Stuck on this please help me

What times what equals = 98?

ira [324]

98/2 = 49

Therefore,

49*2 = 98

Final answer: 49 times 2 equals 98

4 0

3 years ago

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How do I graph it?? HELPP

liraira [26]

Answer: you use desmos. It’s helpful

Good luck !

8 0

2 years ago

The perimeter of a triangle is 57 centimeters. The longest side is 3 centimeter(s) less than the sum of the other two sides. Tw

Setler79 [48]

Let's say x is longest side, y is shortest side and z is the other side. So we have some equations such that

$x+y+z=57$ (by the first sentence)

$x=(y+z) -3$ (by the second sentence)

$2y=x-15$ (by the third sentence)

By the second equation, we have $y+z=x+3$ and if we use it in the first equation, then $x+y+z=x+(x+3)=57$ . So $x=27$

Since $x=27$ , if we use it in the third equation the we have $2y=x-15=27-15=12$ then $y=6$

Therefore $z=24$

5 0

3 years ago

Maggie is looking into joining a gym, She went to Axiom Fitness and to Gold's Gym, and they are offering membership plans that w

zmey [24]

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6 0

3 years ago

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Help me with trigonometry

poizon [28]

Answer:

See below

Step-by-step explanation:

It has something to do with the Weierstrass substitution, 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,

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