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3 years ago

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Thomas took everyone to lunch and there were 11 people at the table. Everyone’s meal cost the same and the total bill was $115.3

9. How much did each person’s meal cost?

2 answers:

NNADVOKAT [17]3 years ago

8 0

Answer: $10.49

Step-by-step explanation:

Each person's meal cost

115.39 / 11 = 10.49

victus00 [196]3 years ago

5 0

Answer:

1269.29

Step-by-step explanation:

Thomas took everyone to lunch and there were 11 people at the table. Everyone’s meal cost the same and the total bill was $115.39. How much did each person’s meal cost?

i would do the equation 11x115.39

1x9 1x3 1x5 1x1 1x1

same thing but 0 to end

full number 126929 but remember to add the decimal point

1269.29

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Simplify each of the expressions. (6y^2 + 4y + 5) – (3 – 7y + y^2)

Ludmilka [50]

<span> (6y^2 + 4y + 5) – (3 – 7y + y^2)
= </span><span> 6y^2 + 4y + 5 – 3 + 7y - y^2
= 5</span>y^2 + 11y + 2

hope it helps

8 0

3 years ago

Read 2 more answers

An aircraft manufacturer wants to determine the best selling price for a new airplane. The company estimates that the initial co

Blizzard [7]

Answer:

$(a)$ $C(x)=500+20x-5x^{\frac{3}{4}}+0.01x^2$

$p(x)=320-7.7p$

$R(x)=(320-7.7p)p=320p-7.7p^2$

$(b)$ $x=82 \text{planes}$

$(c)$ $p=\$30.91 M\;\; \text{per plane}$

$(d)$ maximum profit $=\$ 15.90M$

Step-by-step explanation:

Given that,

The company estimates that the initial cost of designing the aeroplane and setting up the factories in which to build it will be 500 million dollars.

The additional cost of manufacturing each plane can be modelled by the function.

$m(x)=20x-5x^{\frac{3}{4}}+0.01x^2$

$(a)$ Find the cost, demand (or price), and revenue functions.

$C(x)=500+20x-5x^{\frac{3}{4}}+0.01x^2$

$p(x)=320-7.7p$

$R(x)=(320-7.7p)p=320p-7.7p^2$

$(b)$ Find the production level that maximizes profit.

$f=R(x)-C(x)$

$\Rightarrow f=320p-7.7p^2-(500+20x-5x^{\frac{3}{4}}+0.01x^2)$

$\Rightarrow df=320dp-15.4pdp-20dx+5(\frac{3}{4} )x^{\frac{-1}{4} }dx-0.02xdx$

$x=320-7.7p$

$p=\frac{320-x}{7.7}$

$\frac{dp}{dx} = \frac{-1}{7.7}$

$\frac{df}{dx}=\frac{320}{-7.7} -\frac{15.4(320-x) }{7.7(\frac{-1}{7.7} )}-20+5\frac{3}{4} x^{\frac{-1}{4}} -0.02x=0$

$\Rightarrow -41.5584+83.1169-0.2597x-20+3.75x^{\frac{-1}{4} }-0.02x=0$

$\Rightarrow 21.5585+3.75x^{\frac{-1}{4} }-0.279x=0$

$\Rightarrow x=82 \text{planes}$

$(c)$ Find the associated selling price of the aircraft that maximizes profit.

$p=\frac{320-82}{7.7}$

$\Rightarrow p=\$30.91 M\;\; \text{per plane}$

$(d)$ Find the maximum profit.

Manufacturing cost of one plane is:

$m(1)=20-5+0.01$

$=\$15.01 M$

maximum profit $=\$(30.91-15.01)M$

$=\$15.90M$

3 0

3 years ago

Convert 32505 into equivalent number with base 5

velikii [3]

1s -0

5s- 1

25s- 0

125s- 0

625s- 2

3125s- 0

15625s- 2

2020010. I think this is right but would like so confirmation, just taught myself this!

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3 years ago

Some body can help me with a geometric mean maze

Mars2501 [29]

Answer:

See explanation

Step-by-step explanation:

Theorem 1: The length of the altitude to the hypotenuse of a right triangle is the geometric mean of the lengths of the segments of the hypotenuse.

Theorem 2: The length of each leg of a right triangle is the geometric mean of the length of the hypotenuse and the length of the segment of the hypotenuse adjacent to that leg.

1. Start point: By the 1st theorem,

$x^2=25\cdot (49-25)=25\cdot 24=5^2\cdot 2^2\cdot 6\Rightarrow x=5\cdot 2\cdot \sqrt{6}=10\sqrt{6}.$

2. South-East point from the Start: By the 2nd theorem,

$x^2=40\cdot (40+5)=4\cdot 5\cdot 2\cdot 9\cdot 5\Rightarrow x=2\cdot 5\cdot 3\cdot \sqrt{2}=30\sqrt{2}.$

3. West point from the previous: By the 2nd theorem,

$x^2=(32-20)\cdot 32=4\cdot 3\cdot 16\cdot 2\Rightarrow x=2\cdot 4\cdot \sqrt{6}=8\sqrt{6}.$

4. West point from the previous: By the 1st theorem,

$9^2=x\cdot 15\Rightarrow x=\dfrac{81}{15}=\dfrac{27}{5}=5.4.$

5. West point from the previous: By the 2nd theorem,

$10^2=8\cdot (8+x)\Rightarrow 8+x=12.5,\ x=4.5.$

6. North point from the previous: By the 1st theorem,

$x^2=48\cdot 6=6\cdot 4\cdot 2\cdot 6\Rightarrow x=6\cdot 2\cdot \sqrt{2}=12\sqrt{2}.$

7. East point from the previous: By the 2nd theorem,

$x^2=22.5\cdot 30=225\cdot 3\Rightarrow x=15\sqrt{3}.$

8. North point from the previous: By the 1st theorem,

$x^2=7.5\cdot 36=270\Rightarrow x=3\sqrt{30}.$

8. West point from the previous: By the 2nd theorem,

$x^2=12.5\cdot (12.5+13.5)=12.5\cdot 26=25\cdot 13\Rightarrow x=5\sqrt{13}.$

9. North point from the previous: By the 1st theorem,

$12^2=x\cdot 30\Rightarrow x=\dfrac{144}{30}=4.8.$

101. East point from the previous: By the 1st theorem,

$6^2=1.6\cdot (x-1.6)\Rightarrow x-1.6=22.5,\ x=24.1.$

11. East point from the previous: By the 2nd theorem,

$20^2=32\cdot (32-x)\Rightarrow 32-x=12.5,\ x=19.5.$

12. South-east point from the previous: By the 2nd theorem,

$18^2=x\cdot 21.6\Rightarrow x=15.$

13. North point=The end.

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