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3 years ago

HELP ASAP!!!!!!!!!!!!!!

Mathematics

2 answers:

yawa3891 [41]3 years ago

7 0

Answer:

Step-by-step explanation:

just try to make the lines even and get the answer 10

Cloud [144]3 years ago

7 0

Answer:

the answer is √89........

...

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274% as a decimal and fraction?

Dafna11 [192]

2.74 and 137/50 should be correct

4 0

3 years ago

Read 2 more answers

There are 1000 households in a town. Specifically, there are 100 households with one member, 200 households with 2 members, 300

vlada-n [284]

Answer:

3300 people

Step-by-step explanation:

There are 1000 households in a town. Specifically, there are

100 households with one member

= 100 × 1 member

= 100 people

200 households with 2 members

= 200 × 2 members

= 400 people

300 households with 3 members

= 300 × 3 member

= 900 people

200 households with 4 members,

= 200 × 4 members

= 800 people

100 households with 5 members

= 100 × 5 members

= 500 people

100 households with 6 members.

= 100 × 6 member

= 600 people

Thus, the total number of people living in the town is

100 people + 400 people + 900 people + 800 people + 500 people + 600 people

= 3300 people

3 0

3 years ago

Please help me I don't understand and this is due by the end of the day. I looked up ways to do this but I just don't understand

EastWind [94]

Answer:

well you used 6.4 pounds of flour to make bread which means you have 9.6 pounds left so you could make about 14 batches

Step-by-step explanation:

8 0

3 years ago

Find cos y and tan y if csc y = -√6/2 and cot y >0.

fomenos

Answer:

$\cos y = -\dfrac{\sqrt{3} }{3}$

$\tan y = \sqrt{2}$

Step-by-step explanation:

Recall that

$\boxed{\csc y := \dfrac{1}{\sin y}}$

$\boxed{\cot y := \dfrac{\cos y}{\sin y}}$

We know that

$\csc y = \dfrac{-\sqrt{6} }{2}$

Note that according to the definition of $\csc y$ it is true that both sine and cosine are negative, once $\csc y = \dfrac{-\sqrt{6} }{2}$ . Because $\cot y > 0$ , this conclusion is true. We basically have

$\boxed{(-a)(1/-b)=a/b \text{ such that } a,b\in\mathbb{R}_{\geq 0}}$

Sure it is true $\forall y\in\mathbb{R}$ but perhaps this way is better to understand.

In order to find sine, we can use the definition and manipulate the rational expression.

$\csc y = \dfrac{-\sqrt{6} }{2} = \dfrac{-\sqrt{6} / -\sqrt{6} }{2/-\sqrt{6} } = \dfrac{1 }{-\dfrac{2}{\sqrt{6} } }$

Therefore,

$\sin y =-\dfrac{2}{\sqrt{6} }$

Here I just divided numerator and denominator by $-\sqrt{6}$ .

Now, to find cosine we can use the identity

$\boxed{\sin^2y +\cos ^2y =1}$

Thus,

$\left(-\dfrac{2}{\sqrt{6} }\right)^2 + \cos ^2y =1 \implies \dfrac{4}{6 } +\cos ^2y =1$

$\implies \cos ^2y =1 - \dfrac{4}{6 } \implies \cos ^2y =\dfrac{1}{3 } \implies \cos y = \pm \dfrac{\sqrt{1} }{\sqrt{3} } = \pm \dfrac{\sqrt{1} \sqrt{3} }{3} = \pm \dfrac{\sqrt{3} }{3}$

$\cos y = \pm\dfrac{\sqrt{3} }{3}$

Once we have $\cot y > 0$ , we just consider

$\cos y = -\dfrac{\sqrt{3} }{3}$

FInally, for tangent, just consider

$\boxed{\tan y := \dfrac{\sin y}{\cos y}}$

thus,

$\tan y = \dfrac{\sin y}{\cos y} = \dfrac{-\frac{2}{\sqrt{6} }}{-\frac{\sqrt{3} }{3}} = \dfrac{6}{\sqrt{18} } =\dfrac{6}{3\sqrt{2} } =\dfrac{2}{\sqrt{2} } = \sqrt{2}$

5 0

3 years ago

How do I find dy/dx of the following?

Pachacha [2.7K]

Answer:

$\displaystyle\frac{dy}{dx} \ = \ 3x^{2} \ + \ \displaystyle\frac{1}{2\sqrt{x^{3}}} \ - \ \displaystyle\frac{12}{x^{5}}$

Step-by-step explanation:

$y \ = \ x^{3} \ - \ \displaystyle\frac{1}{\sqrt{x}} \ + \ \displaystyle\frac{3}{x^{4}} \\ \\ y \ = \ x^{3} \ - \ x^{-\frac{1}{2}} \ + \ 3x^{-4} \\ \\ \displaystyle\frac{dy}{dx} \ = \ 3x^{3 \ - \ 1} \ - \ \left(-\displaystyle\frac{1}{2}\right)x^{-\frac{1}{2} \ - \ 1} \ + \ \left(-4 \ \times \ 3\right)x^{-4-1}$

$\displaystyle\frac{dy}{dx} \ = \ 3x^{2} \ + \ \displaystyle\frac{1}{2}x^{-\frac{3}{2}} \ - \ 12x^{-5} \\ \\ \displaystyle\frac{dy}{dx} \ = \ 3x^{2} \ + \ \displaystyle\frac{1}{2\sqrt{x^{3}}} \ - \ \displaystyle\frac{12}{x^{5}}$

5 0

2 years ago