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2 years ago

Compare these three ratios using fractions.

Mathematics

2 answers:

Tamiku [17]2 years ago

7 0

I think it would be greatest common factor

Dominik [7]2 years ago

6 0

Answer:

Step-by-step explanation:

<h3>If you are comparing 4/5 and 16/20. Then 4/5 is the largest fraction because 4/5 is an equivalent fraction. The bigger the denominator the smaller the fraction.</h3><h2 /><h2 /><h2 /><h2 /><h2 />

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Rewrite the quadratic function in vertex form.<br> Y=2x^2+4x-1

Fantom [35]

Answer:

$\large\boxed{y=2(x+1)^2-3}$

Step-by-step explanation:

The vertex form of an equation of a parabola:

$y=a(x-h)^2+k$

(h, k) - vertex

We have

$y=2x^2+4x-1=2\left(x^2+2x-\dfrac{1}{2}\right)$

We must use the formula: $(a+b)^2=a^2+2ab+b^2\qquad(*)$

$2\left(x^2+2(x)(1)-\dfrac{1}{2}\right)=2\bigg(\underbrace{x^2+2(x)(1)+1^2}_{(*)}-1^2-\dfrac{1}{2}\bigg)\\\\=2\left((x+1)^2-1-\dfrac{1}{2}\right)=2\left((x+1)^2-\dfrac{3}{2}\right)$

Use the distributive formula a(b + c) = ab + ac

$2(x+1)^2+2\left(-\dfrac{3}{2}\right)=2(x+1)^2-3$

5 0

2 years ago

What are two fraction equivalent to 9/10

Anuta_ua [19.1K]

Two fractions equivalent are 18/20 & 27/30

4 0

2 years ago

Help me with trigonometry

poizon [28]

Answer:

See below

Step-by-step explanation:

It has something to do with the<em> </em><u><em>Weierstrass substitution</em></u>, where we have

$\int\, f(\sin(x), \cos(x))dx = \int\, \dfrac{2}{1+t^2}f\left(\dfrac{2t}{1+t^2}, \dfrac{1-t^2}{1+t^2} \right)dt$

First, consider the double angle formula for tangent:

$\tan(2x)= \dfrac{2\tan(x)}{1-\tan^2(x)}$

Therefore,

$\tan\left(2 \cdot\dfrac{x}{2}\right)= \dfrac{2\tan(x/2)}{1-\tan^2(x/2)} = \tan(x)=\dfrac{2t}{1-t^2}$

Once the double angle identity for sine is

$\sin(2x)= \dfrac{2\tan(x)}{1+\tan^2(x)}$

we know $\sin(x)=\dfrac{2t}{1+t^2}$ , but sure, we can derive this formula considering the double angle identity

$\sin(x)= 2\sin\left(\dfrac{x}{2}\right)\cos\left(\dfrac{x}{2}\right)$

Recall

$\sin \arctan t = \dfrac{t}{\sqrt{1 + t^2}} \text{ and } \cos \arctan t = \dfrac{1}{\sqrt{1 + t^2}}$

Thus,

$\sin(x)= 2 \left(\dfrac{t}{\sqrt{1 + t^2}}\right) \left(\dfrac{1}{\sqrt{1 + t^2}}\right) = \dfrac{2t}{1 + t^2}$

Similarly for cosine, consider the double angle identity

Thus,

$\cos(x)= \left(\dfrac{1}{\sqrt{1 + t^2}}\right)^2- \left(\dfrac{t}{\sqrt{1 + t^2}}\right)^2 = \dfrac{1}{t^2+1}-\dfrac{t^2}{t^2+1} =\dfrac{1-t^2}{1+t^2}$

Hence, we showed $\sin(x) \text { and } \cos(x)$

======================================================

$5\cos(x) =12\sin(x) +3, x \in [0, 2\pi ]$

Solving

$5\,\overbrace{\frac{1-t^2}{1+t^2}}^{\cos(x)} = 12\,\overbrace{\frac{2t}{1+t^2}}^{\sin(x)}+3$

$\implies \dfrac{5-5t^2}{1+t^2}= \dfrac{24t}{1+t^2}+3 \implies \dfrac{5-5t^2 -24t}{1+t^2}= 3$

$\implies 5-5t^2-24t=3\left(1+t^2\right) \implies -8t^2-24t+2=0$

$t = \dfrac{-(-24)\pm \sqrt{(-24)^2-4(-8)\cdot 2}}{2(-8)} = \dfrac{24\pm 8\sqrt{10}}{-16} = \dfrac{3\pm \sqrt{10}}{-2}$

$t=-\dfrac{3+\sqrt{10}}{2}\\t=\dfrac{\sqrt{10}-3}{2}$

Just note that

$\tan\left(\dfrac{x}{2}\right) = \dfrac{3\pm 8\sqrt{10}}{-2}$

and $\tan\left(\dfrac{x}{2}\right)$ is not defined for $x=k\pi , k\in\mathbb{Z}$

6 0

2 years ago

Timothy plans on saving, starting with the $50 he received for his birthday. He is pretty sure he can continue to contribute $50

zimovet [89]

Future value can be v
months is m
v=50+50(m)

7 0

2 years ago

Use the grid to create a model to solve the percent problem. 12 is 60% of what number? Enter your answer in the box

Elenna [48]

Is means equal to and of means multiply. So....your equation would be 12=60%x. Now since u can't multiply by a percent, u gotta do 12=.6x. Now divide by .6 so that it cancels out. Your answer is 20. Hope this helps!

3 0

2 years ago

Read 2 more answers