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3 years ago

A school typically sells 500 yearbooks in a year for $50 dollars each.

Mathematics

2 answers:

Evgen [1.6K]3 years ago

8 0

You divide 500 by $50 to get 10; the school sold 10 yearbooks that year if you’re looking for that as an answer.

HACTEHA [7]3 years ago

3 0

Answer:

Step-by-step explanation:

They should sell each yearbook 5 dollars cheaper because they will make 4,500 dollars more if they did that. With the discount and extra yearboos they will earn $29,250.

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Find a1 if Sn = 89,800, r = 3.4, and n = 10. Round to the nearest hundredth if necessary.

xxMikexx [17]

Answer:

$a_1=1.04$

Step-by-step explanation:

We have a geometric sequence with:

$Sn = 89,800$ , $r = 3.4$ , and $n = 10$

Where

Sn is the sum of the sequence

r is the common ratio

$a_1$ is the first term in the sequence

n is the number of terms in the sequence

The formula to calculate the sum of a finite geometric sequence is:

$S_n=\frac{a_1(1-r^n)}{1-r}$

Then:

$89,800=\frac{a_1(1-(3.4)^{10})}{1-3.4}$

Now we solve for $a_1$

$89,800(1-3.4)=a_1(1-(3.4)^{10})$

$a_1=\frac{89,800(1-3.4)}{1-(3.4)^{10}}\\\\a_1=1.04$

4 0

3 years ago

Please answer quickly thanks :)

Bezzdna [24]

b< 11: The board is shorter than 11 cm

b > 11: The box contains more than 11 books

b < 12: There are fewer than 12 beetles in a jar.

b > 12: The building is taller than 12 ft

5 0

2 years ago

7.6x = 64 is approximately 3.05.<br> is this the solution?

lisov135 [29]

7.6x = 64
x = 64/7.6
x = <span> <span> <span> 8.4210526316 </span> </span> </span>
So, 3.05 is NOT the solution

3 0

3 years ago

The equation t^3=a^2 shows the relationship between a planet’s orbital period, T, and the planet’s mean distance from the sun, A

ankoles [38]

Answer:

2√2

Step-by-step explanation:

We can find the relationship of interest by solving the given equation for A, the mean distance.

<h3>Solve for A</h3>

$T^3=A^2\\\\A=\sqrt{T^3}=T\sqrt{T}\qquad\text{take the square root}$

<h3>Substitute values</h3>

The mean distance of planet X is found in terms of its period to be ...

$D_x=T_x\sqrt{T_x}$

The mean distance of planet Y can be found using the given relation ...

$T_y=2T_x\\\\D_y=T_y\sqrt{T_y}=2T_x\sqrt{2T_x}=(2\sqrt{2})T_x\sqrt{T_x}\\\\D_y=2\sqrt{2}\cdot D_x$

The mean distance of planet Y is increased from that of planet X by the factor ...

2√2

8 0

1 year ago

AB is parallel to ED.

ivolga24 [154]

Answer:

BE = 22.4 cm

Step-by-step explanation:

Δ CAB and Δ CDE are similar , then ratios of corresponding sides are equal, that is

$\frac{CB}{CE}$ = $\frac{AB}{DE}$ , substitute values

$\frac{6.4}{CE}$ = $\frac{8}{20}$ ( cross- multiply )

8 CE = 128 ( divide both sides by 8 )

CE = 16 cm

Then

BE = BC + CE = 6.4 + 16 = 22.4 cm

4 0

2 years ago