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ExtremeBDS [4]
3 years ago
10

One glass of milk provides 305mg of calcium. Maggie drinks one glass of milk every morning. How many grams of calcium does Maggi

e get from her milk in 14 days?
Mathematics
2 answers:
Komok [63]3 years ago
8 0
She drinks 4.27 grams on calcium in 14 days at one glass of milk containing 14 mg per day
Alik [6]3 years ago
3 0

Answer:

4.27grams

Step-by-step explanation:

The initial information we have can be represented in one table:

Glasses of milk        Calcium

           1               ⇒    305mg

and we are told that Maggie drinks one gass of milk every morning.

that means that in 14 days she has drunk 14 glasses of milk,  I will represent  the amount of calcium in 14 glasses of milk by x :

Glasses of milk        Calcium

           1               ⇒    305mg

          14               ⇒    x

We solve this problem by multiplying the cross quantities in the table (305 and 14):

x= (305mg)(14)

x = 4270mg

but we need the answer in grams not in miligrams.

So we make the conversion, since 100mg = 1g

4270mg = 4.27g

The answer is 4.27grams

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Use quadratic formula solve .7x^2 - 28x - 7​
beks73 [17]

Answer:

x = 2 + sqrt(5) or x = 2 - sqrt(5)

Step-by-step explanation using the quadratic formula:

Solve for x over the real numbers:

7 (x^2 - 4 x - 1) = 0

Divide both sides by 7:

x^2 - 4 x - 1 = 0

Add 1 to both sides:

x^2 - 4 x = 1

Add 4 to both sides:

x^2 - 4 x + 4 = 5

Write the left hand side as a square:

(x - 2)^2 = 5

Take the square root of both sides:

x - 2 = sqrt(5) or x - 2 = -sqrt(5)

Add 2 to both sides:

x = 2 + sqrt(5) or x - 2 = -sqrt(5)

Add 2 to both sides:

Answer:  x = 2 + sqrt(5) or x = 2 - sqrt(5)

6 0
3 years ago
Can someone check whether its correct or no? this is supposed to be the steps in integration by parts​
Gwar [14]

Answer:

\displaystyle - \int \dfrac{\sin(2x)}{e^{2x}}\: \text{d}x=\dfrac{\sin(2x)}{4e^{2x}}+\dfrac{\cos(2x)}{4e^{2x}}+\text{C}

Step-by-step explanation:

\boxed{\begin{minipage}{5 cm}\underline{Integration by parts} \\\\$\displaystyle \int u \dfrac{\text{d}v}{\text{d}x}\:\text{d}x=uv-\int v\: \dfrac{\text{d}u}{\text{d}x}\:\text{d}x$ \\ \end{minipage}}

Given integral:

\displaystyle -\int \dfrac{\sin(2x)}{e^{2x}}\:\text{d}x

\textsf{Rewrite }\dfrac{1}{e^{2x}} \textsf{ as }e^{-2x} \textsf{ and bring the negative inside the integral}:

\implies \displaystyle \int -e^{-2x}\sin(2x)\:\text{d}x

Using <u>integration by parts</u>:

\textsf{Let }\:u=\sin (2x) \implies \dfrac{\text{d}u}{\text{d}x}=2 \cos (2x)

\textsf{Let }\:\dfrac{\text{d}v}{\text{d}x}=-e^{-2x} \implies v=\dfrac{1}{2}e^{-2x}

Therefore:

\begin{aligned}\implies \displaystyle -\int e^{-2x}\sin(2x)\:\text{d}x & =\dfrac{1}{2}e^{-2x}\sin (2x)- \int \dfrac{1}{2}e^{-2x} \cdot 2 \cos (2x)\:\text{d}x\\\\& =\dfrac{1}{2}e^{-2x}\sin (2x)- \int e^{-2x} \cos (2x)\:\text{d}x\end{aligned}

\displaystyle \textsf{For }\:-\int e^{-2x} \cos (2x)\:\text{d}x \quad \textsf{integrate by parts}:

\textsf{Let }\:u=\cos(2x) \implies \dfrac{\text{d}u}{\text{d}x}=-2 \sin(2x)

\textsf{Let }\:\dfrac{\text{d}v}{\text{d}x}=-e^{-2x} \implies v=\dfrac{1}{2}e^{-2x}

\begin{aligned}\implies \displaystyle -\int e^{-2x}\cos(2x)\:\text{d}x & =\dfrac{1}{2}e^{-2x}\cos(2x)- \int \dfrac{1}{2}e^{-2x} \cdot -2 \sin(2x)\:\text{d}x\\\\& =\dfrac{1}{2}e^{-2x}\cos(2x)+ \int e^{-2x} \sin(2x)\:\text{d}x\end{aligned}

Therefore:

\implies \displaystyle -\int e^{-2x}\sin(2x)\:\text{d}x =\dfrac{1}{2}e^{-2x}\sin (2x) +\dfrac{1}{2}e^{-2x}\cos(2x)+ \int e^{-2x} \sin(2x)\:\text{d}x

\textsf{Subtract }\: \displaystyle \int e^{-2x}\sin(2x)\:\text{d}x \quad \textsf{from both sides and add the constant C}:

\implies \displaystyle -2\int e^{-2x}\sin(2x)\:\text{d}x =\dfrac{1}{2}e^{-2x}\sin (2x) +\dfrac{1}{2}e^{-2x}\cos(2x)+\text{C}

Divide both sides by 2:

\implies \displaystyle -\int e^{-2x}\sin(2x)\:\text{d}x =\dfrac{1}{4}e^{-2x}\sin (2x) +\dfrac{1}{4}e^{-2x}\cos(2x)+\text{C}

Rewrite in the same format as the given integral:

\displaystyle \implies - \int \dfrac{\sin(2x)}{e^{2x}}\: \text{d}x=\dfrac{\sin(2x)}{4e^{2x}}+\dfrac{\cos(2x)}{4e^{2x}}+\text{C}

5 0
2 years ago
Plz somebody help. ^_^
pickupchik [31]

Answer:

sin θ =  15/17

Step-by-step explanation:

First we have to know how much the hypotenuse measures.

to take out the hypotenuse we will use pitagoras, with the following formula.

h^2 = c1^2 + c2^2

c1 = 8

c2 = 15

h^2 = 8^2 + 15^2

h^2 = 64 + 225

h = √ 289

h = 17

well to start we have to know the relationships between angles, legs and the hypotenuse.

a: adjacent

o: opposite

h: hypotenuse

sin θ = o/h

cos θ= a/h

tan θ = o/a

we want to know the sin of θ

sin θ = o/h

sin θ =  15/17

5 0
3 years ago
An expression is given (3^2)^3•3^6/3^4
Tomtit [17]
Hopes this if it doesn’t I any so sorry if I got it wrong:

Answer: 6561
7 0
3 years ago
Complete the equation describing how x and y are related.
Sav [38]

Answer: The "?" would equal 1.

Step-by-step explanation: This is simply a line with a slope of 1 and a y-intercept of zero.

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3 years ago
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