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3 years ago

Will mark Brainliest. In two or more complete sentences, explain how to solve the cube root equation, ∛x-1+3=0.

Mathematics

1 answer:

inn [45]3 years ago

8 0

∛x-1+3 = 0

Add -3 to both sides to give ∛x-1 = -3.

Now cube both sides to give x - 1 = -27. Add 1 to both sides to find x, which is -26.

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Round off 174 to the nearest hundred

shepuryov [24]

Answer: 200

Step-by-step explanation:

174 rounded to the nearest hundred is:

200

When rounding to the nearest hundred, like we did with 174 above, we use the following rules:

A) We round the number up to the nearest hundred if the last two digits in the number are 50 or above.

B) We round the number down to the nearest hundred if the last two digits in the number are 49 or below.

C) If the last two digits are 00, then we do not have to do any rounding, because it is already to the hundred.

pls mark brainliest!

4 0

3 years ago

Read 2 more answers

True or False. The area of the trapezoid is 24 sq. inches. (A = 1/2h(b1 + b2) or decompose the figure.

Amanda [17]

Answer:

false

Step-by-step explanation:

7 0

3 years ago

Can someone answer this question please answer it correctly if it’s corect I will mark you brainliest

Mamont248 [21]

Answer:

Independent

Step-by-step explanation:

4 0

3 years ago

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In quadratic drag problem, the deceleration is proportional to the square of velocity

Mars2501 [29]

Part A

Given that $a= \frac{dv}{dt} =-kv^2$

Then,

$\int dv= -kv^2\int dt \\ \\ \Rightarrow v(t)=-kv^2t+c$

For $v(0)=v_0$ , then

$v(0)=-kv^2(0)+c=v_0 \\ \\ \Rightarrow c=v_0$

Thus, $v(t)=-kv(t)^2t+v_0$

For $v(t)= \frac{1}{2} v_0$ , we have

$\frac{1}{2} v_0=-k\left( \frac{1}{2} v_0\right)^2t+v_0 \\ \\ \Rightarrow \frac{1}{4} kv_0^2t=v_0- \frac{1}{2} v_0= \frac{1}{2} v_0 \\ \\ \Rightarrow kv_0t=2 \\ \\ \Rightarrow t= \frac{2}{kv_0}$

Part B

Recall that from part A,

$v(t)= \frac{dx}{dt} =-kv^2t+v_0 \\ \\ \Rightarrow dx=-kv^2tdt+v_0dt \\ \\ \Rightarrow\int dx=-kv^2\int tdt+v_0\int dt+a \\ \\ \Rightarrow x=- \frac{1}{2} kv^2t^2+v_0t+a$

Now, at initial position, t = 0 and $v=v_0$ , thus we have

$x=a$

and when the velocity drops to half its value, $v= \frac{1}{2} v_0$ and $t= \frac{2}{kv_0}$

Thus,

$x=- \frac{1}{2} k\left( \frac{1}{2} v_0\right)^2\left( \frac{2}{kv_0} \right)^2+v_0\left( \frac{2}{kv_0} \right)+a \\ \\ =- \frac{1}{2k} + \frac{2}{k} +a$

Thus, the distance the particle moved from its initial position to when its velocity drops to half its initial value is given by

$- \frac{1}{2k} + \frac{2}{k} +a-a \\ \\ = \frac{2}{k} - \frac{1}{2k} = \frac{3}{2k}$

7 0

3 years ago

In a servey of 80 people ,it was found that 60 liked oranges only and 10 liked both oranges and apples.The nuber of people who l

I am Lyosha [343]

Answer:

10 people didn't answer the servey

Step-by-step explanation:

servey=80 people

70= answered the servey

so we should take the difference:-

80–70= 10

Hope I got the right answer!

4 0

3 years ago