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3 years ago

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mr. prowell drank 345.6 milligrams of coffee on Monday, 59 milligrams of coffee on Tuesday, and 193.12 milligrams of coffee on W

ednesday. what is the difference between the day he drank the most coffee and the day he drank the least coffee?

1 answer:

Dmitriy789 [7]3 years ago

3 0

I believe the difference is 295.6 milligrams. Hope I helped!

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92x12 rounder both factors

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90 and 10

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My lil sis-<br> pls answer part a and b

Yakvenalex [24]

Answer:

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3 years ago

Find the domain of the rational expression: x-3/2x-8

miss Akunina [59]

Answer:

$\huge\boxed{x\neq4\to x\in\mathbb{R}\backslash\{4\}}$

Step-by-step explanation:

$\dfrac{x-3}{2x-8}$

We know: the denominator must be different than 0.

Therefore

$2x-8\neq0$ <em>add 8 to both sides</em>

$2x-8+8\neq0+8$

$2x\neq8$ <em>divide both sides by 2</em>

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3 years ago

Matthew invested $3,000 into two accounts. One account paid 3% interest and the other paid 8% interest. He earned 4% interest on

boyakko [2]

<u>Answer:</u>

<em>Mathew invested</em><em> $600 and $2400</em><em> in each account.</em>

<u>Solution:</u>

From question, the total amount invested by Mathew is $3000. Let p = $3000.

Mathew has invested the total amount $3000 in two accounts. Let us consider the amount invested in first account as ‘P’

So, the amount invested in second account = 3000 – P

Step 1:

Given that Mathew has paid 3% interest in first account .Let us calculate the simple interest $(I_1)$ earned in first account for one year,

$\text {simple interest}=\frac{\text {pnr}}{100}$

Where

p = amount invested in first account

n = number of years

r = rate of interest

hence, by using above equation we get $(I_1)$ as,

$I_{1}=\frac{P \times 1 \times 3}{100}$ ----- eqn 1

Step 2:

Mathew has paid 8% interest in second account. Let us calculate the simple interest $(I_2)$ earned in second account,

$I_{2} = \frac{(3000-P) \times 1 \times 8}{100} \text { ------ eqn } 2$

Step 3:

Mathew has earned 4% interest on total investment of $3000. Let us calculate the total simple interest (I)

$I = \frac{3000 \times 1 \times 4}{100} ----- eqn 3$

Step 4:

Total simple interest = simple interest on first account + simple interest on second account.

Hence we get,

$I = I_1+ I_2 ---- eqn 4$

By substituting eqn 1 , 2, 3 in eqn 4

$\frac{3000 \times 1 \times 4}{100} = \frac{P \times 1 \times 3}{100} + \frac{(3000-P) \times 1 \times 8}{100}$

$\frac{12000}{100} = \frac{3 P}{100} + \frac{(24000-8 P)}{100}$

12000=3P + 24000 - 8P

5P = 12000

P = 2400

Thus, the value of the variable ‘P’ is 2400

Hence, the amount invested in first account = p = 2400

The amount invested in second account = 3000 – p = 3000 – 2400 = 600

Hence, Mathew invested $600 and $2400 in each account.

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3 years ago

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Please help!! I’m stuck on this

beks73 [17]

Answer:

B. $x = -4\,\lor \,x = -1$

Step-by-step explanation:

Let $\sqrt{-4-5\cdot x} = x$ , we solve for $x$ by algebraic means.

1) $\sqrt{-4-5\cdot x} = x$ Given

2) $-4-5\cdot x = x^{2}$ Definition of power/Symmetry of equality

3) $x^{2}+5\cdot x + 4 = 0$ Compatibility with addition/Commutative and associative properties/Existence of additive inverse/Modulative property

4) $(x+4)\cdot (x+1) = 0$ $x^{2}+(a+b)\cdot x +a\cdot b = (x + a)\cdot (x+b)$

5) $x = -4\,\lor \,x = -1$ Result

Hence, the correct answer is B.

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3 years ago