

frosja888 [35]

The interval over which the given quadratic equation decreases is: x ∈ (5, ∞).

<h3>How to find the interval of quadratic functions?</h3>

Usually a quadratic graph function decreases either when moving from left to right or moving downwards.

In the given graph, we can see that the coordinate of the vertex is (5, 4) after which the curve goes in the downward direction.

Thus, for the values of x greater than 5, the function decreases and so we conclude that the interval in which the quadratic equation decreases is: (5, ∞).

Read more about Quadratic functions at: brainly.com/question/18030755

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6 0

2 years ago

Read 2 more answers

Let c be the path c(t) = (2t, t2, log(t)), defined for t > 0. find the arc length of c between the points (2, 1, 0) and (12,

V125BC [204]

The path starts at $t=1$ and ends at $t=6$ , so we have an arc length of

$\displaystyle\int_{\mathcal C}\mathrm dS=\int_{t=1}^{t=6}\|\mathbf c'(t)\|\,\mathrm dt$
$=\displaystyle\int_1^6\sqrt{2^2+(2t)^2+\left(\frac1t\right)^2}\,\mathrm dt$
$=\displaystyle\int_1^6\sqrt{4t^2+4+\frac1{t^2}}\,\mathrm dt$
$=\displaystyle\int_1^6\frac1t\sqrt{4t^4+4t^2+1}\,\mathrm dt$
$=\displaystyle\int_1^6\frac1t\sqrt{(2t^2+1)^2}\,\mathrm dt$
$=\displaystyle\int_1^6\frac{2t^2+1}t\,\mathrm dt$
$=\displaystyle\int_1^6\left(2t+\frac1t\right)\,\mathrm dt$
$=t^2+\log t\bigg|_{t=1}^{t=6}$
$=(6^2+\log6)-(1^2+\log1)=35+\log6$

4 0

4 years ago

The solution system to 3y-2x=-9 and y=-2x+5

Alecsey [184]

Answer:

$\boxed{(3,-1)}$

Step-by-step explanation:

Hey there!

Well to find the solution the the given system,

3y - 2x = -9

y = -2x + 5

So to find x lets plug in -2x + 5 for y in 3y - 2x = -9.

3(-2x + 5) - 2x = -9

Distribute

-6x + 15 - 2x = -9

-8x + 15 = -9

-15 to both sides

-8x = -24

Divide -8 to both sides

x = 3

Now that we have x which is 3, we can plug in 3 for x in y = -2x + 5.

y = -2(3) + 5

y = -6 + 5

y = -1

So the solution is (3,-1).

Hope this helps :)

4 0

3 years ago

9x squared -25 a difference of squares

aleksandr82 [10.1K]

Answer:

45xi is the answer

7 0

4 years ago

<img src="https://tex.z-dn.net/?f=%20log_%7B2%7D%28%20%7B0.5%7D%5E%7B3%7D%20%29%20" id="TexFormula1" title=" log_{2}( {0.5}^{3}

erik [133]

Step-by-step explanation:

$log_{2}( {0.5}^{3} ) \\\\=log_{2}( {0.5})^{3} \\ \\ = 3log_{2} {0.5} \\ \\ = 3log_{2} \frac{1}{2} \\ \\ = 3(log_{2} 1 - log_{2} 2) \\ \\ = 3(0 - 1) \\ \\ = 3 \times ( - 1) \\ \\ = - 3 \\ \\ \huge \purple{ \boxed{\therefore \: log_{2}( {0.5})^{3} = - 3}}$

6 0

4 years ago