Math question down below more than one answer

Need help in algebra fast

Gennadij [26K]

Answer:

A

Step-by-step explanation:

This is a strict inequality so the line must be dotted, meaning it can only be A or D.

The inequality says that y is less than -2x+4 so we want the region under the line, making A the correct answer.

5 0

3 years ago

When converted to a household measurement, 9 kilograms is approximately equal to a

Tresset [83]

Answer:

35.274 ounces

Step-by-step explanation:

when converted to a household measurement 9 kg is equal to 1kg in household measurement is equal to 35.274 ounces. 35.274*9=317.466 ounces.

Hope this helps :) 

 Pls mark brainliest :3 

 And have an amazing day <3

6 0

2 years ago

Read 2 more answers

Sam is starting a daily log to record the number of hours he spends practicing his guitar. He begins with his log with 70 minute

myrzilka [38]

P(x) = 70 + 15x
where x = # of days

4 0

2 years ago

If 2 to the power x is equal to 3 Y is equal to 6 to the power z then prove that 1 by X + 1 by Y is equal to 1 by Z

Sladkaya [172]

If $2^{X}=3^{Y}=6^{Z}$ then $\frac{1}{X}+\frac{1}{Y}$ is equal to $\frac{1}{Z}$

What is an exponent?

⇒An exponent refers to the number of times a number is multiplied by itself.

Multiplication of exponents

⇒ It is also known as the law on indices. For multiplication with the same base, we add the powers.

For the same base $a^{m} \times a^{n}=a^{m+n}$

For the different base and same power $a^{m} \times b^{m} = (a\times b)^{m}$

Calculation:

⇒ We have been given that $2^{X}=3^{Y}=6^{Z}$ and by this relation, we have to prove that: $\frac{1}{X}+\frac{1}{Y}$ = $\frac{1}{Y}$

⇒ $2^{X}=3^{Y}=6^{Z}$

$2^{X}=6^{Z}$

$(2)^{X}=(2 \times3)^{Z}$

$2^{X}=3^{Z}\times 2^{Z}$

By taking power of Y both sides

⇒ $(2^{X})^{Y} =(2^{Z})^{Y} \times(3^{Z})^{Y}$

$(2^{X})^{Y} =(2^{Z})^{Y} \times(3^{Y})^{Z}$

$(2^{X})^{Y} =(2^{Z})^{Y} \times(2^{X})^{Z}$

$2^{XY} =2^Z^Y \times2^X^Z$

XY=ZY+XZ

On dividing by 1/XYZ into both sides we will get

⇒ $\frac{1}{Z}=\frac{1}{X}+\frac{1}{Y}$

Hence, it is now proved that if $2^{X}=3^{Y}=6^{Z}$ then $\frac{1}{X}+\frac{1}{Y}$ is equal to $\frac{1}{Z}$

Learn more about the law of indices here :

brainly.com/question/170984

#SPJ9

3 0

2 years ago

Use an appropriate series in (2) in Section 6.1 to find the Maclaurin series of the given function. Write your answer in summati

Norma-Jean [14]

Answer:

$e^{\frac{1}{5}x} = \sum\limits^{\infty}_{k=0} \frac{1}{5}^k \cdot \frac{x^k}{k!}$

Step-by-step explanation:

Poorly formatted question.

The given parameters can be summarized as:

$e^x = \sum\limits^{\infty}_{k=0} \frac{x^k}{k!}$ ----- the series

Required

Determine $e^\frac{1}{5}^x$

We have:

$e^x = \sum\limits^{\infty}_{k=0} \frac{x^k}{k!}$

Substitute $\frac{1}{5}x$ for x

$e^{\frac{1}{5}x} = \sum\limits^{\infty}_{k=0} \frac{(\frac{1}{5}x)^k}{k!}$

Split

$e^{\frac{1}{5}x} = \sum\limits^{\infty}_{k=0} \frac{1}{5}^k \cdot \frac{x^k}{k!}$

3 0

2 years ago