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4 years ago

Write the expression: nine divided by the quantity of x plus y

Mathematics

1 answer:

valentina_108 [34]4 years ago

6 0

Answer:

9 ÷ (x + y)

This is not specific

Step-by-step explanation:

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What is the domain {(6,5), (2,9), (4,6), (5,4), (1,7)

n200080 [17]

Answer:

The set of all first elements of the ordered pairs in a relation R is the domain.

Domain = {1,2,4,5,6}

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3 years ago

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r. Yi buys vegetables at a market. He purchases 6 pounds of potatoes, p, and 3 pounds of onions, n, for $18. Onions cost twice a

ad-work [718]

Answer:

The equation 2n = p should be 2p = n.
The actual cost of the onions is $3.00 per pound.
Potatoes cost $1.50 per pound.

Step-by-step explanation:

The wording "onions cost twice as much as potatoes" is understood to mean the cost per pound of onions (n) is equal to two times the cost per pound of potatoes (2p). Then the appropriate equation would be ...

2p = n

Then the solution is ...

6p +3(2p) = 18

12p = 18

p = 18/12 = 1.50

n = 2p = 2(1.50) = 3.00

The equation should be 2p = n; onions cost $3.00 per pound; potatoes cost $1.50 per pound.

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3 years ago

CAN SOMEONE LLEASE HELP ME IM SO LOST!!

Afina-wow [57]

Answer: 1324 in^2

Step-by-step explanation:

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3 years ago

An unbiased coin is tossed four times. Find the probability of the given event. (Round your answer to four decimal places.)

Tamiku [17]

1/8=0.125
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3 years ago

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Because of their connection with secant lines, tangents, and instantaneous rates, limits of the form ModifyingBelow lim With h

Gre4nikov [31]

Answer:

$\dfrac{1}{2\sqrt{x}}$

Step-by-step explanation:

$f(x) = \sqrt{x} = x^{\frac{1}{2}}$

$f(x+h) = \sqrt{x+h} = (x+h)^{\frac{1}{2}}$

We use binomial expansion for $(x+h)^{\frac{1}{2}}$

This can be rewritten as

$[x(1+\dfrac{h}{x})]^{\frac{1}{2}}$

$x^{\frac{1}{2}}(1+\dfrac{h}{x})^{\frac{1}{2}}$

From the expansion

$(1+x)^n=1+nx+\dfrac{n(n-1)}{2!}+\ldots$

Setting $x=\dfrac{h}{x}$ and $n=\frac{1}{2}$ ,

$(1+\dfrac{h}{x})^{\frac{1}{2}}=1+(\dfrac{h}{x})(\dfrac{1}{2})+\dfrac{\frac{1}{2}(1-\frac{1}{2})}{2!}(\dfrac{h}{x})^2+\tldots$

$=1+\dfrac{h}{2x}-\dfrac{h^2}{8x^2}+\ldots$

Multiplying by $x^{\frac{1}{2}}$ ,

$x^{\frac{1}{2}}(1+\dfrac{h}{x})^{\frac{1}{2}}=x^{\frac{1}{2}}+\dfrac{h}{2x^{\frac{1}{2}}}-\dfrac{h^2}{8x^{\frac{3}{2}}}+\ldots$

$x^{\frac{1}{2}}(1+\dfrac{h}{x})^{\frac{1}{2}}-x^{\frac{1}{2}}=\dfrac{h}{2x^{\frac{1}{2}}}-\dfrac{h^2}{8x^{\frac{3}{2}}}+\ldots$

$\dfrac{x^{\frac{1}{2}}(1+\dfrac{h}{x})^{\frac{1}{2}}-x^{\frac{1}{2}}}{h}=\dfrac{1}{2x^{\frac{1}{2}}}-\dfrac{h}{8x^{\frac{3}{2}}}+\ldots$

The limit of this as $h\to 0$ is

$\lim_{h\to0} \dfrac{f(x+h)-f(x)}{h}=\dfrac{1}{2x^{\frac{1}{2}}}=\dfrac{1}{2\sqrt{x}}$ (since all the other terms involve $h$ and vanish to 0.)

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3 years ago