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3 years ago

HELP ME !!!!! Help me please !!!!

Mathematics

1 answer:

lora16 [44]3 years ago

8 0

Answer: 141.3cm^3

Step-by-step explanation:

Pi((3)^2)(-5)

Pi(9)(5)

Pi(45)

3.14×45

141.3

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Select the correct answer.

laiz [17]

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3 years ago

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Given h(x) = -5x + 3, find h(5).

cluponka [151]

Answer:

h(5) = -22

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Algebra I

Function Notation

Step-by-step explanation:

Step 1: Define

h(x) = -5x + 3

h(5) is x = 5

Step 2: Evaluate

Substitute in x [Function]: h(5) = -5(5) + 3
Multiply: h(5) = -25 + 3
Add: h(5) = -22

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3 years ago

In the portrait mode, your camera will automatically use the smallest aperture possible.

Natali5045456 [20]

False. It will choose a larger aperture for quality.

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3 years ago

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A survey of 2,000 doctors showed that an average of 3 out of 5 doctors use brand X aspirin. How many doctors use brand X aspirin

swat32

You would use cross multiplication:

X/2,000 = 3/5
Multiply: 2,000*3= 6,000
Divide: 6,000/5
Answer: 1,200 doctors use brand x

8 0

3 years ago

Find the tangent line approximation for 10+x−−−−−√ near x=0. Do not approximate any of the values in your formula when entering

Svetllana [295]

Answer:

$L(x)=\sqrt{10}+\frac{\sqrt{10}}{20}x$

Step-by-step explanation:

We are asked to find the tangent line approximation for $f(x)=\sqrt{10+x}$ near $x=0$ .

We will use linear approximation formula for a tangent line $L(x)$ of a function $f(x)$ at $x=a$ to solve our given problem.

$L(x)=f(a)+f'(a)(x-a)$

Let us find value of function at $x=0$ as:

$f(0)=\sqrt{10+x}=\sqrt{10+0}=\sqrt{10}$

Now, we will find derivative of given function as:

$f(x)=\sqrt{10+x}=(10+x)^{\frac{1}{2}}$

$f'(x)=\frac{d}{dx}((10+x)^{\frac{1}{2}})\cdot \frac{d}{dx}(10+x)$

$f'(x)=\frac{1}{2}(10+x)^{-\frac{1}{2}}\cdot 1$

$f'(x)=\frac{1}{2\sqrt{10+x}}$

Let us find derivative at $x=0$

$f'(0)=\frac{1}{2\sqrt{10+0}}=\frac{1}{2\sqrt{10}}$

Upon substituting our given values in linear approximation formula, we will get:

$L(x)=\sqrt{10}+\frac{1}{2\sqrt{10}}(x-0)$

$L(x)=\sqrt{10}+\frac{1}{2\sqrt{10}}x-0$

$L(x)=\sqrt{10}+\frac{\sqrt{10}}{20}x$

Therefore, our required tangent line for approximation would be $L(x)=\sqrt{10}+\frac{\sqrt{10}}{20}x$ .

8 0

3 years ago