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2 years ago

What is the answer to 3k-(-4)=22

Mathematics

2 answers:

mote1985 [20]2 years ago

4 0

Step-by-step explanation:

This will result to 3k+4=22 because minus times minus is plus

Therefore, 3k=22-4, when +4 crosses over the equals to sign it becomes -4

=> 3k=18

=> Divide both sides by 3 to get k

=> k=6

natali 33 [55]2 years ago

4 0

Answer:

The answer is k = 6

Step-by-step explanation:

3k - (-4) = 22

When two negative are close together, they become a positive. So our new equation should look like this:

3k + 4 = 22

Now, lets solve:

3k + 4 = 22

- 4 - 4

3k = 18

/3 /3

k = 6

That's how we got our final answer k = 6

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Find the volume of the triangular prism. 3cm 10cm 7cm

asambeis [7]

Answer:

volume of the triangular prism = 105cm³

Step-by-step explanation:

to find the volume of a triangular prism , given the three sides to be 3cm, 10cm, 7cm,

so, recall the formulae for calculating the volume of a triangular prisim

volume of the triangular prism = 1/2 x a x b x c

volume of the triangular prism = 1/2 x 3cm x 10cm x 7cm

volume of the triangular prism = 3cm x 5cm x 7cm

volume of the triangular prism = 105cm³

therefore the volume of the trangular prism is evaluated to be equals to 105cm³

7 0

3 years ago

64x-16 I need 3 equivalent expressions

Daniel [21]

Answer:

8 (8x - 2)

(64) x X - 16

4 (16 x- 4)

Step-by-step explanation:

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2 years ago

A cooler has a temperature of 32 degrees Fahrenheit. A bottled drink is placed in the cooler with an initial temperature of

FinnZ [79.3K]

Answer:

$k \approx 0.44$

Step-by-step explanation:

Given function:

$f(t) = (ce)^{-kt}+32$

As per question statement:

Initial temperature of bottle is 70 $^\circ F$ .

i.e. when time = 0 minutes, f(t) = 70 $^\circ F$

$70 = ce^{-k\times 0}+32\\\Rightarrow 38 = ce^{0}\\\Rightarrow c = 38$

After t = 3, f(t) = 42 $^\circ F$

$42 = 38 \times e^{-k\times 3}+32\\\Rightarrow 42-32 = 38 \times e^{-3k} \\\Rightarrow 10 = 38 \times e^{-3k} \\\Rightarrow e^{3k} = \dfrac{38}{10}\\\Rightarrow e^{3k} = 3.8\\\\\text{Taking } log_e \text{both the sides:}\\\\\Rightarrow log_e {e^3k} = log_e {3.8}\\\Rightarrow 3k \times log_ee=log_e {3.8} (\because log_pq^r=r \times log_pq)\\\Rightarrow 3k \times 1=log_e {3.8}\\\Rightarrow 3k = 1.34\\\Rightarrow k = \dfrac{1.34}{3}\\\Rightarrow k \approx 0.44$