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3 years ago

Which of the following can prove that figure ABCD is a square?

Mathematics

1 answer:

Klio2033 [76]3 years ago

7 0

Answer:

Step-by-step explanation:

all sides of square is 90degrees

Since AC and BD is the same, when you join the lines together, it will definitely be a square

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Factor the polynomial. 24w8 + 30w6

PilotLPTM [1.2K]

The answer to the question is c

7 0

3 years ago

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Find the limit of the function by using direct substitution. limit as x approaches one of quantity x squared plus three x minus

ivann1987 [24]

Answer:

Limit = -1

Step-by-step explanation:

We are given the following function:

$x^2 + 3x -1$

We are to calculate the limit of this function as x approaches zero.

For that, we will use direct substitution method and substitute the x with 0 in the given function to calculate its limit as follows:

$x ^ 2 + 3 x - 1$

$( 0 ) ^ 2 + 3 ( 0 ) - 1 = - 1$

Therefore, the limit is -1.

6 0

3 years ago

Evaluate arithmetic series:- Step-by-step answer, please!

lorasvet [3.4K]

Let's see

$\\ \rm\Rrightarrow {\displaystyle{\sum^{275}_{k=1}}}(-5k+12)$

$\\ \rm\Rrightarrow (-5(1)+12)+(-5(2)+12)\dots (-5(275)+12)$

$\\ \rm\Rrightarrow 7+5+3+2+1+\dots -1363$

a=7
l=-1363
n=275

Sum:-

$\\ \rm\Rrightarrow S_n=\dfrac{n}{2}[a+l]$

$\\ \rm\Rrightarrow \dfrac{275}{2}(7-1363)$

$\\ \rm\Rrightarrow \dfrac{275}{2}(-1356)$

$\\ \rm\Rrightarrow 275(-678)$

$\\ \rm\Rrightarrow -186450$

4 0

1 year ago

Read 2 more answers

Find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the x-axis. Verify y

mariarad [96]

Answer:

the volume of the solid generated by revolving the region bounded by the graphs of the equations about the x-axis is;

$\frac{\pi }{2} [e^2 - 1 ]$ or 10.036

Step-by-step explanation:

Given the data in the question;

y = $y = e^{(x - 1 )$ , y = 0, x = 1, x = 2.

Now, using the integration capabilities of a graphing utility

y = $y = e_2}^{(x - 1 )_$ , y = 0

Volume = $\pi \int\limits^2_1 ( e^{x-1)^2} - (0)^2 dx$

Volume = $\pi \int\limits^2_1 ( e^{x-1)^2 dx$

Volume = $\pi \int\limits^2_1 e^{2x-2}dx$

Volume = $\frac{\pi }{e^2} \int\limits^2_1 e^{2x}dx$

Volume = $\frac{\pi }{e^2} [\frac{e^{2x}}{2}]^2_1$

Volume = $\frac{\pi }{2e^2} [e^4 - e^2 ]$

Volume = $\frac{\pi }{2} [e^2 - 1 ]$ or 10.036

Therefore, the volume of the solid generated by revolving the region bounded by the graphs of the equations about the x-axis is;

$\frac{\pi }{2} [e^2 - 1 ]$ or 10.036

3 0

3 years ago

The range of three numbers is 17.the mode is 24 and the median is 24 ? What are the three numbers?

vivado [14]

The answer is 7, 24, 24.

The median is the middle number.

The mode is the number which occurs the most.

The range is the difference between the largest number and the smallest number.

Since 24 is the middle number, it is the median.

Since 24 is also the number which occurs the most often, it is also the mode.

To find range, do 24-7 which is equal to seventeen.

Hope it helps

3 0

2 years ago

Read 2 more answers