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3 years ago

If you have 2+2^2 and/or 5^2s when s=14 how much will you have???

Mathematics

2 answers:

QveST [7]3 years ago

4 0

Step-by-step explanation:

2+2^2=6

5^2(14)=25(14)=350

350+6=356

Hope that helps :)lol

never [62]3 years ago

4 0

Answer:

356

Step-by-step explanation:

2+2^2=6

5^2(14)=25(14)=350

350+6=356

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2 years ago

4. Kay and Allen sell cell phones. Last month Kay sold 17 more cell phones than Allen. Together they both sold 117 cell phones.

forsale [732]

Answer:

Allen sold 50 cell phones while Kay sold 67 cell phones.

Step-by-step explanation:

Let the number of cell phones sold by Allen be x. This will imply that the number of cell phones sold by Kay is x + 17.

Furthermore, we are informed that together they sold a total of 117 cell phones. Therefore, we use the following equation to determine the number of cell phones sold by each;

x + x + 17 = 117

2x + 17 = 117

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Therefore, Allen sold 50 cell phones while Kay sold 67 cell phones.

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3 years ago

Read 2 more answers

Drag each tile to the correct box consider the given functions f,g and h place the tiles in order from least to greatest accordi

Alja [10]

Answer:

Step-by-step explanation:

By definition, the average rate of change of a function f over an interval [a,b] is given by

$\dfrac{f(b)-f(a)}{b-a}$

compute the quantity

$\dfrac{f(3)-f(0)}{3}$

for all the three function

Average rate of change of f:

We will simply use the table to check the values for f(3) and f(0):

$\dfrac{f(3)-f(0)}{3}=\dfrac{10-1}{3} = 3$

Average rate of change of g:

We will use the graph to to check the values for g(3) and g(0):

$\dfrac{g(3)-g(0)}{3}=\dfrac{8-1}{3} = \dfrac{7}{3}$

Average rate of change of h:

We can plug the values in the equation to get h(3) and h(0):

$h(3)=3^2+3-6=9+3-6=6,\quad h(0)=0^2+0-6=-6$

And so the average rate of change is

$\dfrac{h(3)-h(0)}{3}=\dfrac{6-(-6)}{3} = 4$

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4 years ago

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Lelechka [254]

Answer:

40 min

Step-by-step explanation:

Sea level is 0 meters. Time when Amir reaches sea level is intersection line and x-axis. It is 40 minutes.

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3 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

——————————

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. =)

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3 years ago