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2 years ago

eugheugh eughhhh I've been listening to my comfort playlist stressing about this test for too long Please just help ;-;

Mathematics

1 answer:

Kaylis [27]2 years ago

6 0

Answer:

2 and 4 are correct

Step-by-step explanation:

1 is incorrect because 4/5 being bigger than 7/10

3 is incorrect because 2/3 is more than 5/9

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For her presentation on the Wonders of the World, Mary baked a square pyramid-shaped cake as pictured below. The slant height of

mihalych1998 [28]

Answer:

$V=196in^3$

Step-by-step explanation:

The volume of a pyramid is:

$V=\frac{A_{b}h}{3}$

where $A_{b}$ is the area of the base and $h$ is the height (the perpendicular measurement between base and highest point, not the slant height)

Since the base is a square, the area is given by:

$A_{b}=l^2$

where $l$ is the length of the side: $l=8in$ , thus:

$A_{b}=(8in)^2\\A_{b}=64in^2$

Now we need to find the height, for this we use the right triangle that forms with half of a square side (8in/2 = 4in), the slant height (10in), and the height.

In this right triangle, the slant height is the hypotenuse, the leg 1 is the unknown height, and leg 2 is half of the square side.

Using pythagoras:

$hypotenuse^2=leg1^2+leg2^2$

substituting our values, and indicating that leg 1 is height h:

$(10in)^2=h^2+(4in)^2$

$100in^2=h^2+16in^2$

and solving for the height:

$h^2=100in^2-16in^2\\h^2=84in^2\\h=\sqrt{84in^2}\\ h=9.165in$

and finally we calculate the volume using this height and the area of the base:

$V=\frac{A_{b}h}{3}$

$V=\frac{(64in^2)(9.165in)}{3} \\V=195.5in^3$

rounding to the nearest cubic inch: $V=196in^3$

4 0

3 years ago

What are all values of x for which the series n 3^n/x^n converges

Eva8 [605]

The interval of the convergence is x < -3 or x > 3 if the series n 3^n/x^n goes infinitely.

<h3>What is convergent of a series?</h3>

A series is convergent if the series of its partial sums approaches a limit; that really is, when the values are added one after the other in the order defined by the numbers, the partial sums getting closer and closer to a certain number.

We can find the interval for the convergent by root test.

Like the Ratio Test, the root Test is used to determine absolute convergence (or not) with factorials, the ratio test is useful.

For the given series:

$\sum^{\infty}_{n=0}\dfrac{3^n}{x^n}$