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3 years ago

6

Wally bakes muffins. in each batch of 12 muffins, he bakes 8 bluberry muffins write in simplest form the fraction of muffins tha

t are not blue berry

1 answer:

Flauer [41]3 years ago

6 0

4/12 = 1/3 muffins are not blueberry

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Let AB = 24, AC = 10 and BC = 26.

nata0808 [166]

Answer:

a) 24

b) 10

c) 12/13

d) 5/13

e) 12/5

Step-by-step explanation:

a) We can see that the leg opposite <C is AB, and we are given AB = 24

b) We can see the leg adjacent to <C is AC, and we are given that AC = 10

c) The trig function sine is equal to

$\frac{opposite}{hypotenuse}$

The opposite, AB, is 24, and the hypotenuse, BC, is 26. We can plug those numbers in:

$sin(c) = \frac{24}{26} = \frac{12}{13}$

d)The trig function cosine is equal to

$\frac{adjacent}{hypotenuse}$

The adjacent, AC, is 10, and the hypotenuse, BC, is 26. We can plug those numbers in:

$cos(c) = \frac{10}{26} = \frac{5}{13}$

d)The trig function tangent is equal to

$\frac{opposite}{adjacent}$

The opposite, AB, is 24, and the adjacent, AC, is 10. We can plug those numbers in:

$tan(c) = \frac{24}{10} = \frac{12}{5}$

4 0

2 years ago

***Please Help***

Neporo4naja [7]

D'(-3,-4)
E'(-3,-10)
F'(3,-6)

4 0

3 years ago

A biker travels at a rate of 20 miles per hour. Calculate

kow [346]

Answer:

60 miles

Step-by-step explanation:

20 miles times 3

3 0

3 years ago

Read 2 more answers

Find the surface area of the solid generated by revolving the region bounded by the graphs of y = x2, y = 0, x = 0, and x = 2 ab

Nikitich [7]

Answer:

See explanation

Step-by-step explanation:

The surface area of the solid generated by revolving the region bounded by the graphs can be calculated using formula

$SA=2\pi \int\limits^a_b f(x)\sqrt{1+f'^2(x)} \, dx$

If $f(x)=x^2,$ then

$f'(x)=2x$

and

$b=0\\ \\a=2$

Therefore,

$SA=2\pi \int\limits^2_0 x^2\sqrt{1+(2x)^2} \, dx=2\pi \int\limits^2_0 x^2\sqrt{1+4x^2} \, dx$

Apply substitution

$x=\dfrac{1}{2}\tan u\\ \\dx=\dfrac{1}{2}\cdot \dfrac{1}{\cos ^2 u}du$

Then

$SA=2\pi \int\limits^2_0 x^2\sqrt{1+4x^2} \, dx=2\pi \int\limits^{\arctan(4)}_0 \dfrac{1}{4}\tan^2u\sqrt{1+\tan^2u} \, \dfrac{1}{2}\dfrac{1}{\cos^2u}du=\\ \\=\dfrac{\pi}{4}\int\limits^{\arctan(4)}_0 \tan^2u\sec^3udu=\dfrac{\pi}{4}\int\limits^{\arctan(4)}_0(\sec^3u+\sec^5u)du$

Now

$\int\limits^{\arctan(4)}_0 \sec^3udu=2\sqrt{17}+\dfrac{1}{2}\ln (4+\sqrt{17})\\ \\ \int\limits^{\arctan(4)}_0 \sec^5udu=\dfrac{1}{8}(-(2\sqrt{17}+\dfrac{1}{2}\ln(4+\sqrt{17})))+17\sqrt{17}+\dfrac{3}{4}(2\sqrt{17}+\dfrac{1}{2}\ln (4+\sqrt{17}))$

Hence,

$SA=\pi \dfrac{-\ln(4+\sqrt{17})+132\sqrt{17}}{32}$

3 0

2 years ago

If A || B and B | y, then ?

aniked [119]

Answere: I believe that the answere is C.

Step-by-step explanation:

Well,since the lines A and B are paralel and the line y is not paralel with any of then y and A are not paralel.Plus there is not a line called x in this particular equazion.If you have any questions , please contact me.

Yours sincerely,

Manos

8 0

2 years ago