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Nady [450]
3 years ago
7

it is recommended that an adult drink 64 fluid ounces of water evey day. Josey already had 700 milliliters of water. How many mo

re liters of water should he drink?
Mathematics
2 answers:
Umnica [9.8K]3 years ago
8 0

Answer:


Step-by-step explanation:

700 millileters= .7liters

64ounces= 1.89liters

1.89-.7= 1.19liters

Josey should drink 1.19 more liters of water

Mnenie [13.5K]3 years ago
6 0

Answer:

1.193 liters


Step-by-step explanation:


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Find the area of the shaded region. Round your answer to the nearest tenth if needed.<br><br> 4B
AlexFokin [52]

Answer:

136.1 inches

Step-by-step explanation:

area of a circle is r^2 x pi

this is 3/4 of a circle

1. so 3/4 x 7.6^2xpi

2. 3/4 x 57.76 x pi

3. .75 x 57.76 x pi

4. 136.1

5 0
3 years ago
a 400 g of a liquid of density 1.6g/cm3 is made from mixing liquids of density 1.2g/cm3 and 1.8g/cm3 find the mass of the liquid
gogolik [260]

Answer:

133.33g

Step-by-step explanation:

Let the:

Mass of 1.2g/cm³ of liquid = x

Mass of 1.8g/cm³ of liquid = y

From our Question above, our system of equations is given as:

x + y = 400........ Equation 1

x = 400 - y

1.2 × x + 1.8 × y = 1.6 × 400

1.2x + 1.8y = 640..... Equation 2

We substitute, 400 - y for x in Equation 2

1.2(400 - y) + 1.8y = 640

480 - 1.2y + 1.8y = 640

- 1.2y + 1.8y = 640 - 480

0.6y = 160

y = 160/0.6y

y = 266.67 g

Solving for x

x = 400 - y

x = 400 - 266.67g

x = 133.33g

Therefore, the mass of the liquid of density 1.2g/cm³ is 133.33g

7 0
3 years ago
Find two power series solutions of the given differential equation about the ordinary point x = 0. compare the series solutions
monitta
I don't know what method is referred to in "section 4.3", but I'll suppose it's reduction of order and use that to find the exact solution. Take z=y', so that z'=y'' and we're left with the ODE linear in z:

y''-y'=0\implies z'-z=0\implies z=C_1e^x\implies y=C_1e^x+C_2

Now suppose y has a power series expansion

y=\displaystyle\sum_{n\ge0}a_nx^n
\implies y'=\displaystyle\sum_{n\ge1}na_nx^{n-1}
\implies y''=\displaystyle\sum_{n\ge2}n(n-1)a_nx^{n-2}

Then the ODE can be written as

\displaystyle\sum_{n\ge2}n(n-1)a_nx^{n-2}-\sum_{n\ge1}na_nx^{n-1}=0

\displaystyle\sum_{n\ge2}n(n-1)a_nx^{n-2}-\sum_{n\ge2}(n-1)a_{n-1}x^{n-2}=0

\displaystyle\sum_{n\ge2}\bigg[n(n-1)a_n-(n-1)a_{n-1}\bigg]x^{n-2}=0

All the coefficients of the series vanish, and setting x=0 in the power series forms for y and y' tell us that y(0)=a_0 and y'(0)=a_1, so we get the recurrence

\begin{cases}a_0=a_0\\\\a_1=a_1\\\\a_n=\dfrac{a_{n-1}}n&\text{for }n\ge2\end{cases}

We can solve explicitly for a_n quite easily:

a_n=\dfrac{a_{n-1}}n\implies a_{n-1}=\dfrac{a_{n-2}}{n-1}\implies a_n=\dfrac{a_{n-2}}{n(n-1)}

and so on. Continuing in this way we end up with

a_n=\dfrac{a_1}{n!}

so that the solution to the ODE is

y(x)=\displaystyle\sum_{n\ge0}\dfrac{a_1}{n!}x^n=a_1+a_1x+\dfrac{a_1}2x^2+\cdots=a_1e^x

We also require the solution to satisfy y(0)=a_0, which we can do easily by adding and subtracting a constant as needed:

y(x)=a_0-a_1+a_1+\displaystyle\sum_{n\ge1}\dfrac{a_1}{n!}x^n=\underbrace{a_0-a_1}_{C_2}+\underbrace{a_1}_{C_1}\displaystyle\sum_{n\ge0}\frac{x^n}{n!}
4 0
3 years ago
Write an expression for "the quotient of 9 and c"
Eva8 [605]

9÷c

Quotient is division.

The quotient of 9 and c would be 9 divided by c

5 0
3 years ago
Please help this is due at 11:59
ANTONII [103]

Answer:

i believe the answer is c

Step-by-step explanation:

tbh i dont have to explain cause you mainly have to just look at the graph

6 0
2 years ago
Read 2 more answers
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