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2 years ago

Which pair shows equivalent expressions?

Mathematics

1 answer:

steposvetlana [31]2 years ago

6 0

Answer:

Step-by-step explanation:

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Plz help 7(4x - 3) = 4x

Vika [28.1K]

Answer:

x= 7/ 8 is the answer

If you wanted to simplify

5 0

3 years ago

Read 2 more answers

I NEED HELP ASAP PLEASE AND THANK YOU<br> What is the axis of symmetry?<br><br> y= -3 (x-2) (x-4)

Lina20 [59]

The axis of symmetry is x=3

3 0

2 years ago

Find the domain and range of the relation: {(–20, 11), (6, –8), (1, –20), (–13, 13)}

user100 [1]

Answer:

D: {-20, -13, 1, 6}

R: {-20, -8, 11, 13}

Step-by-step explanation:

Given the relation, {(–20, 11), (6, –8), (1, –20), (–13, 13)}, all x-values (inputs) make up the domain of the relation while all y-values make up the range of the relation.

Therefore:

Domain: {-20, -13, 1, 6}

Range: {-20, -8, 11, 13}

4 0

3 years ago

The Hudson Bay tides vary between 3 feet and 9 feet. The tide is at its lowest point when time (t) is 0 and completes a full cyc

BaLLatris [955]

Answer:

$y(t)= 6-3cos(\dfrac{2\pi}{14}t )$

Step-by-step explanation:

The function that could model this periodic phenomenon will be of the form

$y(t) = y_0+Acos(wt)$

The tide varies between 3ft and 9ft, which means its amplitude $A$ is

$A =\dfrac{(9-3)ft}{2} \\\\\boxed{A = 3ft}$

and its midline $y_0$ is

$y_o=3+3 \\\\\boxed{y_o= 6ft}$ .

Furthermore, since at $t=0$ the tide is at its lowest ( 3 feet ), we know that the trigonometric function we must use is $-cos(\omega t)$ .

The period of the full cycle is 14 hours, which means

$\omega t =2\pi$

$\omega (t+14)= 4\pi$

giving us

$\boxed{\omega = \dfrac{2\pi}{14}.}$

With all of the values of the variables in place, the function modeling the situation now becomes

$\boxed{y(t)= 6-3cos(\dfrac{2\pi}{14}t ).}$

8 0

3 years ago

Question 3 (Multiples and Factors] Three numbers are given below. Use prime factorisation to determine the HCF and LCM 1848 132

Ilia_Sergeevich [38]

Prime factorization involves rewriting numbers as products

The HCF and the LCM of 1848, 132 and 462 are 66 and 1848 respectively

<h3>How to determine the HCF</h3>

The numbers are given as: 1848, 132 and 462

Using prime factorization, the numbers can be rewritten as:

$1848 = 2^3 * 3 * 7 * 11$

$132 = 2^2 * 3 * 11$

$462 = 2 * 3 * 7 * 11$

The HCF is the product of the highest factors

So, the HCF is:

$HCF = 2 * 3 * 11$

$HCF = 66$

<h3>How to determine the LCM</h3>

In (a), we have:

$1848 = 2^3 * 3 * 7 * 11$

$132 = 2^2 * 3 * 11$

$462 = 2 * 3 * 7 * 11$

So, the LCM is:

$LCM = 2^3 * 3 * 7 * 11$

$LCM = 1848$

Hence, the HCF and the LCM of 1848, 132 and 462 are 66 and 1848 respectively

Read more about prime factorization at:

brainly.com/question/9523814

4 0

2 years ago