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3 years ago

I need help please.

Mathematics

2 answers:

ira [324]3 years ago

7 0

Answer: 3,168

Step-by-step explanation: If You Multiply The Original Cost Of The Textbook By 33 You Will Get A Product Of 3,186

Mandarinka [93]3 years ago

7 0

To find out the cost for the class itself we must understand what the product is and in this case it is the biology book and the prices of which ranges to $96

Now there are 33 students in a class and each on of which wants a textbook

So we multiply 96 by 33 giving us the amount of $3168 for the cost of biology textbook

The answer is $3168

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Dalia is a botanist. She wants to model the growth rate of the patch of ground that is covered by a certain plant as time passes

Verdich [7]

Answer:

The appropriate measure is cm/day.

Step-by-step explanation:

The best unit to measure the height of a plant = cm.

The best measure for time is day.

Rate refers to the ratio between units. Mostly, the difference in some unit when time passes.

She can measure the height of the plant in each day.

Hence, the appropriate unit for Dalla is cm per day.

3 0

3 years ago

Read 2 more answers

Two terms in a geometric sequence are a3=729 and a4=243.

lesantik [10]

<h2>The first term of the given sequence (a) = 6561</h2>

Step-by-step explanation:

Let the first term = a and common difference = d

Given,

$a_{3}$ = 729 and $a_{4}$ = 243

To find, the first term of the given sequence (a) = ?

We know that,

The nth term of a G.P.

$a_{n} =ar^{n-1}$

The 3rd term of a G.P.

$a_{3} =ar^{3-1}$

⇒ $ar^{2}$ = 729 ..............(1)

The 4th term of a G.P.

$a_{4} =ar^{4-1}$

⇒ $ar^{3}$ = 243 ..............(2)

Dividing equation (2) by (1), we get

$\dfrac{ar^{3}}{ar^{2}}$ = $\dfrac{243}{729}$

⇒ $r=\dfrac{1}{3}$

Put $r=\dfrac{1}{3}$ in equation (1), we get

$a(\dfrac{1}{3})^{2}$ = 729

⇒ $a(\dfrac{1}{9})$ = 729

⇒ a = 9 × 729 = 6561

∴ The first term of the given sequence (a) = 6561

8 0

3 years ago

Identify the quotient in the form a + bi. HELP ASAP PLEASE!!

Artist 52 [7]

Solving $\frac{4-5i}{2+3i}$ we get $-\frac{7}{13}-\frac{22i}{13}$

Option A is correct.

Step-by-step explanation:

We need to find the result of $\frac{4-5i}{2+3i}$

We need to remove i from the denominator. For removing i Multiplying and dividing by 2-3i:

$\frac{4-5i}{2+3i}*\frac{2-3i}{2-3i}$

Solving:

$=\frac{4-5i*2-3i}{2+3i*2-3i}\\=\frac{4(2-3i)-5i*(2-3i)}{(2)^2-(3i)^2}\\=\frac{8-12i-10i+15i^2)}{4-9i^2}\\We\,\,i^2=-1\\=\frac{8-12i-10i+15(-1))}{4-9(-1)}\\=\frac{8-12i-10i-15}{4+9}\\=\frac{8-15-12i-10i}{13}\\=\frac{-7-22i}{13}\\=-\frac{7}{13}-\frac{22i}{13}$