MATHEMATICS COLLEGE

A cat has 24 whiskers by its nose. A manatee has 25 times as many whiskers.

How many whiskers does a manatee have?

Answers

Answer 1

Answer:

Answer:

600 whiskers

Step-by-step explanation:

If you multiply 24 by 25 you get 600

Related Questions

6. How many solutions does the given system have?y = 3x - 16x - 2y = -2A. one solutionB. two solutionsC. infinite solutionsD. no solution
To save money you set aside $50. For each following month you set aside 10% more than the previous month.How much money will you save in a year? Explain each step.
I need a reason how sort from least to greatest I don’t understand
A fatigue test was conducted in which the mean stress was 70 MPa, and the stress amplitude was 210 MPa. (a) Compute the maximum and minimum stress levels. (b) Compute the stress ratio. (c) Compute the magnitude of the stress range
What is 9 + 2x = -x + 3

1) Uma pessoa saiu de casa indo as compras com R$240,00 na carteira, sendo que gastou no supermercado 2/3 deste valor, em seguida gastou 14 do que havia sobrado na farmácia. Em uma parada no açougue, deixou mais 72 do que lhe havia sobrado. Após estas despesas qual o valor que lhe havia sobrado. Após estas despesas qual o valor restante na carteira dessa pessoa? A)R$ 40,00 B)R$ 30,00 C)R$ 20,00 D)R$ 1000 E)R$ 5,00

Answers

Answer:

The remaining amount in the person's wallet is $30.

Step-by-step explanation:

We are given that a person left the house to go shopping with $240.00 in his wallet, and he spent 2/3 of this amount in the supermarket, then spent 1/4 of what was left in the pharmacy. At a stop at the butcher shop, he left 1/2 than he had left.

And we have to find the remaining money in the wallet after making all these above expenses.

At the starting, the amount of money in the person's wallet = $240

Amount of money spent in the supermarket = $(2)/(3) \text{ of } \$240$

= $(2)/(3) * 240$

= $2 * 80$ = $160

So, now the amount of money left with him = $240 - $160

= $80

Amount of money spent in the pharmacy = $(1)/(4) \text{ of } \$80$

= $(1)/(4) * 80$ = $20

So, now the amount of money left with him = $80 - $20

= $60

Now, Amount of money spent at the butcher shop = $(1)/(2) \text{ of } \$60$

= $(1)/(2) * 60$ = $30

So, now the amount of money left with him = $60 - $30

= $30

Hence, the amount of money remaining in the person's wallet is $30.

711.9 divided by 8.4 show work?

Answers

Hoped this had helped!!!!

The answer would be 84.75.

Here are 2 images to show it is the proper answer.

I hope this helps!

Determine whether the improper integral converges or diverges, and find the value of each that converges.∫^0_-[infinity] 5e^60x dx

Answers

Answer:

The improper integral converges.

$\displaystyle \int\limits^0_(- \infty) {5e^(60x)} \, dx = (1)/(12)$

General Formulas and Concepts:
Calculus

Limit

Limit Rule [Variable Direct Substitution]: $\displaystyle \lim_(x \to c) x = c$

Differentiation

Derivatives
Derivative Notation

Derivative Property [Multiplied Constant]: $\displaystyle (d)/(dx) [cf(x)] = c \cdot f'(x)$

Derivative Rule [Basic Power Rule]:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Integration

Integrals

Integration Rule [Reverse Power Rule]: $\displaystyle \int {x^n} \, dx = (x^(n + 1))/(n + 1) + C$

Integration Rule [Fundamental Theorem of Calculus 1]: $\displaystyle \int\limits^b_a {f(x)} \, dx = F(b) - F(a)$

Integration Property [Multiplied Constant]: $\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx$

Integration Method: U-Substitution

Improper Integral: $\displaystyle \int\limits^(\infty)_a {f(x)} \, dx = \lim_(b \to \infty) \int\limits^b_a {f(x)} \, dx$

Step-by-step explanation:

Step 1: Define

Identify.

$\displaystyle \int\limits^0_(- \infty) {5e^(60x)} \, dx$

Step 2: Integrate Pt. 1

[Integral] Rewrite [Integration Property - Multiplied Constant]: $\displaystyle \int\limits^0_(- \infty) {5e^(60x)} \, dx = 5 \int\limits^0_(- \infty) {e^(60x)} \, dx$
[Integral] Rewrite [Improper Integral]: $\displaystyle \int\limits^0_(- \infty) {5e^(60x)} \, dx = \lim_(a \to - \infty) 5 \int\limits^0_(a) {e^(60x)} \, dx$

Step 3: Integrate Pt. 2

Identify variables for u-substitution.

Set u: $\displaystyle u = 60x$
[u] Differentiate [Derivative Properties and Rules]: $\displaystyle du = 60 \ dx$
[Bounds] Swap: $\displaystyle \left \{ {{x = 0 \rightarrow u = 0} \atop {x = a \rightarrow u = 60a}} \right.$

Step 4: Integrate Pt. 3

[Integral] Rewrite [Integration Property - Multiplied Constant]: $\displaystyle \int\limits^0_(- \infty) {5e^(60x)} \, dx = \lim_(a \to - \infty) (1)/(12) \int\limits^0_(a) {60e^(60x)} \, dx$
[Integral] Apply Integration Method [U-Substitution]: $\displaystyle \int\limits^0_(- \infty) {5e^(60x)} \, dx = \lim_(a \to - \infty) (1)/(12) \int\limits^0_(60a) {e^(u)} \, du$
[Integral] Apply Exponential Integration: $\displaystyle \int\limits^0_(- \infty) {5e^(60x)} \, dx = \lim_(a \to - \infty) (1)/(12) e^u \bigg| \limits^0_(60a)$
Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]: $\displaystyle \int\limits^0_(- \infty) {5e^(60x)} \, dx = \lim_(a \to - \infty) (1 - e^(60a))/(12)$
[Limit] Evaluate [Limit Rule - Variable Direct Substitution]: $\displaystyle \int\limits^0_(- \infty) {5e^(60x)} \, dx = (1 - e^(60(-\infty)))/(12)$
Rewrite: $\displaystyle \int\limits^0_(- \infty) {5e^(60x)} \, dx = (1)/(12) - (1)/(12e^(60(\infty)))$
Simplify: $\displaystyle \int\limits^0_(- \infty) {5e^(60x)} \, dx = (1)/(12)$