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3 years ago

What is the partial quotient of 85÷1,995

2 answers:

8 0

I believe ur answer is 0.42, if u multiple 1995 to 85 than is gonna be 43 but u say 85 multiple 1995 so the answers is 0.42

avanturin [10]3 years ago

4 0

I believe it's 23 as your answer

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A jar pinto beans and black beans in a ratio of 1:1, and 300 of the beans are pinto beans. How many beans are there in the jar?

Serhud [2]

Answer: 600 beans

Step-by-step explanation:

Since the beans are in a ratio 1:1, you can write it like this: x/x. Since we know x=300, we can do this: x/x=300/?. We know x is again 300(given) so, x/x=300/300. Total is adding the 2 beans up which is 300+300=600

8 0

4 years ago

What is the simplified form of 15 times x to the fifth power over 20 times y to the sixth power times the fraction 6 times y to

Tanya [424]

The answer is 9 x cubed over 10 y squared!

6 0

2 years ago

Read 2 more answers

An airplane travels 4020 kilometers against the wind in 6 hours and 4800 kilometers with the wind in the same amount of time. Wh

Ahat [919]

Answer:

Rate of plane 735 km/hr

Rate of wind 65 km/hr

Step-by-step explanation:

Calculation to determine the rate of the plane in still air

Let Va represent the velocity of the airplane

Let Vw represent the velocity of the wind

When flying with the wind:

(Va+Vw)*(6 hours) = 4800

6Va + 6Vw = 4800

6Vw = 4800 - 6Va

Vw=4800/6-Va

Vw = 800 - Va

When flying against the wind:

(Va-Vw)*(6 hours) = 4020 km

6Va - 6Vw = 4020

Substitute 800-Va for Vw and solve for Va:

6Va - 6(800-Va) = 4020

6Va -4800 + 6Va = 4020

12Va = 8820

Va=8820/12

Va = 735 km/hr

Therefore the rate of the plane in still air is 735 km/hr

Calculation to determine the rate of the wind

Rate of wind:

Vw = 800 - Va

Vw= 800 -735

Vw= 65 km/hr

Therefore the rate of the wind is 65 km/hr

5 0

3 years ago

PLZ HELP ITS ALLREADY OVER DUE

Alex Ar [27]

24.96 yd. Just have to multiply length times width.

8 0

3 years ago

Read 2 more answers

The twice–differentiable function f is defined for all real numbers and satisfies the following conditions: f(0)=3 f′(0)=5 f″(0)

Vladimir79 [104]

$g(x)=e^{ax}+f(x)\implies g'(x)=ae^{ax}+f'(x)\implies g''(x)=a^2e^{ax}+f''(x)$

Given that $f'(0)=5$ and $f''(0)=7$ , it follows that

$g'(0)=a+5$

$g''(0)=a^2+7$

###

$h(x)=\cos(kx)f(x)+\sin x\implies h'(x)=-k\sin(kx)f(x)+\cos(kx)f'(x)+\cos x$

When $x=0$ , we have

$h(0)=\cos0f(0)+\sin0=f(0)=3$

The slope of the line tangent to $h(x)$ at (0, 3) has slope $h'(0)$ ,

$h'(0)=-k\sin0f(0)+\cos0f'(0)+\cos0=5+1=6$

Then the tangent line at this point has equation

$y-3=6(x-0)\implies y=6x+3$

###

Differentiating both sides of

$4x^2+y^2=48+2xy$

with respect to $x$ yields

$8x+2y\dfrac{\mathrm dy}{\mathrm dx}=2y+2x\dfrac{\mathrm dy}{\mathrm dx}$

$\implies(2y-2x)\dfrac{\mathrm dy}{\mathrm dx}=2y-8x$

$\implies\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{y-4x}{y-x}$

On this curve, when $x=2$ we have

$4(2)^2+y^2=48+2(2)y\implies y^2-4y-32=(y-8)(y+4)=0\implies y=8$

(ignoring the negative solution because we don't care about it)

The tangent to this curve at the point $(x,y)$ has slope $\dfrac{\mathrm dy}{\mathrm dx}$ . This tangent line is horizontal when its slope is 0. This happens for

$\dfrac{y-4x}{y-x}=0\implies y-4x=0\implies y=4x$

and when $x=2$ , there is a horizontal tangent line to the curve at the point (2, 8).

5 0

3 years ago