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3 years ago

Find the sum: 4√3+11√12 15√15 15√3 26√3 48√3

Mathematics

2 answers:

Andreyy893 years ago

8 0

Answer:

$26\sqrt{3}$

Step-by-step explanation:

Simplify them:

$11\sqrt{12} = 22\sqrt{3}$

$22\sqrt{3}+4\sqrt{3}=26\sqrt{3}$

sasho [114]3 years ago

3 0

Answer:

26√3

Step-by-step explanation:

4√3+11√12

We know that sqrt(ab) = sqrt(a) sqrt(b)

4√3+11 $\sqrt{4} \sqrt{3}$

4√3+11* $2 \sqrt{3}$

4√3+22 $\sqrt{3}$

Combine like terms

26√3

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3 years ago

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Fudgin [204]

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3 years ago

Jacob transformed quadrilateral FGHJ to F'G'H'J'.

Rudik [331]

Answer:

A. Reflection across the line x = 1

Step-by-step explanation:

Please find the attachment.

We have been given that Jacob transformed quadrilateral FGHJ to F'G'H'J'. We are asked to find which transformation Jacob used to reflect FGHJ to F'G'H'J'.

Since we know that while reflecting a figure, the line of reflection will lie between the original figure and reflected figure. Each point of the reflected figure will have the same distance from the line of reflection as the corresponding point of the original figure.

Now let us see our given choices one by one.

A. Reflection across the line x = 1.

Upon looking at point G and G' we can see that both points are equidistant (2 units) from the line x=1. Other corresponding points of both quadrilaterals are also equidistant from line x=1, therefore, Jacob used the reflection across the line x=1 to transform quadrilateral FGHJ to F'G'H'J'.

B. Reflection across the line y = 1 .

If Jacob had reflected quadrilateral across line y=1, the points of quadrilateral F'G'H'J' will lie in second and third quadrant. We can see from our graph that F'G'H'J' lies in 1st quadrant, therefore, option B is not a correct choice.

C. Reflection across the line y-axis.

If Jacob had reflected quadrilateral across y-axis, the coordinates of points of quadrilateral F'G'H'J' will be G'(1,4), H'(1,2), J'(4,0) and F'(2,5). Therefore, option B is not a correct choice.

D. Reflection across the x -axis.

If Jacob had reflected quadrilateral across x-axis, the points of quadrilateral F'G'H'J' will lie in third quadrant. We can see from our graph that F'G'H'J' lies in 1st quadrant, therefore, option D is not a correct choice.

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3 years ago

A(-5,-4) ——> A’ is a glide reflection where the translation is (x,y)—->(x+6,y), and the line of reflection is y=3. What ar

Nastasia [14]

Solution:

The Point in the coordinate plane is A(-5,-4).

Perpendicular or shortest Distance from line y=3 that is (-5,3) to point (-5,-4) is

$=\sqrt{(-5+5)^2+(3+4)^2}\\\\=7$

When it is reflected through the line, y=3, the coordinate of point A (-5,-4) changes to (-5,3+7)= B(-5,10).

Now, the Point B is translated by the rule , (x,y)—->(x+6,y),

So,the point B is translated to, (-5+6,10)=(1,10)

Option C: (1,10) is the glide reflection of point A(-5,-4).

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3 years ago

Read 2 more answers

(1 point) consider the function f(t)=⎧⎩⎨⎪⎪⎪⎪0,−5,−6,6,t<00≤t<11≤t<7t≥7;f(t)={0,t<0−5,0≤t<1−6,1≤t<76,t≥7; 1. wr

sergij07 [2.7K]

$f(t)=\begin{cases}0&\text{for }t$

Recall that

$u(t)=\begin{cases}0&\text{for }t$

Take it one piece at a time. For $t\ge0$ , we can scale $u(t)$ by -5:

$-5u(t)=\begin{cases}0&\text{for }t$

If we shift the argument by 1 and scale by -5, we have

$-5u(t-1)=\begin{cases}0&\text{for }t$

so if we subtract this from $-5u(t)$ , we'll end up with

$-5u(t)+5u(t-1)=\begin{cases}0&\text{for }t$

For the next piece, we can add another scaled and shifted step like

$-6u(t-1)+6u(t-7)=\begin{cases}0&\text{for }t$

so that

$-5u(t)+5u(t-1)-6u(t-1)+6u(t-7)=\begin{cases}0&\text{for }t$

For the last piece, we add one more term:

$6u(t-7)=\begin{cases}0&\text{for }t$

and so putting everything together, we get $f(t)$ :

$f(t)\equiv-5u(t)+5u(t-1)-6u(t-1)+6u(t-7)+6u(t-7)$
$f(t)\equiv-5u(t)-u(t-1)+12u(t-7)$

5 0

3 years ago