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3 years ago

Math please save me

Mathematics

1 answer:

Naily [24]3 years ago

4 0

The answer is 18 beacause 6+6+2+4=18

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By exercising regularly, Abby was able to improve her running right this year. Last year, Abby ran a mile in 12 minutes. This ye

OleMash [197]

Answer:

She decreased her running time at a rate of 16.67%

Step-by-step explanation:

In this question, we are asked basically to calculate the percentage decrease in Abby’s running time.

Mathematically, the percentage decrease equals: (new running time - old running time)/old running time * 100%

We input the values and proceed as follows:

Percentage decrease = (10-12)/12 * 100

-2/12 * 100/1 = -1/6 * 100 = -16.67%

Since it’s a decrease, we just simply say that her running time decrease by 16.67%

4 0

3 years ago

There are 5 4/5 cups of milk left in a carton write 5 4/5 as an improper fraction

olga_2 [115]

29/5 should be the correct answer

5 0

3 years ago

May someone help me with this question

Dafna11 [192]

X + 13 = -4
x = -4 - 13
x = - 17
I am thinking it is the first one..solution is -17

3 0

3 years ago

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

3 years ago

What is 5,841 divided by 62

baherus [9]

Answer:

The answer is 94.2096774194

5 0

3 years ago

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