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Zigmanuir [339]
3 years ago
9

11 Select the expression equivalent to

Mathematics
1 answer:
Annette [7]3 years ago
4 0

Answer:

B

Step-by-step explanation:

evry step is in the picture

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Find the measurement indicated in each parallelogram.<br> No links
podryga [215]

Answer:

D

Step-by-step explanation:

QG=GS

x+5=2x-2

x=7

QG=7+5=12

4 0
3 years ago
I need help I have to turn it in tomorrow
m_a_m_a [10]

Hello.

I'm not sure which question you need help for, so I'm going to answer both.

TOP: The improper fraction for the mixed number is 16/3. Multiply the whole number and denominator, then add the numerator.

BOTTOM: The number rounded to the nearest hundredth is 765.88.

7 0
4 years ago
Read 2 more answers
What is the volume of a rectangular prism that has a length, width, and height of 12 m, 14 m, and 23 m, respectively?
AveGali [126]
Find the volume with this formula
v = l × w × h

Given from the question
l = 12 m
w = 14 m
h = 23 m

Plug in the numbers into the formula
v = l × w × h
v = 12 × 14 × 23
v = 3,864

The volume of the rectangular prism is 3,864 m³
6 0
4 years ago
What is the image point of (-4, 8) after a translation right 4 units and up 2 units?
Vitek1552 [10]

Given:

Point = ( -4,8).

To find:

The image point of (-4, 8) after a translation right 4 units and up 2 units.

Solution:

We know that, if a figure translated 4 units right and 2 units up, then the rule of translation is

(x,y)\to (x+4,y+2)

The point is (-4,8).

Putting x=-4 and y=8, we get

(-4,2)\to (-4+4,8+2)

(-4,2)\to (0,10)

Therefore, the image of point (-4,8) is (0,10).

4 0
3 years ago
How do you use the limit comparison test on this particular series?<br>Calculus series tests​
barxatty [35]

Compare \dfrac1{\sqrt{n^2+1}} to \dfrac1{\sqrt{n^2}}=\dfrac1n. Then in applying the LCT, we have

\displaystyle\lim_{n\to\infty}\left|\frac{\frac1{\sqrt{n^2+1}}}{\frac1n}\right|=\lim_{n\to\infty}\frac n{\sqrt{n^2+1}}=1

Because this limit is finite, both

\displaystyle\sum_{n=1}^\infty\frac1{\sqrt{n^2+1}}

and

\displaystyle\sum_{n=1}^\infty\frac1n

behave the same way. The second series diverges, so

\displaystyle\sum_{n=0}^\infty\frac1{\sqrt{n^2+1}}=1+\sum_{n=1}^\infty\frac1n

is divergent.

4 0
4 years ago
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