All categories

3 years ago

Simplify the expression

Mathematics

2 answers:

horrorfan [7]3 years ago

7 0

The answer would be C

love history [14]3 years ago

6 0

The answer is C.) 27f-21

You might be interested in

Joe has $500 in his bank account at the start of the summer. Every week he withdraws $40 for food. How many weeks can he withdra

Ksivusya [100]

Answer

he can withdraw the money around 7 weeks which means he withdraw $280 and in his bank account he will have $220.

8 0

3 years ago

Select the numeral form of this number written in scientific notation. 5 × 10^3

FinnZ [79.3K]

$5 \times 10^3 = 5,000$

7 0

3 years ago

What is the value of t?

max2010maxim [7]

Answer:p = -1.25

Step-by-step explanation:

Simplifying

3 + 17p = -32 + -11p

Solving

3 + 17p = -32 + -11p

Solving for variable 'p'.

Move all terms containing p to the left, all other terms to the right.

Add '11p' to each side of the equation.

3 + 17p + 11p = -32 + -11p + 11p

Combine like terms: 17p + 11p = 28p

3 + 28p = -32 + -11p + 11p

Combine like terms: -11p + 11p = 0

3 + 28p = -32 + 0

3 + 28p = -32

Add '-3' to each side of the equation.

3 + -3 + 28p = -32 + -3

Combine like terms: 3 + -3 = 0

0 + 28p = -32 + -3

28p = -32 + -3

Combine like terms: -32 + -3 = -35

28p = -35

Divide each side by '28'.

p = -1.25

Simplifying

p = -1.25

3 0

3 years ago

3 and 5/16 as a decimal please explain

JulsSmile [24]

3.3125 you can find this answer by using a calculator

5 0

3 years ago

Rewrite the statement in mathematical notation. (Let y be the distance from the top of the ladder to the floor, x be the distanc

In-s [12.5K]

Answer:

$\frac{dy}{dt}=\frac{6y}{x}\text{ ft per sec}$

Step-by-step explanation:

Let L be the length of the ladder,

Given,

x = the distance from the base of the ladder to the wall, and t be time.

y = distance from the base of the ladder to the wall,

So, by the Pythagoras theorem,

$L^2 = y^2 + x^2$

$\implies L = \sqrt{y^2 + x^2}$ ,

Differentiating with respect to time (t),

$\frac{dL}{dt}=\frac{d}{dt}(\sqrt{x^2 + y^2})$

$=\frac{1}{2\sqrt{x^2 + y^2}}\frac{d}{dt}(x^2 + y^2)$

$=\frac{1}{2\sqrt{x^2 + y^2}}(2x\frac{dx}{dt}+2y\frac{dy}{dt})$

$=\frac{1}{\sqrt{x^2 +y^2}}(x\frac{dx}{dt}+y\frac{dy}{dt})$

Here,

$\frac{dy}{dt}=-6\text{ ft per sec}$

Also, $\frac{dL}{dt} = 0$ ( Ladder length = constant ),

$\implies \frac{1}{\sqrt{x^2 +y^2}}(x(-6)+y\frac{dy}{dt})=0$

$-6x + y\frac{dy}{dt}=0$

$y\frac{dy}{dt}=6x$

$\implies \frac{dy}{dt}=\frac{6y}{x}\text{ ft per sec}$

Which is the required notation.

8 0

3 years ago