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Vanyuwa [196]
3 years ago
15

Can someone help me on this?

Mathematics
1 answer:
sineoko [7]3 years ago
5 0
D= -3,-2
C= 1,-2
B= -1,4
A= -3,0
You might be interested in
Hi could you please help me
Eddi Din [679]

Answer:

if their numbers is 96

and ratio is 9: 15 then

9×4: 15×4

36+60=96

I hope u understand

8 0
3 years ago
Read 2 more answers
Which statements are true about the interquartile range? Select all that apply. Subtract the lowest and highest values to find t
katen-ka-za [31]

Answer:

The correct options are:

Interquartile ranges are not significantly impacted by outliers.

Lower and upper quartiles are needed to find the interquartile range.

The data values should be listed in order before trying to find the interquartile range.


The option Subtract the lowest and highest values to find the interquartile range is incorrect because the difference between lowest and highest values will give us range.

The option A small interquartile range means the data is spread far away from the median is incorrect because a small interquartile means data is nor spread far away from the median

6 0
3 years ago
Read 2 more answers
Solve x – 2y = 6 3x – 6y = 18
VashaNatasha [74]
X will equal 0 and y will equal -3
6 0
3 years ago
PLEASE TELL ME ITS ANSWER
Dafna11 [192]

Step-by-step explanation:

6) x+y=60

x/y=7/5------(2)

x=(7/5)y

(7/5)y + y =60

(12/5)y=60

y=60(5/12)

y= 25

x=(7/5)y=(7/5)×25 =35

the two numbers are 35 and 25

7)distance walked by A = 4t

distance walked by B = 5t

4t+5t=27

9t=27

t=27/9=3hours

7 0
3 years ago
Please help me right away.
statuscvo [17]

So, with rational equations, we have three different cases. If the numerator has degree m and the denominator degree n, if m>n, the rational equation has an oblique(slant) asymptote. If m=n, the asymptote is the quotient of the leading coefficient of the numerator divided by the leading coefficient of the denominator. If m<n, the rational equation has an asymptote at 0. Since m>n in this problem, we must perform polynomial division.

\frac{x^3+2x-8}{x^2+x}= 3x-\frac{x+8}{x^2+x}

Since the remainder tends to 0 as it approaches infinity, we have a slant asymptote at y=3x.

3 0
3 years ago
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