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3 years ago

Can anybody help me please

Mathematics

1 answer:

Alexxandr [17]3 years ago

5 0

Step-by-step explanation:

I think you are clear. You can follow me to get more answers. Thanks.

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What choices are correct?

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Answer:

its b

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NEED HELP ASAP!!!Of all the students who tried out for volleyball, 83% made the team what fraction of the students who tried out

zvonat [6]

Answer:I think it’s 1/5 but if u have another attemp and it’s wrong somehow do 1/3

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7t-8t+4=5t-2<br> Solve for t

astra-53 [7]

Answer:

t=1

Step-by-step explanation:

7t-8t+4=5t-2

-1t+4=5t-2

-1t+6=5t

6=6t

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Check

7t-8t+4=5t-2

7(1)-8(1)+4=5(1)-2

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3 years ago

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Add the two complex numbers 6 + 7i and 10 + 2i. Write your answer as a complex number in standard form, a + bi.

lina2011 [118]

1. The complex number 6 + 7i has

real part 6;
imaginary part 7.

2. The complex number 10 + 2i has

real part 10;
imaginary part 2.

3. When we add two complex numbers, we add real parts and imaginary parts separately:

(6+7i)+(10+2i)=(6+10)+(7+2)i=16+9i.

Answer: 16+9i

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4 years ago

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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