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notsponge [240]
3 years ago
15

Find the minimum maximum and domain and range of the functions h(x) = -x² + 4x - 2​

Mathematics
1 answer:
Svet_ta [14]3 years ago
3 0

Domain (-infinity,infinity)

Min(?

max(2,2)

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Two angles are supplementary if their sum is 180 degrees. One angle measures two times the measure of a smaller angle. If x repr
Serjik [45]
The smaller angle is 60 degrees and the larger angle is 120 degrees.
x+2x = 180, 3x=180, so x = 60 and 2x = 120.
8 0
3 years ago
Find an equation of the line that passes through the points (1,2) and (2,3).
klemol [59]

Answer:

y=x+1

Step-by-step explanation:

m=(y2-y1)/(x2-x1)

m=(3-2)/(2-1)

m=1/1

m=1

y-y1=m(x-x1)

y-2=1(x-1)

y-2=x-1

y=x-1+2

y=x+1

6 0
3 years ago
Read 2 more answers
Point p is the midpoint of DE. DP =3x + 2 and DE = 10x - 12. What is the length of DP
shepuryov [24]
If P <span>is the midpoint of DE then 
DE = 2(DP)
10x - 12 = 2(3x + 2)
10x - 12 = 6x + 4
10x - 6x = 4 + 12
4x = 16
x = 16/4
x = 4

DP = 3x + 2 = 3 * 4 + 2 = 12 + 2 = 14 units</span>
6 0
3 years ago
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To find the equation of a line, we need the slope of the line and a point on the line. Since we are requested to find the equati
murzikaleks [220]

Answer:

x-4y+4=0

f(x)=\sqrt x and x=4

Step-by-step explanation:

We are given that a curve

y=\sqrt x

We have to find the equation of tangent at point (4,2) on the given curve.

Let y=f(x)

Differentiate w.r.t x

f'(x)=\frac{dy}{dx}=\frac{1}{2\sqrt x}

By using the formula \frac{d(\sqrt x)}{dx}=\frac{1}{2\sqrt x}

Substitute x=4

Slope of tangent

m=f'(x)=\frac{1}{2\sqrt 4}=\frac{1}{2\times 2}=\frac{1}{4}

In given question

m=\lim_{x\rightarrow a}\frac{f(x)-f(a)}{x-a}

\frac{1}{4}=\lim_{x\rightarrow 4}\frac{f(x)-f(4)}{x-4}

By comparing we get a=4

Point-slope form

y-y_1=m(x-x_1)

Using the formula

The equation of tangent at point (4,2)

y-2=\frac{1}{4}(x-4)

4y-8=x-4

x-4y-4+8=0

x-4y+4=0

8 0
3 years ago
Simplify x2 + 12x + 35/3x + 15
kodGreya [7K]

x+7
——— is the simplified version
3
7 0
3 years ago
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