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2 years ago

Simple math. Question 11.

Mathematics

2 answers:

Alborosie2 years ago

8 0

Answer: By 9.1%

Step-by-step explanation:

8_murik_8 [283]2 years ago

8 0

Answer:

8.3333....

Step-by-step explanation:

275/300=1/12

1/12=0.083333...

0.083333...=8.3333%

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The weight of an object on the moon varies directly with its weight on the earth. If an object weighing 95 lbs on the moon weigh

Luda [366]

ANSWER

450lb

EXPLANATION

If the weight, m of an object on the moon varies directly as the weight of an object e, on the earth, then we can write the mathematical statement,.
$m \propto \: e$
We introduce the constant of proportionality to obtain,

$m = ke$

When, m=95, e=570,

This implies that,

$95 = 570k$
$k = \frac{95}{570}$

$k = \frac{1}{6}$

The equation now becomes,

$m = \frac{1}{6} e$

We want to find e, when m=2700,

$m = \frac{1}{6} \times 2700$

$m =450lb$

8 0

3 years ago

At 2 PM, a thermometer reading 80°F is taken outside where the air temperature is 20°F. At 2:03 PM, the temperature reading yiel

Alexeev081 [22]

Use Law of Cooling:
$T(t) = (T_0 - T_A)e^{-kt} +T_A$
T0 = initial temperature, TA = ambient or final temperature

First solve for k using given info, T(3) = 42
$42 = (80-20)e^{-3k} +20 \\ \\ e^{-3k} = \frac{22}{60}=\frac{11}{30} \\ \\ k = -\frac{1}{3} ln (\frac{11}{30})$

Substituting k back into cooling equation gives:
$T(t) = 60(\frac{11}{30})^{t/3} + 20$

At some time "t", it is brought back inside at temperature "x".
We know that temperature goes back up to 71 at 2:10 so the time it is inside is 10-t, where t is time that it had been outside.
The new cooling equation for when its back inside is:
$T(t) = (x-80)(\frac{11}{30})^{t/3} + 80 \\ \\ 71 = (x-80)(\frac{11}{30})^{\frac{10-t}{3}} + 80$
Solve for x:
$x = -9(\frac{11}{30})^{\frac{t-10}{3}} + 80$
Sub back into original cooling equation, x = T(t)
$-9(\frac{11}{30})^{\frac{t-10}{3}} + 80 = 60 (\frac{11}{30})^{t/3} +20$
Solve for t:
$60 (\frac{11}{30})^{t/3} +9(\frac{11}{30})^{\frac{-10}{3}}(\frac{11}{30})^{t/3} = 60 \\ \\ (\frac{11}{30})^{t/3} = \frac{60}{60+9(\frac{11}{30})^{\frac{-10}{3}}} \\ \\ \\ t = 3(\frac{ln(\frac{60}{60+9(\frac{11}{30})^{\frac{-10}{3}}})}{ln (\frac{11}{30})}) \\ \\ \\ t = 4.959$

This means the exact time it was brought indoors was about 2.5 seconds before 2:05 PM

8 0

3 years ago

Leah loves chicken wings and is comparing the deals at three different restaurants. Buffalo Bills has 88 wings for \$7$7. Buffal

Ksivusya [100]

Answer:

Buffalo mild wings offers the lowest price per wing ($0.83).

Step-by-step explanation:

Let us find unit price per buffalo wing of each restaurant.

$\text{Unit rate for buffalo bills}=\frac{\$7}{8\text{ wings}}$

$\text{Unit rate for buffalo bills}=\frac{\$0.875}{\text{ wing}}$

Upon rounding our answer to nearest cent we will get,

$\text{Unit rate for buffalo bills}=\frac{\$0.88}{\text{ wing}}$

Therefore, buffalo bills offers each wing for $0.88.

$\text{Unit rate for buffalo mild wings}=\frac{\$10}{12\text{ wings}}$

$\text{Unit rate for buffalo mild wings}=\frac{\$0.83333}{\text{ wing}}$

Upon rounding our answer to nearest cent we will get,

$\text{Unit rate for buffalo mild wings}=\frac{\$0.83}{\text{ wing}}$

Therefore, buffalo mild wings offers each wing for $0.83.

$\text{Unit rate for wingers}=\frac{\$17}{20\text{ wings}}$

$\text{Unit rate for wingers}=\frac{\$0.85}{\text{ wing}}$

Therefore, wingers offers each wing for $0.85.

We can see that buffalo mild wings offers the lowest price per wing that is $0.83 per wing.

3 0

3 years ago

A closed can, in a shape of a circular, is to contain 500cm^3 of liquid when full. The cylinder, radius r cm and height h cm, is

Gemiola [76]

To express the height as a function of the volume and the radius, we are going to use the volume formula for a cylinder: $V= \pi r^2h$
where
$V$ is the volume
$r$ is the radius
$h$ is the height

We know for our problem that the cylindrical can is to contain 500cm^3 when full, so the volume of our cylinder is 500cm^3. In other words: $V=500cm^3$ . We also know that the radius is r cm and height is h cm, so $r=rcm$ and $h=hcm$ . Lets replace the values in our formula:
$V= \pi r^2h$
$500cm^3= \pi (rcm^2)(hcm)$
$500cm^3=h \pi r^2cm^3$
$h= \frac{500cm^3}{ \pi r^2cm^3}$
$h= \frac{500}{ \pi r^2}$

Next, we are going to use the formula for the area of a cylinder: $A=2 \pi rh+2 \pi r^2$
where
$A$ is the area
$r$ is the radius
$h$ is the height

We know from our previous calculation that $h= \frac{500}{ \pi r^2}$ , so lets replace that value in our area formula:
$A=2 \pi rh+2 \pi r^2$
$A=2 \pi r(\frac{500}{ \pi r^2})+2 \pi r^2$
$A= \frac{1000}{r} +2 \pi r^2$
By the commutative property of addition, we can conclude that:
$A=2 \pi r^2+\frac{1000}{r}$

7 0

3 years ago

Find the slope intercept form of the equation of the line through the points(2,-4) parallel to the line 8x+2y+2=0

Margarita [4]

Answer:

y = -4x + 4

Step-by-step explanation:

8x+2y+2=0

turn that into slope intercept form:

y = -4x - 1

a parallel line has the same slope, but different y-intercept

y = -4x + b

put (2,-4) into the equation

-4 = -4(2) + b

you get b=4

final product:

y = -4x + 4

3 0

3 years ago