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velikii [3]
3 years ago
7

If there is 1/3 cups of yogurts for each smoothie. how many cups of yogurts would be needed to make 7 smoothies

Mathematics
1 answer:
Inessa [10]3 years ago
8 0
7/3 cups or 2 1/3 cups of yogurt
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Cindy paid $1.74 for 0.25 pound of tuna salad. What was the cost of 1 pound of tuna salad?​
MrMuchimi

Answer:

$6.96

Step-by-step explanation:

0.25 pound = $1.74

for 1 pound multiply by 4, since there are four 0.25 (quarters) to 1 pound.

1.74*4 = 6.96

5 0
3 years ago
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Which sentence represents the inequality -2b + 7 > 27.5?
Nady [450]

Answer:

Negative two times a number, plus seven, is greater than twenty-seven and five tenths.

3 0
4 years ago
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What is the GREATEST COMMON FACTOR of 20x2 , 10x , and 15?
irina [24]

Answer:

5

Step-by-step explanation:

x can only go into 20x2 and 10x and 5 is the Gcf of these three numbers

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<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
3 years ago
draw an example of a shape that has at least one right angle one angle less than a right angle and one angle greater than a righ
MaRussiya [10]
That is not possible to make a triangle with a right angle and a acute angle because a acute triangle only has acute angles and a right triangle has a obtuse angle in it
3 0
3 years ago
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