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Ilia_Sergeevich [38]
3 years ago
10

What is the range of the function represented by these ordered pairs? {(–2, 1), (0, 0), (3, –1), (–1, 7), (5, 7)}

Mathematics
1 answer:
qaws [65]3 years ago
5 0
The range is the set of y coordinates (or outputs). For any point (x,y) the y coordinate is the second number listed. For example, with (8,7) the y coordinate is 7.

So we simply list all of the y coordinates of each ordered pair. Those values are: 1, 0, -1, 7, 7

Let's order them as you would see on a number line

-1, 0, 1, 7, 7

then let's toss out the extra 7 to get

-1, 0, 1, 7

So the range is the set {-1, 0, 1, 7}
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Find the sum of the first 20 terms of an arithmetic progression of which the third term is 55 and the last term is -98
ryzh [129]

The sum of first 20 arithmetic series S_{20}=\frac{-3475}{16}

Given:

Arithmetic series for 3rd term is 55

Arithmetic series for 7th term is -98

To find:

The sum of first 20 Arithmetic series

<u>Step by Step Explanation: </u>

Solution:

Formula for calculating arithmetic series

Arithmetic series=a+(n-1) d

Arithmetic series for 3rd term a_{3}=a_{1}+(3-1) d

a_{1}+2 d=55

Arithmetic series for 19th term is

a_{19}=a_{1}+(19-1) d=-98

a_{19}+18 d=-98

Subtracting equation 2 from 1

\left[a_{19}+18 d=-98\right]+\left[a_{1}+2 d=55\right]

16d=-98-55

16d=-153

d=\frac{-153}{16}

Also we knowa_{1}+2 d=55

a_{1}+2(-153 / 16)=55

a_{1}+(-153 / 8)=55

a_{1}=55+(153 / 8)

a_{1}=440+153 / 8

a_{1}=553 / 8

First 20 terms of an AP  

a_{n=} a_{1}+(n-1) d

a_{20}=553 / 8+19(-153 / 16)

a_{20}=553 / 8+19(-153 / 16)

a_{20}=\{553 * 2 / 8 * 2\}-2907 / 16

a_{20}=[1106 / 16]-[2907 / 16]

a_{20}=-1801 / 16

Sum of 20 Arithmetic series is

S_{n}=n\left(a_{1}+a_{n}\right) / 2

Substitute the known values in the above equation we get

S_{20}=\left[\frac{20\left(\left(\frac{558}{8}\right)+\left(\frac{-1801}{16}\right)\right)}{2}\right]

S_{20}=\left[\frac{\left.20\left(\frac{1106}{16}\right)+\left(\frac{-1801}{16}\right)\right)}{2}\right]

S_{20}=10 \frac{(-695 / 16)}{2}

S_{20}=5\left[\frac{-695}{16}\right]

S_{20}=\frac{-3475}{16}

Result:

Thus the sum of first 20 terms in an arithmetic series is S_{20}=\frac{-3475}{16}

7 0
3 years ago
Please help! Prove: Ray E B bisects ∠AEC.
Norma-Jean [14]

Answer:  Angle Bisector Theorem states that if a ray bisect an angle of a triangle,then it divides the opposite side into segments that are proportional to the other two side.

if this helped pls mark brainliest.

3 0
3 years ago
Read 2 more answers
Simplify fully<br> A) <br> X^7 / X^3 <br><br><br> B)<br> Y^5 / Y^4
DENIUS [597]

Hi,

\frac{x^{a} }{x^{b} } =x^{a-b}

a)

\frac{x^{7} }{x^{3} }= x^{7-3} = x^{4}

b)

\frac{y^{5} }{y^{4} } = y^{5-4} = y^{1} = y

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8 0
2 years ago
Please show work<br><br>I'll give brainliest ​
timurjin [86]

Answer:

7

Step-by-step explanation:

a right angle is 90 degrees so

90=76+2x        subtract 76 from both sides

-76 -76

________

14=2x                divide 2 from both sides

_   ___

2     2

7=x

Check:

76+(2*7)=90 correct

                                               

8 0
3 years ago
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Write the equation of the circle with center (3,2) and with (9,3) being a point on the circle
miss Akunina [59]

Answer:

(x-3)^2 + (y-2)^2 = 37

Step-by-step explanation:

To write the equation of a circle, use the formula (x-h)^2+(y-k)^2 = r^2 where the center of the circle is (h, k).

This means the equation is (x-3)^2 + (y-2)^2 = r^2.

Find the radius r by finding the distance between (3,2) and (9,3) using the distance formula.

d = \sqrt{(9-3)^2 + (3-2)^2} =\sqrt{6^2 + 1^2} =\sqrt{36+1} =\sqrt{37}

Since the radius is √37 and therefore r^2 = 37.

The equation is (x-3)^2 + (y-2)^2 = 37.

6 0
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