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pochemuha
3 years ago
10

If sin(y degrees) = cos(x degrees), which of the following statements is true?

Mathematics
1 answer:
grin007 [14]3 years ago
6 0

Answer:

Angle y and Angle x are complementary angles

Step-by-step explanation:

we know that

If

sin(y°)=cos(x°)

then

Angle y and Angle x are complementary angles

so

∠y+∠x=90°

∠x=90°-∠y

sin(y°)=cos(90°-y°)

therefore

The sine of an angle is equal to the cosine  of its complementary angle

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Evaluate: -(-1) + (-7) + (2) - (-4) - (-2) A) -2 B) 0 C) 1 D) 2'
Masja [62]

Answer:

2

Step-by-step explanation:

-(-1) + (-7) + (2) - (-4) - (-2) =

= 1 + (-7) + 2 + 4 + 2

= -6 + 2 + 4 + 2

= -4 + 4 + 2

= 0 + 2

= 2

6 0
3 years ago
A tank with a capacity of 1000 L is full of a mixture of water and chlorine with a concentration of 0.02 g of chlorine per liter
faltersainse [42]

At the start, the tank contains

(0.02 g/L) * (1000 L) = 20 g

of chlorine. Let <em>c</em> (<em>t</em> ) denote the amount of chlorine (in grams) in the tank at time <em>t </em>.

Pure water is pumped into the tank, so no chlorine is flowing into it, but is flowing out at a rate of

(<em>c</em> (<em>t</em> )/(1000 + (10 - 25)<em>t</em> ) g/L) * (25 L/s) = 5<em>c</em> (<em>t</em> ) /(200 - 3<em>t</em> ) g/s

In case it's unclear why this is the case:

The amount of liquid in the tank at the start is 1000 L. If water is pumped in at a rate of 10 L/s, then after <em>t</em> s there will be (1000 + 10<em>t</em> ) L of liquid in the tank. But we're also removing 25 L from the tank per second, so there is a net "gain" of 10 - 25 = -15 L of liquid each second. So the volume of liquid in the tank at time <em>t</em> is (1000 - 15<em>t </em>) L. Then the concentration of chlorine per unit volume is <em>c</em> (<em>t</em> ) divided by this volume.

So the amount of chlorine in the tank changes according to

\dfrac{\mathrm dc(t)}{\mathrm dt}=-\dfrac{5c(t)}{200-3t}

which is a linear equation. Move the non-derivative term to the left, then multiply both sides by the integrating factor 1/(200 - 5<em>t</em> )^(5/3), then integrate both sides to solve for <em>c</em> (<em>t</em> ):

\dfrac{\mathrm dc(t)}{\mathrm dt}+\dfrac{5c(t)}{200-3t}=0

\dfrac1{(200-3t)^{5/3}}\dfrac{\mathrm dc(t)}{\mathrm dt}+\dfrac{5c(t)}{(200-3t)^{8/3}}=0

\dfrac{\mathrm d}{\mathrm dt}\left[\dfrac{c(t)}{(200-3t)^{5/3}}\right]=0

\dfrac{c(t)}{(200-3t)^{5/3}}=C

c(t)=C(200-3t)^{5/3}

There are 20 g of chlorine at the start, so <em>c</em> (0) = 20. Use this to solve for <em>C</em> :

20=C(200)^{5/3}\implies C=\dfrac1{200\cdot5^{1/3}}

\implies\boxed{c(t)=\dfrac1{200}\sqrt[3]{\dfrac{(200-3t)^5}5}}

7 0
3 years ago
Explain how to use the inverse operations to solve for x in the equation x^2=4
olchik [2.2K]
Divide 2 and 4 is 2 so 2•2=4
3 0
3 years ago
Read 2 more answers
What is the sign of f on the interval -2
TEA [102]

Answer:

f is sometimes positive and sometimes negative.

Step-by-step explanation:

f(x)=(x-3)(x+2)(x+4)(x+4)(x-1)(2x-9)

Take x=-1\in(-2,\frac{9}{2})\ as-2

f(-1)=(-1-3)(-1+2)(-1+4)(-1-1)(-2-9)\\\\=(-4)(1)(3)(-2)(-11)\\\\=-264\\\\f(-1)

Take x=2\in(-2,\frac{9}{2})\ as\ -2

f(2)=(2,-3)(2+2)(2+4)(2-1)(2\times2-9)\\\\=(-1)(4)(6)(1)(-5)\\\\=120\\\\f(2)>0

Hence f(x) for x=-1 and f(x)>0 for x=2

So f is sometimes positive and sometimes negative in the interval.

3 0
3 years ago
Thanks for the help
Brrunno [24]
The answer is B.

In the question it says "the intercept of 6" which means the equation has to have +6 in the answer, and there is only one answer choice containing that.
4 0
3 years ago
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