All categories

2 years ago

Evaluate (a + b)3 – 12 when a = 2 and b = 1.

Mathematics

1 answer:

dimulka [17.4K]2 years ago

4 0

Answer:

the answer is -3

Step-by-step explanation:

(2+1)*3-12 is -3

You might be interested in

Savings are what percentage of monthly net income? Round your answer to the nearest whole percent.

cricket20 [7]

To find the percent of the monthly income that is savings, you will divide the amount saved by the total amount of income.

300/1387 = 0.216

The approximate percentage of income that is savings is 22%.

7 0

2 years ago

Which is greater? 4 tenths and 2 hundredths or 2 tenths and 4 hundredths

Orlov [11]

4 tenths and 2 hundredths is written .42

2 tenths and 4 hundreds is written .24

.42 is greater than .24

.42 > .24

Hope this helps.

8 0

3 years ago

Find the components of the vertical force Bold Upper FFequals=left angle 0 comma negative 10 right angle0,−10 in the directions

quester [9]

Solution :

Let $v_0$ be the unit vector in the direction parallel to the plane and let $F_1$ be the component of F in the direction of $v_0$ and $F_2$ be the component normal to $v_0$ .

Since, $|v_0| = 1,$

$(v_0)_x=\cos 60^\circ= \frac{1}{2}$

$(v_0)_y=\sin 60^\circ= \frac{\sqrt 3}{2}$

Therefore, $v_0 = \left$

From figure,

$|F_1|= |F| \cos 30^\circ = 10 \times \frac{\sqrt 3}{2} = 5 \sqrt3$

We know that the direction of $F_1$ is opposite of the direction of $v_0$ , so we have

$F_1 = -5\sqrt3 v_0$

$=-5\sqrt3 \left$

$= \left$

The unit vector in the direction normal to the plane, $v_1$ has components :

$(v_1)_x= \cos 30^\circ = \frac{\sqrt3}{2}$

$(v_1)_y= -\sin 30^\circ =- \frac{1}{2}$

Therefore, $v_1=\left< \frac{\sqrt3}{2}, -\frac{1}{2} \right>$

From figure,

$|F_2 | = |F| \sin 30^\circ = 10 \times \frac{1}{2} = 5$

∴ $F_2 = 5v_1 = 5 \left< \frac{\sqrt3}{2}, - \frac{1}{2} \right>$

$=\left$

Therefore,

$F_1+F_2 = \left< -\frac{5\sqrt3}{2}, -\frac{15}{2} \right> + \left< \frac{5 \sqrt3}{2}, -\frac{5}{2} \right>$

$= = F$

3 0

2 years ago

Find a vector parametric equation r⃗(t) for the line through the points P=(1,0,−2) and Q=(1,5,1) for each of the given condition

const2013 [10]

So the question ask to find and calculate the vector parametric equation r(t) for the line through the points P=(1,0,-2) and Q(1,5,1) for each given condition. And the possible vector parametric equation is <1,2,-2>+t/4<1,5,3>. I hope you are satisfied with my answer and feel free to ask for more

7 0

2 years ago

F(t)=t^2+19t+60<br> What are the zeros of the function and what is the vertex of the parabola?

SVEN [57.7K]

Answer: $-4,-15;\ \left(-\dfrac{19}{2},-\dfrac{121}{4}\right)$

Step-by-step explanation:

Given

Function is $F(t)=t^2+19+60$

Zeroes of the function are

$t^2+19t+60=0\\\\\Rightarrow t=\dfrac{-19\pm\sqrt{19^2-4\times 1\times 60}}{2\times 1}\\\\\Rightarrow t=\dfrac{-19\pm \sqrt{121}}{2}\\\\\Rightarrow t=\dfrac{-19\pm 11}{2}\\\\\Rightarrow t=-4,-15$

Using completing the square method

$y=t^2+2\times \dfrac{19}{2}t+\dfrac{19^2}{2^2}-\dfrac{19^2}{2^2}+60\\\\y=\left(t+\dfrac{19}{2}\right)^2+60-\dfrac{361}{4}\\\\y=\left(t+\dfrac{19}{2}\right)^2-\dfrac{121}{4}\\\\y+\dfrac{121}{4}=\left(t+\dfrac{19}{2}\right)^2\\\\\left(y-(-\dfrac{121}{4}\right)=\left(t-(-\dfrac{19}{2})\right)^2\\\\\text{The vertex is }\left(-\dfrac{19}{2},-\dfrac{121}{4}\right)$

4 0

2 years ago