All categories

3 years ago

A 40 oz jar of peanut butter costs 5.25. how much does a 100 oz jar cost

Mathematics

2 answers:

Karolina [17]3 years ago

7 0

Answer:

13.13

Step-by-step explanation:

5.25 x 100/40 is 13.125 which rounds up to 13.13

Readme [11.4K]3 years ago

4 0

A 100 oz jar would cost 13.125$
or rounded to the nearest hundredth, 13.13$

You might be interested in

Marcy can walk 12 blocks in 5 minutes. If each block is 50 feet long, how many feet will she walk during the 30 minutes that she

a_sh-v [17]

Answer:

3600ft.

Step-by-step explanation:

12blocks x 50ft/block = 600 feet/5min.

30min/5min = 6

600 x 6 = 3600 ft.

5 0

3 years ago

Read 2 more answers

It has been reported that men are more likely than women to participate in online auctions. In a recent survey, 65% of repondent

kupik [55]

Answer:

Hence, the Female and not participated is $\frac{28}{100}=0.28\%$

Step-by-step explanation:

Let Total respondents $=100$

Participated in online action $(P)=65$

Men $(m)=45$ and Women $(F)=100-45=55$

Male and participated $(M\; \text{and}\;P)=38$

Female and participated $(F\; \text{and}\;P)=65-38=27$

Male and not participated is $45-38=7$

Female and not participated is $55-27=28$

Therefore, the Female and not participated is $\frac{28}{100}=0.28\%$

Hence, the correct option is $D$ .

5 0

3 years ago

What is the value of this expression (4.08)(0.7)?

elena-s [515]

7 0

3 years ago

Read 2 more answers

If(√14/√7-2)-(√14/√7+2)=a√7+b√2 find the values of a and b where a and b are rational numbers

seraphim [82]

Answer:

a = 4/3 and b = 0

============================

<h2>Given expression:</h2>

$\dfrac{\sqrt{14} }{\sqrt{7}-2} -\dfrac{\sqrt{14} }{\sqrt{7}+2}$

<h2>Simplify it in steps:</h2>

Bring both fractions into common denominator:

$\dfrac{\sqrt{14} (\sqrt{7}+2)}{(\sqrt{7}-2)(\sqrt{7}+2)} - \dfrac{\sqrt{14} (\sqrt{7}-2)}{(\sqrt{7}-2)(\sqrt{7}+2)}$

Simplify:

$\dfrac{\sqrt{14} ((\sqrt{7}+2) - (\sqrt{7}-2))}{(\sqrt{7}-2)(\sqrt{7}+2)} =$

$\dfrac{\sqrt{14} (\sqrt{7}+2 - \sqrt{7}+2)}{(\sqrt{7}-2)(\sqrt{7}+2)} =$

$\dfrac{4\sqrt{14} }{(\sqrt{7}-2)(\sqrt{7}+2)} =$

$\dfrac{4\sqrt{14} }{(\sqrt{7})^2-2^2} =$

$\dfrac{4\sqrt{14} }{7-4} =$

$\dfrac{4}{3} \sqrt{14} }$

Compare the result with given expression to get:

a = 4/3 and b = 0

4 0

2 years ago

Derivative, by first principle<br><img src="https://tex.z-dn.net/?f=%20%5Ctan%28%20%5Csqrt%7Bx%20%7D%20%29%20" id="TexFormula1"

vampirchik [111]

$\displaystyle\lim_{h\to0}\frac{\tan\sqrt{x+h}-\tan x}h$

Employ a standard trick used in proving the chain rule:

$\dfrac{\tan\sqrt{x+h}-\tan x}{\sqrt{x+h}-\sqrt x}\cdot\dfrac{\sqrt{x+h}-\sqrt x}h$

The limit of a product is the product of limits, i.e. we can write

$\displaystyle\left(\lim_{h\to0}\frac{\tan\sqrt{x+h}-\tan x}{\sqrt{x+h}-\sqrt x}\right)\cdot\left(\lim_{h\to0}\frac{\sqrt{x+h}-\sqrt x}h\right)$

The rightmost limit is an exercise in differentiating $\sqrt x$ using the definition, which you probably already know is $\dfrac1{2\sqrt x}$ .

For the leftmost limit, we make a substitution $y=\sqrt x$ . Now, if we make a slight change to $x$ by adding a small number $h$ , this propagates a similar small change in $y$ that we'll call $h'$ , so that we can set $y+h'=\sqrt{x+h}$ . Then as $h\to0$ , we see that it's also the case that $h'\to0$ (since we fix $y=\sqrt x$ ). So we can write the remaining limit as

$\displaystyle\lim_{h\to0}\frac{\tan\sqrt{x+h}-\tan\sqrt x}{\sqrt{x+h}-\sqrt x}=\lim_{h'\to0}\frac{\tan(y+h')-\tan y}{y+h'-y}=\lim_{h'\to0}\frac{\tan(y+h')-\tan y}{h'}$

which in turn is the derivative of $\tan y$ , another limit you probably already know how to compute. We'd end up with $\sec^2y$ , or $\sec^2\sqrt x$ .

So we find that

$\dfrac{\mathrm d\tan\sqrt x}{\mathrm dx}=\dfrac{\sec^2\sqrt x}{2\sqrt x}$

7 0

3 years ago