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julia-pushkina [17]
3 years ago
10

A broken calculator can double any number or permute its digits (but 0 cannot go up front). Can we start with number 1 and obtai

n 78 by a series of these operations
Mathematics
1 answer:
stira [4]3 years ago
8 0

Answer: it's impossible to obtain 78

Step-by-step explanation: Since the calculator can double any number or permute its digits except 0 

starting with number 1 

1, 2, 4, 8, 16, 32, 64, 128

Therefore, it's impossible to obtain 78 by a series of these operations

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8 0
3 years ago
I ordered 8 pizzas for my son’s birthday party. Each pizza had 8 slices. If my son and his friends ate 47 slices of pizza at the
Andrews [41]

Answer:

17

Step-by-step explanation:

8*8 = 64

64 - 47 = 17

6 0
3 years ago
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dmitriy555 [2]

Answer: Yes, Raoul is correct because the initial value, also known as the y-intercept of function A is 0 and function B is 3.

Step-by-step explanation: The initial value is when x=0.

3 0
3 years ago
Read 2 more answers
Given f(X) =-X+ 6 write an equation for f(x + 3).
luda_lava [24]

Answer:

3 - x

Step-by-step explanation:

f(x) = -x + 6

f(x + 3)

Substitute (x + 3) in for x;

f(x + 3) = -(x + 3) + 6

= - x - 3 + 6

= -x + 3

3 0
3 years ago
Evaluate Dx / ^ 9-8x - x2^
Solnce55 [7]
It depends on what you mean by the delimiting carats "^"...

Since you use parentheses appropriately in the answer choices, I'm going to go out on a limb here and assume something like "^x^" stands for \sqrt x.

In that case, you want to find the antiderivative,

\displaystyle\int\frac{\mathrm dx}{\sqrt{9-8x-x^2}}

Complete the square in the denominator:

9-8x-x^2=25-(16+8x+x^2)=5^2-(x+4)^2

Now substitute x+4=5\sin y, so that \mathrm dx=5\cos y\,\mathrm dy. Then

\displaystyle\int\frac{\mathrm dx}{\sqrt{9-8x-x^2}}=\int\frac{5\cos y}{\sqrt{5^2-(5\sin y)^2}}\,\mathrm dy

which simplifies to

\displaystyle\int\frac{5\cos 
y}{5\sqrt{1-\sin^2y}}\,\mathrm dy=\int\frac{\cos y}{\sqrt{\cos^2y}}\,\mathrm dy

Now, recall that \sqrt{x^2}=|x|. But we want the substitution we made to be reversible, so that

x+4=5\sin y\iff y=\sin^{-1}\left(\dfrac{x+4}5\right)

which implies that -\dfrac\pi2\le y\le\dfrac\pi2. (This is the range of the inverse sine function.)

Under these conditions, we have \cos y\ge0, which lets us reduce \sqrt{\cos^2y}=|\cos y|=\cos y. Finally,

\displaystyle\int\frac{\cos y}{\cos y}\,\mathrm dy=\int\mathrm dy=y+C

and back-substituting to get this in terms of x yields

\displaystyle\int\frac{\mathrm dx}{\sqrt{9-8x-x^2}}=\sin^{-1}\left(\frac{x+4}5\right)+C
4 0
3 years ago
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