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pentagon [3]
3 years ago
7

20n^2t / 5n^t what is the answer? and y?

Mathematics
1 answer:
Anastasy [175]3 years ago
6 0
I hope this helps you

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What is 4b+4=22 linear equation​
o-na [289]

Answer:

b=4.5

Step-by-step explanation:

4b+4=22

4b=22-4

4b=18

b=4.5

4 0
3 years ago
The graph of f ′ (x), the derivative of f(x), is continuous for all x and consists of five line segments as shown below. Given f
Nataliya [291]

The maxima of f(x) occur at its critical points, where f '(x) is zero or undefined. We're given f '(x) is continuous, so we only care about the first case. Looking at the plot, we see that f '(x) = 0 when x = -4, x = 0, and x = 5.

Notice that f '(x) ≥ 0 for all x in the interval [0, 5]. This means f(x) is strictly increasing, and so the absolute maximum of f(x) over [0, 5] occurs at x = 5.

By the fundamental theorem of calculus,

\displaystyle f(5) = f(0) + \int_0^5 f'(x) \, dx

The definite integral corresponds to the area of a trapezoid with height 2 and "bases" of length 5 and 2, so

\displaystyle \int_0^5 f'(x) \, dx = \frac{5+2}2 \times 2 = 7

\implies \max\{f(x) \mid 0\le x \le5\} = f(5) = f(0) + 7 = \boxed{13}

8 0
2 years ago
B) T is due north of C, calculate the bearing of B from C
choli [55]

Answer:

(a) 52°

(b) 322°

Step-by-step explanation:

(a) The details of the circle are;

The diameter of the circle = AOC

The center of the circle = Point O

The point the line AT cuts the circle = Point B

The point the tangent PT touches the circle = Point C

Angle ∠COB = 76°

We have that angle AOB and angle COB are supplementary angles, therefore;

∠AOB + ∠COB = 180°

∠AOB = 180° - ∠COB

∴ ∠AOB = 180° - 76° = 104°

∠AOB = 104°

OA = OB = The radius of the circle

Therefore, ΔAOB  =  An isosceles triangle

∠OAB = ∠OBA by base angles of an isosceles triangle are equal

∠AOB + ∠OAB + ∠OBA = 180° by angle summation property

∴ ∠AOB + ∠OAB + ∠OBA = ∠AOB + ∠OAB + ∠OAB = ∠AOB + 2×∠OAB = 180°

∠OAB = (180° - ∠AOB)/2

∴ ∠OAB = (180° - 104°)/2 = 38°

∠TAC = ∠OAB = 38° by reflexive property

AOC is perpendicular to tangent PT at point C, by tangent to a circle property, therefore;

∠TCA = 90° and ΔTCA = A right triangle

∠TAC + ∠ATC + ∠TCA = 180° by angle sum property

∠ATC = 180° - (∠TAC + ∠TCA)

∴ ∠ATC = 180° - (38° + 90°) = 52°

Angle ATC = 52°

(b) In ΔABC, ∠ABC = Angle subtended by the diameter = 90°

∴ ΔABC = A right triangle

∠ABC and ∠TBC are supplementary angles, therefore;

∠ABC + ∠TBC = 180°

∠TBC = 180° - ∠ABC

∴ ∠TBC = 180° - 90° = 90°

∠TCB = 180° - (∠TBC + ∠ATC)

∴ ∠TCB = 180° - (90° + 52°) = 38°

The bearing of B from C = (360° - 38°) = 322°.

7 0
2 years ago
How to solve axis of symmetry for quadratics
rjkz [21]
-b/2a will give you the axis of symmetry.
5 0
3 years ago
Read 2 more answers
I need help with this
madreJ [45]

Answer:

x = 14

Step-by-step explanation:

The sum of the 3 angles in a triangle = 180°, hence

4x + 14 + 2x + 15 + 5x - 3 = 180

11x + 26 = 180 ( subtract 26 from both sides )

11x = 154 ( divide both sides by 11 )

x = 14

7 0
2 years ago
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