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VLD [36.1K]
3 years ago
5

What slope does a vertical line have

Mathematics
2 answers:
Anika [276]3 years ago
5 0
The slope on a vertical line will be 0

alexandr1967 [171]3 years ago
4 0
Slope is 0 on a vertical line
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Each of the students im Romi's class raised at least $25 during the jump-a-thon. What is the least amount of money the class rai
EastWind [94]
If each of the 28 students made at least $25, you would multiply 28 and 25 together to obtain the least amount of money the class raised. That gets, 28x25 = 700. The class made at least $700.
8 0
3 years ago
Solve for x. (1 point) −3x + 2b > 8 x < the quantity negative 2 times b plus 8 all over negative 3 x > the quantity neg
masha68 [24]

Given:

-3x+2b>8

To find:

The value of x.

Solution:

We have,

-3x+2b>8

To find the value of x, we need to isolate x on one side.

Subtract 2b from both sides.

-3x>8-2b

Divide both sides by -3. On multiplying or dividing an inequality by a negative number, we need to change the sign of inequality.

\dfrac{-3x}{-3}

x

The required inequality for x is x.

Therefore, the correct option is A.

8 0
3 years ago
Match the areas of the following squares with their correct side lengths.
9966 [12]

x=2

z=45

 d=6

hope this helps


6 0
3 years ago
A lake polluted by bacteria is treated with an antibacterial chemical. Aftertdays, thenumber N of bacteria per milliliter of wat
jeyben [28]

Answer:

N(1)=50 is a minimum

N(15)=4391.7 is a maximum

Step-by-step explanation:

<u>Extrema values of functions </u>

If the first and second derivative of a function f exists, then f'(a)=0 will produce values for a called critical points. If a is a critical point and f''(a) is negative, then x=a is a local maximum, if f''(a) is positive, then x=a is a local minimum.  

We are given a function (corrected)

N(t) = 20(t^2-lnt^2)+ 30

N(t) = 20(t^2-2lnt)+ 30

(a)

First, we take its derivative

N'(t) = 20(2t-\frac{2}{t})

Solve N'(t)=0

20(2t-\frac{2}{t})=0

Simplifying

2t^2-2=0

Solving for t

t=1\ ,t=-1

Only t=1 belongs to the valid interval 1\leqslant t\leqslant 15

Taking the second derivative

N''(t) = 20(2+\frac{2}{t^2})

Which is always positive, so t=1 is a minimum

(b)

N(1)=20(1^2-2ln1)+ 30

N(1)=50 is a minimum

(c) Since no local maximum can be found, we test for the endpoints. t=1 was already determined as a minimum, we take t=15

(d)

N(15)=20(15^2-2ln15)+ 30

N(15)=4391.7 is a maximum

7 0
3 years ago
Solve y=x2 +7 fоr x.<br> ОА. x = t /у +7<br> ОВ. x= y +7<br> Ос. x=y-7<br> OD. х = + y -7
elena55 [62]

Answer:

x = ± \sqrt{y-7}

Step-by-step explanation:

Given

y = x² + 7 ( subtract 7 from both sides )

y - 7 = x² ( take the square root of both sides, thus

x = ± \sqrt{y-7}

5 0
2 years ago
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