What are the answers?

What is the missing number? 9/21 ?/105 A. 5 B. 45 C. 89 D. 93

4vir4ik [10]

B.

Hope this helps!!!

6 0

3 years ago

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HELPPP MEHHHHHHHHH!!!:(

Lady bird [3.3K]

Answer:

1

Step-by-step explanation:

Rounding 5 to the nearest tenth would give you 1 and adding 1 to 9 would give you 10 which would 10 tenths which is 1

6 0

3 years ago

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Anna has a total of 36 goldfish in her aquarium, some are orange and some are black. The number of orange goldfish is 7 more tha

andriy [413]

Answer:

x + y = 36.....Equation 1

x = 7 + 2y...... Equation 2

Step-by-step explanation:

Let the number of:

Orange goldfish = x

Black gold fish = y

Anna has a total of 36 goldfish in her aquarium, some are orange and some are black.

x + y = 36.....Equation 1

The number of orange goldfish is 7 more than twice the number of black goldfish.

x = 7 + 2y...... Equation 2

We can rewrite this as:

x - 2y = 7..... Equation 2

The system of equations below that can be used to describe the number of black and orange goldfish in Anna's aquarium is :

x + y = 36.....Equation 1

x = 7 + 2y...... Equation 2

3 0

3 years ago

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Chang knows one side of a triangle is 13 cm. Which set of two sides is possible for the lengths of the other two sides of this t

natima [27]

Answer:

8cm and 2cm

I would appreciate if my answer is chosen as a brainliest answer

4 0

2 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.)

8 0

3 years ago