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sesenic [268]
2 years ago
10

Solve the inequality 6h−5(h−1)≤7h−11 and write the solution in interval notation. Use improper fractions if necessary.

Mathematics
1 answer:
Kobotan [32]2 years ago
6 0

Answer:

h \geq 2\frac{2}[3}

Step-by-step explanation:

We solve the inequality similarly to how we would solve an equalitu.

6h - 5(h-1) \leq 7h - 11

6h - 5h + 5 \leq 7h - 11

h - 7h \leq -11 - 5/

-6h \leq -16

Multiplying everything by -1

6h \geq 16

Simplifying by 2

3h \geq 8

h \geq \frac{8}{3}

8 divided by 3 is 2 with rest two. So as a improper fraction, the answer is:

h \geq 2\frac{2}[3}

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saveliy_v [14]

We have been given an equation ae^{ct}=d. We are asked to solve the equation for t.

First of all, we will divide both sides of equation by a.

\frac{ae^{ct}}{a}=\frac{d}{a}

e^{ct}=\frac{d}{a}

Now we will take natural log on both sides.

\text{ln}(e^{ct})=\text{ln}(\frac{d}{a})

Using natural log property \text{ln}(a^b)=b\cdot \text{ln}(a), we will get:

ct\cdot \text{ln}(e)=\text{ln}(\frac{d}{a})

We know that \text{ln}(e)=1, so we will get:

ct\cdot 1=\text{ln}(\frac{d}{a})

ct=\text{ln}(\frac{d}{a})

Now we will divide both sides by c as:

\frac{ct}{c}=\frac{\text{ln}(\frac{d}{a})}{c}

t=\frac{\text{ln}(\frac{d}{a})}{c}

Therefore, our solution would be t=\frac{\text{ln}(\frac{d}{a})}{c}.

5 0
2 years ago
On a coordinate plane, a triangle has points (negative 5, 1), (2, 1), (2, negative 1).
daser333 [38]

Answer:

Step-by-step explanation:

Given a triangle has points:

(-5,1),(2,1), (2,-1)

Let us label the points:

A(2,1),

B(-5,1) and  

C(2,-1)

To find:

Distance between (−5, 1) and (2, −1) i.e. BC.

Horizontal leg AB and

Vertical leg, AC.

Solution:

Please refer to the attached diagram for the labeling of the points on xy coordinate plane.

We can simply use Distance formula here, to find the distance between two coordinates.

Distance formula :

D = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

For BC:

x_2 = 2\\x_1 = -5\\y_2 = -1\\y_1 = 1

BC = \sqrt{(2--5)^2+(-1-1)^2} = \sqrt{63}

Horizontal leg, AC:

x_2 = -5\\x_1 = 2\\y_2 = 1\\y_1 = 1

AC = \sqrt{(2-(-5))^2+(1-1)^2} = 7

Vertical Leg,  AB:

x_2 = 2\\x_1 = 2\\y_2 = -1\\y_1 = 1

AB = \sqrt{(2-2)^2+(-1-1)^2} = 2

5 0
2 years ago
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7 0
2 years ago
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dexar [7]

Answer:

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Hence,

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3 0
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