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2 years ago

Helpppp plsssSssssssss

Mathematics

1 answer:

blsea [12.9K]2 years ago

4 0

Answer:

<h3>your answer will be A</h3>

$x = \sqrt{52}$

Step-by-step explanation:

hope it helps you

have a great day

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A cafeteria sells 6 times as many bottles of juice. if the cafeteria sells 750 bottles of water. how many bottles of juice did t

faltersainse [42]

Answer:

<h2>4,500 bottles of juice.</h2>

Step-by-step explanation:

<h3>Clue word: Times means multiply</h3><h3>If you multiply 750 x 6 you will get 4,500.</h3>

5 0

1 year ago

Graph ARST with vertices R(6, 6), S(3, -6), and T(0, 3) and its image after a

padilas [110]

Answer:

The answer is the second figure and the vertices of Δ R'S'T' are:

R' = (-6 , 6) , S' = (-3 , -6) , T' = (0 , 3)

Step-by-step explanation:

* Lets revise some transformation

- If point (x , y) reflected across the x-axis

∴ Its image is (x , -y)

- If point (x , y) reflected across the y-axis

∴ Its image is (-x , y)

- If point (x , y) reflected across the line y = x

∴ Its image is (y , x)

- If point (x , y) reflected across the line y = -x

∴ Its image is (-y , -x)

- Now we can solve the problem

∵ R = (6 , 6) , S = (3 , -6) , T = (0 , 3), they are the vertices of ΔRST

- The triangle RST is reflected over the y-axis

- According to the rule above the signs of x-coordinates will change

∵ R = (6 , 6)

∴ Its image is (-6 , 6)

∵ S = (3 , -6)

∴ Its image is (-3 , -6)

∵ T = (0 , 3)

∴ Its image is (0 , 3)

* Now lets look to the figure to find the correct answers

- The image of Δ RST is ΔR'S'T'

∵ The vertices of the image of ΔRST are:

R' = (-6 , 6) , S' = (-3 , -6) , T' = (0 , 3)

* The answer is the second figure

7 0

2 years ago

Which of the following is a like radical to RootIndex 3 StartRoot 7x EndRoot?

lana [24]

Answer:

A. 4 (RootIndex 3 StartRoot 7 x EndRoot) or $4(\sqrt[3]{7x})$

Step-by-step explanation:

Given:

A radical whose value is, $r_1=\sqrt[3]{7x}$

Now, we need to find the like radical for $r_1$ .

Let the like radical be $r_2$ .

As per the definition of like radicals, like radicals are those that can be expressed as multiples of each other.

So, if two radicals $r_1\ and\ r_2$ are like radicals, then

$r_1 = n \times r_2
$

Where, 'n' is a real number.

Here, $r_1=\sqrt[3]{7x}$

Now, let us check all the options .

Option A:

4 (RootIndex 3 StartRoot 7 x EndRoot) or $r_2=4\sqrt[3]{7x}$

Now, we observe that $r_2$ is a multiple of $r_1$ because

$r_2=4\times \sqrt[3]{7x}\\\\ r_2=4\times r_1..............(r_1=\sqrt[3]{7x})$

Therefore, option A is correct.

Option B:

StartRoot 7 x EndRoot or $r_2=\sqrt{7x}$

As the above radical is square root and not a cubic root, this option is incorrect.

Option C:

x (RootIndex 3 StartRoot 7 EndRoot) or $r_2=x\sqrt[3]{7}$

As the term inside the cubic root is not same as that of $r_1$ , this option is also incorrect.

Option D:

7 StartRoot x EndRoot or $r_2=7\sqrt{x}$

As the above radical is square root and not a cubic root, this option is incorrect.

Therefore, the like radical is option (A) only.

4 0

2 years ago

Read 2 more answers

What is (4x+1) degrees and 17 degrees show your work simplify.

Alenkasestr [34]

Answer:

Let's simplify step-by-step.

4x+1+17

Combine Like Terms:

=4x+1+17

=(4x)+(1+17)

=4x+18

Answer:

=4x+18

5 0

2 years ago

PLS HELP!!!

stira [4]

Answer: Option 4

Step-by-step explanation:

$\mathbf{u}=\langle 70 \cos 230^{\circ}, 70 \sin 230^{\circ} \rangle\\\\\mathbf{v}=\langle 60\cos 140^{\circ}, 60 \sin 140^{\circ} \rangle\\\\\mathbf{w}=\langle 80 \cos 30^{\circ}, 80 \sin 30^{\circ} \rangle\\\\\\\therefore \mathbf{u}+\mathbf{v}+\mathbf{w}=\langle 70 \cos 230^{\circ}+60 \cos 140^{\circ}+80 \cos 30^{\circ}, 70 \sin 230^{\circ}+60 \sin 140^{\circ}+80 \sin 30^{\circ} \rangle\\\\ \approx \langle -21.675, 24.944 \rangle$

$\implies |\mathbf{u}+\mathbf{v}+\mathbf{w}| \approx \sqrt{(-21.675)^2 + (24.944)^2} \approx 33.046$

Direction: The vector is in the second quadrant, so the direction is about $180^{\circ}-\arctan \left(\frac{24.944}{-21.675} \right) \approx 131^{\circ}$

8 0

7 months ago