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3 years ago

What is the answer to this question

Mathematics

1 answer:

riadik2000 [5.3K]3 years ago

7 0

Answer: Your Answer to this question is C. Can you mark me as Brainliest and a thanks plss!!

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Answer my question please now :L

LUCKY_DIMON [66]

Answer:kdkejtjejejbf he neendhejrhwh

4 0

2 years ago

Read 2 more answers

For what value of a is the volume of the tetrahedron formed by the coordinate planes and the plane (x/a) (y/10) (z/6)

kotykmax [81]

This question is incomplete, the complete question is;

For what value of a is the volume of the tetrahedron formed by the coordinate planes and the plane (x/a) + (y/10) + (z/6) = 1 equal to 10?

Answer: the value of a is 1

Step-by-step explanation:

Given that;

Volume of tetrahedron bounded by plane (x/a) + (y/10) + (z/6) = 1

and coordinate plane is; V = 1/6|abc|

(x/a) + (y/10) + (z/6) = 1

volume = 10

10 = 1/6 | a × 10 × 6 |

60 = a × 10 × 6

60 = 60a

a = 60 / 60

a = 1

Therefore the value of a is 1

3 0

2 years ago

PLEASE HELP FAST!!!

Leni [432]

Answer:

Hello,

I just think that it is A

Why you may ask?

Because it has a Zero and that is an undefined number than

Step-by-step explanation:

Please I want brainest

6 0

2 years ago

What is the area of the figure? Use 3.14 for π. Round to the nearest hundredth if necessary.

vfiekz [6]

Answer:

81.12 cm²

Step-by-step explanation:

The area for a semicircle can be figured with this formula: (πr²)/2. The triangle's area formula is bh/2.

(3.14 * 4²)/2 + (14 * 8)/2 (Work the operations in parentheses first.)

(3.14 * 16)/2 + 112/2

50.24/2 + 112/2 (Divide 50.24 by 2. Do the same for 112.)

25.12 + 56 (Add 25.12 & 56.)

81.12 cm²

The area of the figure is 81.12 cm².

7 0

2 years ago

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

2 years ago