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ss7ja [257]
3 years ago
13

How do I simplify 2( √ 4)^-2​

Mathematics
2 answers:
padilas [110]3 years ago
7 0

This is the answer the picture below hope this help

Afina-wow [57]3 years ago
4 0

Answer:

the answer is 1/2 ( I may be wrong)

Step-by-step explanation:

Just type it in your calculator

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Geometry question on finding x.
Dmitrij [34]

Answer:

x =63^o

Step-by-step explanation:

<u>Given </u><u>:</u><u>-</u><u> </u>

  • \text{ $\overline{MN} $ and $\overline{ML}$ are tangents .}

We know that the tangents are perpendicular to the radius at the point of contact . And here OLMN is a quadrilateral. Also we know that , the angle sum property a triangle is 360° .

<u>According</u><u> to</u><u> question</u><u> </u><u>:</u><u>-</u><u> </u>

\implies x + 117^o + 90^o +90^o = 360^o \\\\\implies x + 297^o = 360^o \\\\\implies x = 360^o - 297^o \\\\\implies x = 63^o

<u>Hence</u><u> the</u><u> </u><u>value</u><u> of</u><u> x</u><u> </u><u>is </u><u>6</u><u>3</u><u>°</u><u> </u><u>.</u>

6 0
3 years ago
What is the solution to the equation
Firdavs [7]
3c/5-1/2=-31/2 Add 1/2 both sides
      +1/2 +1/2
3c/5=-30/2 Then simplify the fraction
(5)3c/5=-15(5) Multiply 5 both sides
3c=75 Finally, divide by 3 both sides
c=25
So the answer to your question is E. 
7 0
3 years ago
Jason eats 10 ounces of candy in 5 days how long does it take Jason to eat one pound of candy
tester [92]

there are 16 ounces to a pound

10/5 = 2 ounces per day

16/2 = 8

 so it took 8 days to eat 1 ppound

6 0
3 years ago
Please help i have no clue how to do this please solve and explain
Mariulka [41]

The answer is c. 9z^7/y^6

6 0
3 years ago
The number of withdrawals a bank processes in a day follows a random variable X. The number of deposits in a day is represented
Serhud [2]

Answer:

Check the explanation

Step-by-step explanation:

Number of transactions in a day is sum of number of withdrawals and number of deposits. So,

Number of transactions in a day, Z = X + Y

Moment Generating function of Z is,

T+1

Expected number of transactions in a day = E[Z]

= \frac{\mathrm{d} }{\mathrm{d} t}_{t=0}M_Z(t) = (r+1)\left ( \frac{p}{1-qe^t} \right )^{r} * \frac{-p}{(1-qe^t)^2} * (-qe^t)    for t = 0

= (r+1)\left ( \frac{p}{1-qe^0} \right )^{r} * \frac{-p}{(1-qe^0)^2} * (-qe^0)

= (r+1)\left ( \frac{p}{1-q} \right )^{r} * \frac{-p}{(1-q)^2} * (-q)

= (r+1)\left ( \frac{p}{p} \right )^{r} * \frac{-p}{(p)^2} * (-q)

= \frac{(r+1)q}{p}

4 0
3 years ago
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