Find a pattern for the sum of 4 times a number and twice the sum of the same number and 3

Okay can someone pls help me i literally am so confused

masha68 [24]

View answer attached below

5 0

3 years ago

Can someone give me the answer for this? thanks!!

vlabodo [156]

Answer:

Step-by-step explanation:

No

3 0

3 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

3 years ago

A total of $76,000 is to be invested, some in bonds and some in certificates of deposit (CDs). If the amount invested in bonds

Anestetic [448]

Answer:

The amount invested as certificates of deposit is $ 35,000 and The amount invested as bonds is $ 41,000

Step-by-step explanation:

Given as :

The total amount to be invested as bonds and certificates of deposit (CD) = $76,000

Let the amount invested as bonds = $ x + $6000

And The amount invested as CD = $ x

Now, According to question

The amount invested as CD + The amount invested as bonds = $ x + $ x + $6000

Or, $ x + $ x + $6000 = $76,000

Or, $ 2x = $76,000 - $6000

So, $ 2x = $70,000

∴ x = $\frac{70000}{2}$ = $ 35,000

And $ x + $6000 = $ 35,000 + $6000 = $ 41,000

Hence The amount invested as certificates of deposit is $ 35,000 and The amount invested as bonds is $ 41,000 Answer

4 0

2 years ago

How many solutions ? ( 1 solution , no solution , infinite solutions ) I WILL MAKE U BRAINLIST NEED ASAP

tangare [24]

Answer:

#4 = ∞

#5= no solution

#6= no solution

Step-by-step explanation:

GOOD LUCK!!!!

4 0

2 years ago