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gtnhenbr [62]
3 years ago
13

Find the slope of the line.

Mathematics
2 answers:
Luda [366]3 years ago
6 0
(-2,5) (2,0)
Just use the slope formula to find slope.

m = (y2-y1)/(x2-x1)
m = (0-5)/(2+2)
m = -5/4

Hence, the slope of the line is -5/4.
VLD [36.1K]3 years ago
3 0

We can use the points (-2, 5) and (2, 0) to solve.

Slope formula: y2-y1/x2-x1

= 0-5/2-(-2)

= -5/4

Hope This Helped! Good Luck!

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William has a lemonade stand .Yesterday he made $17.55in lemonade sales and today he made $14.02 sales. How much does he have no
Zolol [24]

Answer: 31.57

Step-by-step explanation: 17.55 + 14.02 =31.57

5 0
3 years ago
X+x-z; use x=3 & z=4​
nadezda [96]

Answer:

2

Step-by-step explanation:

Order of operations is not as applicable in this situation because we're just dealing with addition and subtraction.  This is a simple case of plugging in values for the given expression.

x = 3

z = 4

So it becomes...

3 + 3 - 4

6 - 4

2

So with the given expression of x + x - z and the given values of x = 3 and z = 4, the expression should sum to: 2

6 0
3 years ago
At what point does the curve have maximum curvature? Y = 4ex (x, y) = what happens to the curvature as x → ∞? Κ(x) approaches as
MAXImum [283]

<u>Answer-</u>

At x= \frac{1}{2304e^4-16e^2} the curve has maximum curvature.

<u>Solution-</u>

The formula for curvature =

K(x)=\frac{{y}''}{(1+({y}')^2)^{\frac{3}{2}}}

Here,

y=4e^{x}

Then,

{y}' = 4e^{x} \ and \ {y}''=4e^{x}

Putting the values,

K(x)=\frac{{4e^{x}}}{(1+(4e^{x})^2)^{\frac{3}{2}}} = \frac{{4e^{x}}}{(1+16e^{2x})^{\frac{3}{2}}}

Now, in order to get the max curvature value, we have to calculate the first derivative of this function and then to get where its value is max, we have to equate it to 0.

 {k}'(x) = \frac{(1+16e^{2x})^{\frac{3}{2} } (4e^{x})-(4e^{x})(\frac{3}{2}(1+e^{2x})^{\frac{1}{2}})(32e^{2x})}{(1+16e^{2x} )^{2}}

Now, equating this to 0

(1+16e^{2x})^{\frac{3}{2} } (4e^{x})-(4e^{x})(\frac{3}{2}(1+e^{2x})^{\frac{1}{2}})(32e^{2x}) =0

\Rightarrow (1+16e^{2x})^{\frac{3}{2}}-(\frac{3}{2}(1+e^{2x})^{\frac{1}{2}})(32e^{2x})

\Rightarrow (1+16e^{2x})^{\frac{3}{2}}=(\frac{3}{2}(1+e^{2x})^{\frac{1}{2}})(32e^{2x})

\Rightarrow (1+16e^{2x})^{\frac{1}{2}}=48e^{2x}

\Rightarrow (1+16e^{2x})}=48^2e^{2x}=2304e^{2x}

\Rightarrow 2304e^{2x}-16e^{2x}-1=0

Solving this eq,

we get x= \frac{1}{2304e^4-16e^2}

∴ At  x= \frac{1}{2304e^4-16e^2} the curvature is maximum.




6 0
3 years ago
In triangle MPQ points N and R are located on sides MP and QP with NR drawn. which set of measurements below would justify that
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6 0
2 years ago
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kap26 [50]

Answer:

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Step-by-step explanation:

Confidence Interval = mean + or - Error margin (E)

mean = 37

sd = 10

n = 44

degree of freedom = n - 1 = 44 - 1 = 43

t-value corresponding to 43 degrees of freedom and 95% confidence level is 2.0165

E = t × sd/√n = 2.0165×10/√44 = 3.04

Lower limit = mean - E = 37 - 3.04 = 33.96

Upper limit = mean + E = 37 + 3.04 = 40.04

95% confidence interval is between 33.96 and 40.04

7 0
3 years ago
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