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Lena [83]
3 years ago
6

The table shows the lowest temperatures recorded on five days in April 2020, but the value for 4/23/20 has been lost . if X Is t

he lost value , and we know that X was the middle value of the five days , which choice could be the value of X?
A. -1.9 °C
B. -0.9 °C
C. -0.1 °C
D. -0.4 °C

Mathematics
2 answers:
fenix001 [56]3 years ago
8 0
The answer is d -0.4 c
Snowcat [4.5K]3 years ago
6 0
D
I think but it makes most SENCE
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Find the break-Even point for the following fuctions c(x)=10x and r(x)=210-4x
AlekseyPX

Answer: The break-even point is x = 15

Step-by-step explanation:

We have the equations:

c(x) = 10*x

r(x) = 210 - 4*x

I suppose that c(x) is the cost equation, and r(x) is the revenue equation.

"break-even" means that you don't lose nor win anything, then the difference between the revenue and the cost is 0.

Then we need to solve:

r(x) - c(x) = 0

(210 - 4*x) - (10*x) = 0

210 - 4*x - 10*x = 0

210 - 14*x = 0

210 = 14*x

210/14 = x

15

This means that for x = 15, you break even.

4 0
3 years ago
A. 2√5<br>B. 2√13<br>C. 5<br>D. 2​
sasho [114]

Answer:

A. 2√5

Step-by-step explanation:

hypotenuse (h) = 6

base (b) = 4

perpendicular (p) = ?

We know by using Pythagoras theorem

p = √h² - b²

= √ 6² - 4²

= √ 20

= 2√5

Hope it will help :)

8 0
3 years ago
Compare the process of solving |x – 1| + 1 &lt; 15 to that of solving |x – 1| + 1 &gt; 15.
kicyunya [14]
In the first case we'd subtract 1 from both sides, obtaining |x-1|<14.

In the second case we'd also subtract 1 from both sides, and would obtain
 |x-1|>14.

What would the graphs look like?

In the first case, the graph would be on the x-axis with "center" at x=1.  From this center count 14 units to the right, and then place a circle around that location (which would be at x=15).  Next, count 14 units to the left of this center, and place a circle around that location (which would be -13).  Draw a line segment connecting the two circles.  Notice that all of the solutions are between -13 and +15, not including these endpoints.

In the second case, x has to be greater than 15 or less than -13.  Draw an arrow from x=1 to the left, and then draw a separate arrow from 15 to the right.  None of the values in between are solutions.

4 0
3 years ago
Read 2 more answers
Is the answer no solution?
Nutka1998 [239]
Cross multiply and solve
1(y^2-9)=6(y-3)

y^2-9= 6y-18
y^2-6y-9+18=0
y^2-6y+9=0

(y-3)(y-3)=0
(y-3)=0
y=3

When you plug y in however, it makes the denominator 0, making both fractions undefined.

Final answer: No real solution
4 0
3 years ago
Solve for a in the figure shown.
aleksandrvk [35]

Answer: 58 is the answer

hope this help and here my steps use a mark or a pin

and line it with the line if it's wrong sorry for the wrong

answer

7 0
3 years ago
Read 2 more answers
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