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miss Akunina [59]
3 years ago
5

Which number is both a rational number and an integer? A. B. 13/14 C. 194 D. 13.5

Mathematics
1 answer:
Misha Larkins [42]3 years ago
7 0

Answer:

C. 194

Step-by-step explanation:

An integer can be a positive or negative whole number, but not a decimal. If this number is an integer, then it is also a rational number. So, 194 fits this description.

Maybe this image can help you.

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Y is a differentiable function of x. Choose the alternative that is the derivative dy / dx.
murzikaleks [220]

Differentiating both sides of

x^3-y^3=1

with respect to <em>x</em> yields (using the chain rule)

3x^2 - 3y^2 \dfrac{\mathrm dy}{\mathrm dx} = 0

Solve for d<em>y</em>/d<em>x</em> :

3x^2 - 3y^2 \dfrac{\mathrm dy}{\mathrm dx} = 0 \\\\ 3y^2\dfrac{\mathrm dy}{\mathrm dx} = 3x^2 \\\\ \dfrac{\mathrm dy}{\mathrm dx} = \dfrac{3x^2}{3y^2} = \dfrac{x^2}{y^2}

The answer is then D.

3 0
2 years ago
Use a number line to add and subtract.<br><br> +7 + (-3)=
Karolina [17]

Answer:

4

Step-by-step explanation:

You have positive number and it is bigger that negative so your result will be positive too:

7-3=4

4 0
3 years ago
What is the equations of the line passing through points (-3,7) and (9,-1)?
diamong [38]

Answer:


y = -2/3x + 5 or A.

8 0
2 years ago
A croissant shop produces two products: bear claws (B) and almond filled croissants (C). Each bear claw requires 6 oz of flour,
Kruka [31]

Answer:

letter A: B = 400; C = 1000; Max Z = $380

Step-by-step explanation:

bear claws (B) need 6 oz of flour (f), 1 oz of yeast (y), 2 ts of paste (p)

croissants (C) need 3 oz of flour (f), 1 oz of yeast (y), 4 ts of paste (p)

putting that information in equations, we have:

B = 6f + y + 2p

C = 3f + y + 4p

The total number of resources (R) are:

R = 6600f + 1400y + 4800p

Let's call M our total profit, "k1" the number of B produced and "k2" the number of C produced.

So, we can state that:

M = k1*0.2 + k2*0.3

The number of resources R2 will demand is calculated like this:

R2 = k1*B + k2*C = (6k1+3k2)f + (k1+k2)y + (2k1+4k2)p

using R2 and R, we can make some inequations:

6k1+3k2 <= 6600 -> 2k1+k2 <= 2200

k1+k2 <= 1400

2k1+4k2 <= 4800 -> k1+2k2 <= 2400

if we try to maximize k2 (as it worths more), we will have k1 = 0 and k2 = 1200 (limited by p), but looking at the resources R, we will still have resources to use (f and y). Looking at B and C expressions, we see that removing one C gives enough 'p' to make 2 B, which is a good trade (as 2B worths 0.4 cents, and 1C worths 0.3 cents). we have 200 'y' remaining, so doing this 200 times give us k1 = 400 and k2 = 1000, and the only resource remaining will be some of 'f'.

calculating the profit M, we have:

M = 400*0.2 + 1000*0.3 = 380$

the right answer is letter A.

3 0
3 years ago
-8 1/16 as decimal number
Anon25 [30]

Answer:

-8.0625

Step-by-step explanation:

5 0
3 years ago
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