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2 years ago

13

Find the area of a circle with a

1 answer:

podryga [215]2 years ago

3 0

Answer:

A=pir²

Step-by-step explanation:

pi=3.14

r=15

r²=15²=225

A=225× 3.14= 706.5

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A patient’s systolic blood pressure drops from 210 mmHg to 100 mmHg in 10 minutes after they took nitroglycerin. Calculate the p

denpristay [2]

Answer:

7

Step-by-step explanation:

8 0

2 years ago

The point P(7, −2) lies on the curve y = 2/(6 − x). (a) If Q is the point (x, 2/(6 − x)), use your calculator to find the slope

NARA [144]

Answer:

a) (i) $m = 2.22$ , (ii) $m = 2$ , (iii) $m = 2$ , (iv) $m = 2$ , (v) $m = 1.82$ , (vi) $m = 2$ , (vii) $m = 2$ , (viii) $m = 2$ ; b) $m \approx 2$ ; c) The equation of the tangent line to curve at P (7, -2) is $y = 2\cdot x + 12$ .

Step-by-step explanation:

a) The slope of the secant line PQ is represented by the following definition of slope:

$m = \frac{\Delta y}{\Delta x} = \frac{y_{Q}-y_{P}}{x_{Q}-x_{P}}$

(i) $x_{Q} = 6.9$ :

$y_{Q} =\frac{2}{6-6.9}$

$y_{Q} = -2.222$

$m = \frac{-2.222 + 2}{6.9-7}$

$m = 2.22$

(ii) $x_{Q} = 6.99$

$y_{Q} =\frac{2}{6-6.99}$

$y_{Q} = -2.020$

$m = \frac{-2.020 + 2}{6.99-7}$

$m = 2$

(iii) $x_{Q} = 6.999$

$y_{Q} =\frac{2}{6-6.999}$

$y_{Q} = -2.002$

$m = \frac{-2.002 + 2}{6.999-7}$

$m = 2$

(iv) $x_{Q} = 6.9999$

$y_{Q} =\frac{2}{6-6.9999}$

$y_{Q} = -2.0002$

$m = \frac{-2.0002 + 2}{6.9999-7}$

$m = 2$

(v) $x_{Q} = 7.1$

$y_{Q} =\frac{2}{6-7.1}$

$y_{Q} = -1.818$

$m = \frac{-1.818 + 2}{7.1-7}$

$m = 1.82$

(vi) $x_{Q} = 7.01$

$y_{Q} =\frac{2}{6-7.01}$

$y_{Q} = -1.980$

$m = \frac{-1.980 + 2}{7.01-7}$

$m = 2$

(vii) $x_{Q} = 7.001$

$y_{Q} =\frac{2}{6-7.001}$

$y_{Q} = -1.998$

$m = \frac{-1.998 + 2}{7.001-7}$

$m = 2$

(viii) $x_{Q} = 7.0001$

$y_{Q} =\frac{2}{6-7.0001}$

$y_{Q} = -1.9998$

$m = \frac{-1.9998 + 2}{7.0001-7}$

$m = 2$

b) The slope at P (7,-2) can be estimated by using the following average:

$m \approx \frac{f(6.9999)+f(7.0001)}{2}$

$m \approx \frac{2+2}{2}$

$m \approx 2$

The slope of the tangent line to the curve at P(7, -2) is 2.

c) The equation of the tangent line is a first-order polynomial with the following characteristics:

$y = m\cdot x + b$

Where:

$x$ - Independent variable.

$y$ - Depedent variable.

$m$ - Slope.

$b$ - x-Intercept.

The slope was found in point (b) (m = 2). Besides, the point of tangency (7,-2) is known and value of x-Intercept can be obtained after clearing the respective variable:

$-2 = 2 \cdot 7 + b$

$b = -2 + 14$

$b = 12$

The equation of the tangent line to curve at P (7, -2) is $y = 2\cdot x + 12$ .

7 0

2 years ago

Help, please and thank you❤️

Tomtit [17]

Answer:

your answer should be 56.52 .

Step-by-step explanation:

this reason because is because radius is usually half of the diameter so just multiply 28.26 by 2

4 0

2 years ago

If f(x)=4x-5 and g(x)=3-2x find f(x)+g(x)

Akimi4 [234]

(fg)(x) = [f(x)][g(x)]
(fg)(x) = (4x - 5)(3) = 12x - 15

8 0

2 years ago

Find the values of the trigonometric functions of θ from the information given.

OLga [1]

Answer:

sin(θ) = -(4√15)/17
cos(θ) = 7/17 . . . . . . . given
tan(θ) = -(4√15)/7
csc(θ) = -(17√15)/60
sec(θ) = 17/7
cot(θ) = -(7√15)/60

Step-by-step explanation:

The relationship between sine and cosine is ...

sin² + cos² = 1

Solving for sine gives ...

sin = ±√(1 -cos²)

In this problem, we want the negative root.

sin(θ) = -√(1 -(7/17)²) = -√(240/289) = -(4√15)/17

tan(θ) = sin(θ)/cos(θ) = ((-4√15)/17)/(7/17) = -(4√15)/7

___

And the inverse functions are ...

sec(θ) = 1/cos(θ) = 17/7

csc(θ) = 1/sin(θ) = -17/(4√15) = -(17√15)/60

cot(θ) = 1/tan(θ) = -(7√15)/60

_____

Of course, you're aware that 1/√15 = (√15)/15.

3 0

2 years ago