3 more than 5 times the number of dogs

5,286÷3 how many digits will the classroom number have

docker41 [41]

<span>5,286÷3 how many digits will the classroom number have
Notice that we have 4 digits as dividend with a place value of thousands as the highest and to be divided with our divisor that have only 1 digit with a place value of ones.
Now, let’s see how many digit will our quotient have:
=> 5 286 / 3
=> 1 762 is the quotient, it has still 4 digit with a place value of thousands.
To check simply multiply our quotient and divisor.</span>

5 0

3 years ago

Rigid transformations how to know when a figure is dilation transformation, rotation and dilation?

Naddika [18.5K]

Answer:

the shape could be congruent or similar to its preimage. There are basically four types of transformations: Rotation; Translation; Dilation; Reflection; Definition of Transformations. Transformations could be rigid (where the shape or size of preimage is not changed) and non-rigid (where the size is changed but the shape remains the same).

Step-by-step explanation:

7 0

3 years ago

What is the sum of all values of m that satisfy 2m (squared) -16m+8=0?

vodomira [7]

<h2>Steps:</h2>

So for this, we will be completing the square to solve for m. Firstly, subtract 8 on both sides:

$2m^2-16m=-8$

Next, divide both sides by 2:

$m^2-8m=-4$

Next, we want to make the left side of the equation a perfect square. To find the constant of this perfect square, divide the m coefficient by 2, then square the quotient. In this case:

-8 ÷ 2 = -4, (-4)² = 16

Add 16 to both sides of the equation:

$m^2-8m+16=12$

Next, factor the left side:

$(m-4)^2=12$

Next, square root both sides of the equation:

$m-4=\pm \sqrt{12}$

Next, add 4 to both sides of the equation:

$m=4\pm \sqrt{12}$

Now, while this is your answer, you can further simplify the radical using the product rule of radicals:

Product rule of radicals: √ab = √a × √b

√12 = √4 × √3 = 2√3.

$m=4\pm 2\sqrt{3}$

<h2>Answer:</h2>

In exact form, your answer is $m=4\pm \sqrt{12}\ \textsf{OR}\ m=4\pm 2\sqrt{3}$

In approximate form, your answers are (rounded to the hundreths) $m=7.46, 0.54$

6 0

3 years ago

Questions attached as screenshot below:Please help me I need good explanations before final testI pay attention

Nikitich [7]

The acceleration of the particle is given by the formula mentioned below:

$a=\frac{d^2s}{dt^2}$

Differentiate the position vector with respect to t.

$\begin{gathered} \frac{ds(t)}{dt}=\frac{d}{dt}\sqrt[]{\mleft(t^3+1\mright)} \\ =-\frac{1}{2}(t^3+1)^{-\frac{1}{2}}\times3t^2 \\ =\frac{3}{2}\frac{t^2}{\sqrt{(t^3+1)}} \end{gathered}$

Differentiate both sides of the obtained equation with respect to t.

$\begin{gathered} \frac{d^2s(t)}{dx^2}=\frac{3}{2}(\frac{2t}{\sqrt[]{(t^3+1)}}+t^2(-\frac{3}{2})\times\frac{1}{(t^3+1)^{\frac{3}{2}}}) \\ =\frac{3t}{\sqrt[]{(t^3+1)}}-\frac{9}{4}\frac{t^2}{(t^3+1)^{\frac{3}{2}}} \end{gathered}$

Substitute t=2 in the above equation to obtain the acceleration of the particle at 2 seconds.