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2 years ago

Solve 7e^(2x)-5>58. Round to the nearest hundredth.

Mathematics

1 answer:

7nadin3 [17]2 years ago

5 0

Answer:

$x\gtrapprox1.1$

Step-by-step explanation:

$7e^{2x}-5>58$

$7e^{2x}>63$

$e^{2x}>9$

$2x>ln(9)$

$x>\frac{ln(9)}{2}$

$x\gtrapprox1.1$

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Need help finding a constant k at which the function is continuous everywhere.

saveliy_v [14]

Answer:

k = -6/35

Step-by-step explanation:

To make the function continuous

kx^2 = x+k

These must be equal where the function is defined for two different intervals

This is at the point x=-6 so let x=-6

k(-6)^2 = -6+k

36k = -6+k

Subtract k from each side

36k-k = -6+k-k

35k = -6

Divide by 35

35k/35 = -6/35

k = -6/35

3 0

3 years ago

Read 2 more answers

A wheel of diameter 50.0 cm starts from rest and rotates with a constant angular acceleration of 2.00 rad/s^2. At the instant th

Kruka [31]

R = 25 cm = 0.25 m
α = 2.00 rad /s²
T(heta) = 4 π
ω = T(heta)/ t
ω = α t
α t = 4 π / t
t ² = 2 π
t = √ ( 2π ) = 2.5 s
ω = α t = 2 · 2.5 = 5 rad/s
v = ω r = 5 · 0.25 = 1.25 m/s
(1) a (rad) = ω² r = 25 · 0.25 = 6.25 m/s²
(2) a (rad) = v² / r = ( 1.25 )² / 0.25 = 0.15625 / 0.25 = 6.25 m/s²

8 0

3 years ago

Which set of parametric equations over the interval 0 ≤ t ≤ 1 defines a line segment with initial point (–5, 3) and terminal poi

Ksivusya [100]

Given:

A line segment with initial point (–5, 3) and terminal point (1, –6).

To find:

The set of parametric equations over the interval 0 ≤ t ≤ 1 which defines the given line segment.

Solution:

Initial point is (–5, 3). So,

$x(0)=-5,y(0)=3$