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2 years ago

What is the height ? Please help, I need it! Urgent !

Mathematics

1 answer:

Vika [28.1K]2 years ago

4 0

Answer:

207.24 inch^2

Step-by-step explanation:

A=2*3.14*9+2.3.14*3*8

=56.52+150.72

=207.24 square inches

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Is this a function please help I’m failing

igomit [66]

Answer:

Yes

Step-by-step explanation:

The relation is a function. For a relation to be a function there must be a unique x value for each y value. So this means x's can not repeat, and in this relation, the x-values never repeat. Therefore this is a function.

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2 years ago

Hayden has 7 shells . Elly has 5 fewer shells than Hayden . Beth has 2 shells . How many shells do they have in all?

sleet_krkn [62]

Answer:

11 i think

Step-by-step explanation:

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2 years ago

John is hiking in the San Juan National Forest. Every 4 hours his elevation increases between 140 and 210 feet. 36 hours ago, Jo

Ksju [112]

Answer D. It is the closets answer to my calculations so im sorry if im wrong

8 0

2 years ago

Help me please i really don’t get this

LekaFEV [45]

Answer:

The answer is D

Step-by-step explanation:

Coplanar means that they are on the same closed area(or plane)- P, M, C, N are all on the same Plane.

3 0

2 years ago

Read 2 more answers

Find all solutions of the equation: 2cos^2x-cosx=1

Art [367]

If you're using the app, try seeing this answer through your browser: brainly.com/question/3166243

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Solve the trigonometric equation:

$\mathsf{2\,cos^2\,x-cos\,x=1}\\\\ \mathsf{2\,cos^2\,x-cos\,x-1=0}$

Make a substitution:

$\mathsf{cos\,x=t\qquad (-1\le t\le 1)}$

and the equation becomes

$\mathsf{2t^2-t-1=0}$

Rewrite conveniently – t as + t – 2t, and then factor the left-hand side by grouping:

$\mathsf{2t^2+t-2t-1=0}\\\\ \mathsf{t\cdot (2t+1)-1\cdot (2t+1)=0}$

Factor out 2t + 1:

$\mathsf{(2t+1)\cdot (t-1)=0}\\\\ \begin{array}{rcl} \mathsf{2t+1=0}&~\textsf{ or }~&\mathsf{t-1=0}\\\\ \mathsf{2t=1}&~\textsf{ or }~&\mathsf{t=1}\\\\ \mathsf{t=\dfrac{\,1\,}{2}}&~\textsf{ or }~&\mathsf{t=1} \end{array}$

Substitute back for t = cos x:

$\begin{array}{rcl}\mathsf{cos\,x=\dfrac{\,1\,}{2}}&~\textsf{ or }~&\mathsf{cos\,x=1}\\\\ \mathsf{cos\,x=cos\,60^\circ}&~\textsf{ or }~&\mathsf{cos\,x=cos\,0} \end{array}$

Therefore,

$\begin{array}{rcl} \mathsf{x=\pm\,60^\circ+k\cdot 360^\circ}&~\textsf{ or }~&\mathsf{cos\,x=0+k\cdot 360^\circ} \end{array}$

where k is an integer.

Solution set:

$\mathsf{S=\left\{x\in\mathbb{R}:~~x=-\,60^\circ+k\cdot 360^\circ~~or~~x=60^\circ+k\cdot 360^\circ~~or~~x=k\cdot 360^\circ,~~k\in\mathbb{Z}\right\}}$

I hope this helps. =)

3 0

2 years ago