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3 years ago

Graph y=-|x|+10 PLS TELL ME THE CORD NUMBERS I WILL GIVE YOU MORE POINTS PLS AWNSER THIS QUICKLY!

Mathematics

1 answer:

meriva3 years ago

5 0

Answer: answer is in the ss i put

Source:trust me bro

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Ahat [919]

Answer:

Step-by-step explanation:

7 0

2 years ago

Kayla says that the point labeled C in the diagram below is the center. Raymond says that point C is the radius. A sphere. The c

liq [111]

Answer:

Kayla is correct; the center is a fixed point in the middle of the sphere.

Step-by-step explanation:

Kayla is correct. Raymond is wrong because a point cannot be a radius: a radius is a line segment from the center to the surface.

The center is the fixed point in the middle of the sphere.

6 0

3 years ago

Read 2 more answers

Find the values of A B C AND D of <br> 4x^2 (2x^3+5x)= Ax^B +Cx^D

iren2701 [21]

The values are $A=8, B=5, C= 20$ and $D=3$

Explanation:

The expression is $4x^{2} (2x^{3} +5x)=Ax^{B} +Cx^{D}$

Simplifying, we get,

$8x^{5} +20x^3=Ax^{B} +Cx^{D}$

Since, both sides of the expression are equal, we can equate the corresponding values of A, B, C and D.

Thus, we get,

$8 x^{5}=A x^{B}$ ⇒ $A=8$ and $B=5$

Also, equating, $20 x^{3}=C x^{D}$ , we get,

$C=20$ and $D=3$

Thus, the values are $A=8, B=5, C= 20$ and $D=3$

3 0

3 years ago

5. Trista has of a yard of ribbon to use for making decorative frames. If one frame used

Lunna [17]

0 yards remaining if she used one yard

5 0

3 years ago

Use any of the methods to determine whether the series converges or diverges. Give reasons for your answer.

Aleks [24]

Answer:

It means $\sum_{n=1}^\inf} = \frac{7n^2-4n+3}{12+2n^6}$ also converges.

Step-by-step explanation:

The actual Series is::

$\sum_{n=1}^\inf} = \frac{7n^2-4n+3}{12+2n^6}$

The method we are going to use is comparison method:

According to comparison method, we have:

$\sum_{n=1}^{inf}a_n\ \ \ \ \ \ \ \ \sum_{n=1}^{inf}b_n$

If series one converges, the second converges and if second diverges series, one diverges

Now Simplify the given series:

Taking"n^2"common from numerator and "n^6"from denominator.

$=\frac{n^2[7-\frac{4}{n}+\frac{3}{n^2}]}{n^6[\frac{12}{n^6}+2]} \\\\=\frac{[7-\frac{4}{n}+\frac{3}{n^2}]}{n^4[\frac{12}{n^6}+2]}$

$\sum_{n=1}^{inf}a_n=\sum_{n=1}^{inf}\frac{[7-\frac{4}{n}+\frac{3}{n^2}]}{[\frac{12}{n^6}+2]}\ \ \ \ \ \ \ \ \sum_{n=1}^{inf}b_n=\sum_{n=1}^{inf} \frac{1}{n^4}$

Now:

$\sum_{n=1}^{inf}a_n=\sum_{n=1}^{inf}\frac{[7-\frac{4}{n}+\frac{3}{n^2}]}{[\frac{12}{n^6}+2]}\\ \\\lim_{n \to \infty} a_n = \lim_{n \to \infty} \frac{[7-\frac{4}{n}+\frac{3}{n^2}]}{[\frac{12}{n^6}+2]}\\=\frac{7-\frac{4}{inf}+\frac{3}{inf}}{\frac{12}{inf}+2}\\\\=\frac{7}{2}$