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kompoz [17]
4 years ago
9

Which number is an irrational number?Help Plzzzzz​

Mathematics
2 answers:
xeze [42]4 years ago
8 0

Answer:

  _  

1.25 im pretty sure

Step-by-step explanation:

Aleonysh [2.5K]4 years ago
6 0

The square root of 15 is irrational.

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What is −10x + 1 + 7x = 37
seropon [69]

Answer:

x = -12

Step-by-step explanation:

−10x + 1 + 7x = 37

Combine like terms

-3x+1 = 37

Subtract 1 from each side

-3x +1-1 = 37-1

-3x = 36

Divide each side by -3

-3x/-3 = 36/-3

x = -12

8 0
4 years ago
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if ZA and ZB are supplementary angles. if mZA = (3x-23) and mZB= (2x-12) then find the measure of ZB​
Brilliant_brown [7]

Answer:

mZB= 74

Step-by-step explanation:

ZB+ZA=180

3X-23+2X-12=180

5X-35=180

5X=180+35

5X=215  (then divide both sides by 5 to x be left alone.)

X=43....so ZB=2(43)-12

                     =86-12=74

3 0
3 years ago
The sign of the product of –35 and –625 is .
harina [27]

Answer:

21875

Step-by-step explanation:

-35 x (-625) = 21875

7 0
3 years ago
Ill give brainlest pls help
Katarina [22]

Answer:

It is C

Step-by-step explanation:

To get the new coordinates you would move the shape up the y-axis 8 times, and right on the x-axis 8 times as well.

8 0
3 years ago
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A 4 meter tall bear is walking in the woods and sees honey flavoured chips strapped to a flashlight hanging from a tree 7 meters
otez555 [7]

Answer:

The length of the shadow is decreasing 8 metres per second.

Step-by-step explanation:

In the attached figure, we draw the situation of the bear approaching the tree.

We can use the Tales to make some calculations.

The proportion between the height of the bear (h) and his shadow (S) is equal to the proportion between the height of the flashlight (H) and the sum of the distance of the bear (d) and the bear shadow.

This can be written as:

\frac{h}{S} =\frac{H}{S+d}

If we clear S we have:

S*H=S*h+d*h\\\\(H-h)*S=d*h\\\\S=(\frac{h}{H-h}) d

We know that the distance is a function of time and it is reduced, as the bear approaches the tree, as a rate of 6 m/s.

We can express then the rate of variation of the shadow S as:

\Delta S=S_2-S_1=(\frac{h}{H-h}) (d_2-d_1)=(\frac{h}{H-h}) \Delta d\\\\\\\Delta d=-6\\\\H=7\\\\h=4\\\\\\\Delta S=(\frac{4}{7-4}) *(-6)=\frac{4}{3} *(-6)=-8

The length of the shadow is decreasing 8 metres per second.

6 0
3 years ago
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