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3 years ago

some 8,000 crocus flowers required ro make 2 oumces of saffron. how many flowers are needed to make 16 ounces of saffron

Mathematics

1 answer:

Elena L [17]3 years ago

6 0

8000/ 2 = x/16
cross multiply
2x = 8000 * 16
2x = 128000
x = 128000/2
x = 64000 crocus flowers

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Please look at the assignment be creative and come up with your own ratio and example in your own words. ALSO PLEASE HURRY!

svetoff [14.1K]

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2 years ago

Check all that apply. If csc =13/5 , then:

nirvana33 [79]

Answer:

The answer are the options B and C

$sin(x)=\frac{5}{13}$

$tan(x)=\frac{5}{12}$

Step-by-step explanation:

we know that

$csc(x)=\frac{1}{sin(x)}$

In this problem we have

$csc(x)=\frac{13}{5}$

substitute

$\frac{13}{5}=\frac{1}{sin(x)}$

solve for sin(x)

$sin(x)=\frac{5}{13}$

$cos^{2}(x)+sin^{2}(x)=1$

substitute the value of sin(x) and solve for cos(x)

$cos^{2}(x)+(\frac{5}{13})^{2}=1$

$cos^{2}(x)=1-(\frac{5}{13})^{2}$

$cos^{2}(x)=(\frac{13^{2}-5^{2}}{13^{2}})$

$cos^{2}(x)=(\frac{144}{13^{2}})$

$cos^{2}(x)=(\frac{12^{2}}{13^{2}})$

$cos(x)=(\frac{12}{13})$

The function tangent is equal to

$tan(x)=\frac{sin(x)}{cos(x)}$

substitute the values

$tan(x)=\frac{(5/13)}{(12/13)}=5/12$

8 0

3 years ago

Read 2 more answers

What is the square root of -2i?

Studentka2010 [4]

Answer:

1-i and -1+i

Step-by-step explanation:

We are to find the square roots of $z=0-2i$ . First, convert from Cartesian to polar form:

$r=\sqrt{a^2+b^2}\\r=\sqrt{0^2+(-2)^2}\\r=\sqrt{0+4}\\r=\sqrt{4}\\r=2$

$\theta=tan^{-1}(\frac{b}{a})\\ \theta=tan^{-1}(\frac{-2}{0})\\\theta=\frac{3\pi}{2}$

$z=2(\cos\frac{3\pi}{2}+i\sin\frac{3\pi}{2})$

Next, use the formula $\displaystyle \sqrt[n]{r}\biggr[\cis\biggr(\frac{\theta+2\pi k}{n}\biggr)\biggr]$ where $\displaystyle k=0,1,2,...\:,n-1$ to find the square roots:

When k=1

$\displaystyle \sqrt[2]{2}\biggr[cis\biggr(\frac{\frac{3\pi}{2}+2\pi(1)}{2}\biggr)\biggr]$

$\displaystyle \sqrt{2}\biggr[cis\biggr(\frac{3\pi}{4}+\pi\biggr)\biggr]$

$\sqrt{2}\biggr(cis\frac{7\pi}{4}\biggr)$

$\sqrt{2}(\cos\frac{7\pi}{4}+i\sin\frac{7\pi}{4})\\ \\\sqrt{2}(\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i)\\ \\1-i$

When k=0

$\displaystyle \sqrt[2]{2}\biggr[cis\biggr(\frac{\frac{3\pi}{2}+2\pi(0)}{2}\biggr)\biggr]$

$\sqrt{2}\biggr(cis\frac{3\pi}{4}\biggr)$

$\sqrt{2}(\cos\frac{3\pi}{4}+i\sin\frac{3\pi}{4})\\ \\\sqrt{2}(-\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i)\\ \\-1+i$

Thus, the square roots of -2i are 1-i and -1+i

4 0

2 years ago

What times what equals 52

hoa [83]

4 times 13 is equal to 52

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3 years ago

Read 2 more answers

What is the period of the function y=3(tan 6x)?

Ostrovityanka [42]

With all trig functions, the period can be calculated with the equation PB = 2π.

P = period
B = anything in the tangent part of the equation except the x. (in this case, it would be 6)

When we plug in 6 for B, we get 6P = 2π.

Divide both sides by 6 and you get P = 2π / 6

OR

P = π/3

4 0

3 years ago