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3 years ago

15

At First Class Pizza, 12% of the pizzas made yesterday were pepperoni pizzas. If 27 pizzas have pepperoni, how many pizzas were

made in all? please answer ASAP:(

1 answer:

gogolik [260]3 years ago

3 0

Answer: 225

Step-by-step explanation:

Let the total number of pizza made be represented by x.

Therefore, the situation in the question can be written as:

12% × x = 27

12/100 × x = 27

0.12 × x = 27

0.12x = 27

Divide both side by 0.12

0.12x/0.12 = 27/0.12

x = 225

The total pizza made was 225.

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15 points each answer!<br> Will mark brainliest! Absurd answers will be reported.

marin [14]

Answer:

C x = 8

Step-by-step explanation:

x = 8 is the domain of the given function.

4 0

3 years ago

Let w(s,t)=f(u(s,t),v(s,t)) where u(1,0)=−6,∂u∂s(1,0)=5,∂u∂1(1,0)=7 v(1,0)=−8,∂v∂s(1,0)=−8,∂v∂t(1,0)=6 ∂f∂u(−6,−8)=−1,∂f∂v(−6,−8

Blababa [14]

$w(s,t)=f(u(s,t),v(s,t))$

From the given set of conditions, it's likely that you are asked to find the values of $\dfrac{\partial w}{\partial s}$ and $\dfrac{\partial w}{\partial t}$ at the point $(s,t)=(1,0)$ .

By the chain rule, the partial derivative with respect to $s$ is

$\dfrac{\partial w}{\partial s}=\dfrac{\partial f}{\partial u}\dfrac{\partial u}{\partial s}+\dfrac{\partial f}{\partial v}\dfrac{\partial v}{\partial s}$

and so at the point $(1,0)$ , we have

$\dfrac{\partial w}{\partial s}\bigg|_{(s,t)=(1,0)}=\dfrac{\partial f}{\partial 
u}\bigg|_{(u,v)=(-6,-8)}\dfrac{\partial u}{\partial s}\bigg|_{(s,t)=(1,0)}+\dfrac{\partial f}{\partial 
v}\bigg|_{(u,v)=(-6,-8)}\dfrac{\partial v}{\partial s}\bigg|_{(s,t)=(1,0)}$
$\dfrac{\partial w}{\partial s}\bigg|_{(s,t)=(1,0)}=(-1)(5)+(2)(-8)=-21$

Similarly, the partial derivative with respect to $t$ would be found via

$\dfrac{\partial w}{\partial t}\bigg|_{(s,t)=(1,0)}=\dfrac{\partial f}{\partial 
u}\bigg|_{(u,v)=(-6,-8)}\dfrac{\partial u}{\partial t}\bigg|_{(s,t)=(1,0)}+\dfrac{\partial f}{\partial 
v}\bigg|_{(u,v)=(-6,-8)}\dfrac{\partial v}{\partial t}\bigg|_{(s,t)=(1,0)}$
$\dfrac{\partial w}{\partial t}\bigg|_{(s,t)=(1,0)}=(-1)(7)+(2)(6)=5$

6 0

3 years ago

Please help asap!! :)

DochEvi [55]

Answer:

$C\approx 5.14$

Step-by-step explanation:

The circumference is the distance around a complete circle: in essence, the perimeter of a circle. The formula to find the circumference of a circle is the following:

$C=(d)(\pi)$

Where (d) represents the diameter, the largest cord that can be drawn in a circle, it goes from one end to another, passing through the center of the circle. The parameter ( $\pi$ ) represents the number (3.1415...). However, one is given a semicircle, thus one must divide the result in two, and then add the value of the diameter to find the perimeter of the given object. Therefore the equation for the circumference of this circle is as follows:

$C=\frac{(d)(\pi)}{2}+(d)$

Substitute in the given value,

$C=\frac{(2)(3.1415)}{2}+(2)$

Simplify,

$C=3.1415...+2$

$C=5.1415$

$C\approx 5.14$

3 0

3 years ago

I really need help guys!!!

kifflom [539]

What exactly do you need help with??

8 0

3 years ago

On Martin's first stroke, his golf ball traveled 4/5 of the distance to the hole. On his second stroke, the ball traveled 79 met

Sav [38]

Answer:

0.395 kilometre

Step-by-step explanation:

Given:

On Martin's first stroke, his golf ball traveled 4/5 of the distance to the hole.

On his second stroke, the ball traveled 79 meters and went into the hole.

<u>Question asked:</u>

How many kilometres from the hole was Martin when he started?

<u>Solution:</u>

Let distance from Martin starting point to the hole in meters = $x$

On Martin's first stroke, ball traveled = $\frac{4}{5} \ of \ total \ distance\ to\ the\ hole$

$=\frac{4}{5} \times x=\frac{4x}{5}$

On his second stroke, the ball traveled and went to the hole = 79 meters

Total distance from starting point to the hole = Ball traveled from first stroke + Ball traveled from second stroke

$x=\frac{4x}{5} +79\\ \\ Subtracting\ both\ sides\ by \ \frac{4x}{5}\\ \\ x- \frac{4x}{5}= \frac{4x}{5}- \frac{4x}{5}+79\\ \\ \frac{5x-4x}{5} =79\\ \\ By \ cross\ multiplication\\ \\ x=79\times5\\ \\ x=395\ meters$

Now, convert it into kilometre:

1000 meter = 1 km

1 meter = $\frac{1}{1000}$

395 meters = $\frac{1}{1000}\times395=0.395\ kilometre$

Thus, there are 0.395 kilometre distance from Martin starting point to the hole.

8 0

3 years ago