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Hunter-Best [27]

3 years ago

8

Order the integers 4, –6, 8, and –4 from least to greatest.

1 answer:

Murljashka [212]3 years ago

5 0

Answer:

-6, -4, 4, 8

Step-by-step explanation:

hope it helps <3

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The perimeter it a square field is 224. How long is

geniusboy [140]

Since each side of a sqare is equal and there are four sides to the square
$224 = 4x$
X being the side length.
Then you would simiply divide by 4 to isolate x
$x = 56$
A side length is 56

8 0

4 years ago

A bag contains 6 black marbles and 4 white marbles. Sally takes out a black marble and does not put it back. What is the probabi

Neko [114]

Answer:

$\frac{5}{9}$

Step-by-step explanation:

$Total\ marbles=10\\\\Black=6\\\\White=4$

After removing one black marble:

$Total\ remaining\ marbles=9\\\\Remaining\ Black=5\\\\P(Black)=\frac{Remaining\ Black\ marbles}{Total\ remaining\ marbles}\\\\P(Black)=\frac{5}{9}$

6 0

3 years ago

SOMEBODY HELP MEEE!!!!

VladimirAG [237]

Solution= x=5
Alernative form= 21 x - 105= 0<span />

8 0

3 years ago

Find the nth term of 8, 2, -4, -10,... then find A50

kumpel [21]

Answer:

An=8+(n-1)(-6)

=8-6n+6

=14-6n

A50=14-6(50)

= -286

7 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