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scoray [572]
2 years ago
6

One cup holds 100 grams of sugar. How many cups will hold half a kilogram of sugar?

Mathematics
2 answers:
Flauer [41]2 years ago
8 0

Answer:

5 cups

Step-by-step explanation:

Half a kilogram is 500 grams.

500 grams is 5 times bigger than 100 grams.

Which means it takes 5 times more cups to hold 500 grams.

100 grams = 1 cup

500 grams = 5 cups

vitfil [10]2 years ago
7 0

Answer:

5 cups

Step-by-step explanation:

1 kilogram is 1000 grams

1/2 of a kilogram is 500 grams

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zlopas [31]
Answer: x = 6
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2 years ago
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melomori [17]

Answer:

x = 15

Step-by-step explanation:

The easiest way to solve this is to realise that a triangle takes up half the area of a rectangle of the same width and height.

We are told that the width of the triangle is 10, and that the line of length 10 is perpendicular to the longest side of the triangle.  Because of that we know that x can be multiplied by ten to get the area of the rectangle that is twice the area of the triangle.

We are also told that the triangle's area is 75 units.  With all of that put together, we can say:

a = (10 * x) / 2\\75 = 5 * x\\x = 15

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3 years ago
Write the inverse function for the function, ƒ(x) =1/2x + 4. Then, find the value of ƒ^ -1(4). Type your answers in the box.
Kipish [7]

Answer:

f⁻¹(x) = 2x - 8

f⁻¹(4) = 2 × 4 - 8

f⁻¹(4) = 0

Step-by-step explanation:

f(x) = \frac{1}{2} x + 4\\x = \frac{1}{2} f^{-1}(x) + 4\\x - 4 = \frac{1}{2} f^{-1}(x)\\2x - 8 = f^{-1}(x)\\f^{-1}(x) = 2x - 8

Let's test it

f^{-1}(f(x)) = 2(f(x)) - 8\\f^{-1}(f(x)) = 2( \frac{1}{2}x + 4) - 8\\f^{-1}(f(x)) = \frac{2}{2}x + 8 - 8\\f^{-1}(f(x)) = x

So we do indeed have the inverse function, so using that we can plug in the values requested:

f⁻¹(x) = 2x - 8

f⁻¹(4) = 2 × 4 - 8

f⁻¹(4) = 0

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3 years ago
Solve 5y'' + 3y' – 2y = 0, y(0) = 0, y'(0) = 2.8 y(t) = 0 Preview
mario62 [17]

Answer:  The required solution is

y(t)=-\dfrac{7}{3}e^{-t}+\dfrac{7}{3}e^{\frac{1}{5}t}.

Step-by-step explanation:   We are given to solve the following differential equation :

5y^{\prime\prime}+3y^\prime-2y=0,~~~~~~~y(0)=0,~~y^\prime(0)=2.8~~~~~~~~~~~~~~~~~~~~~~~~(i)

Let us consider that

y=e^{mt} be an auxiliary solution of equation (i).

Then, we have

y^prime=me^{mt},~~~~~y^{\prime\prime}=m^2e^{mt}.

Substituting these values in equation (i), we get

5m^2e^{mt}+3me^{mt}-2e^{mt}=0\\\\\Rightarrow (5m^2+3y-2)e^{mt}=0\\\\\Rightarrow 5m^2+3m-2=0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{since }e^{mt}\neq0]\\\\\Rightarrow 5m^2+5m-2m-2=0\\\\\Rightarrow 5m(m+1)-2(m+1)=0\\\\\Rightarrow (m+1)(5m-1)=0\\\\\Rightarrow m+1=0,~~~~~5m-1=0\\\\\Rightarrow m=-1,~\dfrac{1}{5}.

So, the general solution of the given equation is

y(t)=Ae^{-t}+Be^{\frac{1}{5}t}.

Differentiating with respect to t, we get

y^\prime(t)=-Ae^{-t}+\dfrac{B}{5}e^{\frac{1}{5}t}.

According to the given conditions, we have

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From equation (ii), we get

B=\dfrac{7}{3}.

Thus, the required solution is

y(t)=-\dfrac{7}{3}e^{-t}+\dfrac{7}{3}e^{\frac{1}{5}t}.

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Alik [6]

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