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2 years ago

Evaluate 8^

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Mathematics

2 answers:

Minchanka [31]2 years ago

7 0

8^5/3 = 10922.666666 (repeating)

almond37 [142]2 years ago

6 0

Answer:

Step-by-step explanation:

32 is the correct answer

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On a number line there are six spaces. On the first is the number 2, on the last space is the number 42, on the third space is x

trapecia [35]

1st space is 2
6th space (last) is 42
in betwen that is 5 spaces to go
find distance from 2 to 42
42-2=40

40=5 spaces
divide 5
8=1 space
each space goes upby 8

2=1st space
2+8=2nd space
2+8+8=3rd space=18=x

x=18

5 0

3 years ago

I rlly rlly rlly need help! Complete the following proof. Given: Points R, S, T, Q on circle O Prove: m\overarc RS + m\overarc S

harina [27]

Answer:

$m(\widehat {RS})+m(\widehat{ST})+m(\widehat{TQ}) + m(major\ arc RQ) = 360\ degree$

Step-by-step explanation:

As we can see in the figure that

The R, S, T ,and Q are the points on the circle O.

Also

The measurement of the circular arc is equivalent to the measurement of the angle at the center of the arc

So by this

$m(\widehat{RS})=m(\angle ROS)$

$m(\widehat{ST})=m(\angle SOT)$

$m(\widehat{TQ})=m(\angle TOQ)$

m(major arc RQ) = m(∠QOR)

So,

$m(\widehat {RS})+m(\widehat{ST})+m(\widehat{TQ}) + m(major\ arc RQ) = m(\angle ROS)+m(\angle SOT)+m(\angle TOQ)+m(\angle QOR)$

And as we know that

All angles sum = 360°

Therefore

$m(\widehat {RS})+m(\widehat{ST})+m(\widehat{TQ}) + m(major\ arc RQ) = 360\ degree$

8 0

3 years ago

Divide the rational expressions and express in simplest form. When typing your answer for the numerator and denominator be sure

Veseljchak [2.6K]

Dividing by a fraction is equivalent to multiply by its reciprocal, then:

$\begin{gathered} \frac{3y^2-7y-6}{2y^2-3y-9}\div\frac{y^2+y-2}{2y^2+y-3^{}}= \\ =\frac{3y^2-7y-6}{2y^2-3y-9}\cdot\frac{2y^2+y-3}{y^2+y-2}= \\ =\frac{(3y^2-7y-6)(2y^2+y-3)}{(2y^2-3y-9)(y^2+y-2)} \end{gathered}$

Now, we need to express the quadratic polynomials using their roots, as follows:

$ay^2+by+c=a(y-y_1)(y-y_2)$

where y1 and y2 are the roots.

Applying the quadratic formula to the first polynomial:

$\begin{gathered} y_{1,2}=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ y_{1,2}=\frac{7\pm\sqrt[]{(-7)^2-4\cdot3\cdot(-6)}}{2\cdot3} \\ y_{1,2}=\frac{7\pm\sqrt[]{121}}{6} \\ y_1=\frac{7+11}{6}=3 \\ y_2=\frac{7-11}{6}=-\frac{2}{3} \end{gathered}$

Applying the quadratic formula to the second polynomial:

$\begin{gathered} y_{1,2}=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ y_{1,2}=\frac{-1\pm\sqrt[]{1^2-4\cdot2\cdot(-3)}}{2\cdot2} \\ y_{1,2}=\frac{-1\pm\sqrt[]{25}}{4} \\ y_1=\frac{-1+5}{4}=1 \\ y_2=\frac{-1-5}{4}=-\frac{3}{2} \end{gathered}$

Applying the quadratic formula to the third polynomial:

$\begin{gathered} y_{1,2}=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ y_{1,2}=\frac{3\pm\sqrt[]{(-3)^2-4\cdot2\cdot(-9)}}{2\cdot2} \\ y_{1,2}=\frac{3\pm\sqrt[]{81}}{4} \\ y_1=\frac{3+9}{4}=3 \\ y_2=\frac{3-9}{4}=-\frac{3}{2} \end{gathered}$

Applying the quadratic formula to the fourth polynomial:

$\begin{gathered} y_{1,2}=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ y_{1,2}=\frac{-1\pm\sqrt[]{1^2-4\cdot1\cdot(-2)}}{2\cdot1} \\ y_{1,2}=\frac{-1\pm\sqrt[]{9}}{2} \\ y_1=\frac{-1+3}{2}=1 \\ y_2=\frac{-1-3}{2}=-2 \end{gathered}$

Substituting into the rational expression and simplifying:

$\begin{gathered} \frac{3(y-3)(y+\frac{2}{3})2(y-1)(y+\frac{3}{2})}{2(y-3)(y+\frac{3}{2})(y-1)(y+2)}= \\ =\frac{3(y+\frac{2}{3})}{2(y+2)}= \\ =\frac{3y+2}{2y+4} \end{gathered}$

8 0

9 months ago

In the following figure, the square ABCD is inscribed in the circle centered at point P. What is the area of the circle?

Ainat [17]

Answer:

$A_{Circle} \approx 39.24\: units^2$

Step-by-step explanation:

Square ABCD is inscribed in a circle with center P such that BC = 5 units.

BD will be diagonal of the square as well as diameter of the circle.

$BD = BC\times\sqrt 2$

$\implies BD = 5\sqrt 2$

-> Diameter of the circle $(d) = 5\sqrt 2$

-> Radius of the circle $(r) = 2.5\sqrt 2=3.535\: units$

$A_{Circle} =\pi(r)^2$

$\implies A_{Circle} =3.14(3.535)^2$

$\implies A_{Circle} =39.2381465$

$\implies A_{Circle} \approx 39.24\: units^2$

8 0

2 years ago

Find the exponential function that satisfies the given conditions: Initial value = 64, decreasing at a rate of 0.5% per week

mrs_skeptik [129]

Answer:

Step-by-step explanation:

An exponential function is of the form

$y=ab^x$

where a is the initial value and b is the growth/decay rate. Our initial value is 64. That's easy to plug in. It goes in for a. So the first choice is out. Considering b now...

If the rate is decreasing at .5% per week, this means it still retains a rate of

100% - .5% = 99.5%

which is .995 in decimal form.

b is a rate of decay when it is greater than 0 but less than 1; b is a growth rate when it is greater than 1. .995 is less than 1 so it is a rate of decay. The exponential function is, in terms of t,

$f(t) = 64(.995)^t$

4 0

2 years ago