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Margarita [4]
3 years ago
13

Which is the better by 8-foot of cord for $2.80or 9-foot piece of cord for $2.88​

Mathematics
2 answers:
Dmitrij [34]3 years ago
4 0

Answer:

9-foot piece is better deal by $0.03 per foot

Step-by-step explanation:

2.80/8 = $0.35 per foot (for 8-foot piece)

2.88/9 = $0.32 per foot (for 9-foot piece)

lara31 [8.8K]3 years ago
3 0

Answer:

9 foot peice of cord

Step-by-step explanation:

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A basin of a water fountain is cube shaped and has a volume of 91.125 cubic feet
sertanlavr [38]

Answer: what's 9 + 10 = 21 :) ur stuipood that's ur answer only here for the points sorry dude.

Step-by-step explanation:

8 0
3 years ago
The school that Scott goes to is selling tickets to a fall musical. On the first day of ticket sales the school sold 12 adult ti
harina [27]

Answer:

The price of one adult ticket = $13

The price of one student ticket = $4

Step-by-step explanation:

Let the price of 1 adult ticket = x

Let the price of 1 student ticket = y

On the first day of ticket sales the school sold 12 adult tickets and 10 student tickets for a total of $196.

12 × x + 10 × y = $196

12x + 10y = 196....... Equation 1

The school took in $59 on the second day by selling 3 adult tickets and 5 student tickets

3 × x + 5 × y = $59

3x + 5y = 59.......... Equation 2

Using Elimination method

We eliminate y, by Multiplying equation 1 by 5 and Equation 2 by 10

12x + 10y = 196....... Equation 1 × 5

3x + 5y = 59.......... Equation 2 × 10

60x + 50y = 980....... Equation 3

30x + 50y = 590......... Equation 4

Subtract Equation 4 from Equation 3

30x = 390

x = 390/30

x = $13

Therefore, the price of one adult ticket = $13

Remember: x = $13

3x + 5y = 59.......... Equation 2

Substitute

3(13) + 5y = 59

39 + 5y = 59

5y = 59 - 39

5y = 20

y = 20/5

y = $4

Therefore, the price of one student ticket = $4

8 0
3 years ago
Step 1: pick a number
Bogdan [553]
If I pick n, it becomes 50(2n+5)+1764-2001 this is the expression
7 0
3 years ago
I need help with questions #7 and #8 plz
katen-ka-za [31]

Answer:

7. A = 40.8 deg; B = 60.6 deg; C = 78.6 deg

8. A = 20.7 deg; B = 127.2 deg; C = 32.1 deg

Step-by-step explanation:

Law of Cosines

c^2 = a^2 + b^2 - 2ab \cos C

You know the lengths of the sides, so you know a, b, and c. You can use the law of cosines to find C, the measure of angle C.

Then you can use the law of cosines again for each of the other angles. An easier way to solve for angles A and B is, after solving for C with the law of cosines, solve for either A or B with the law of sines and solve for the last angle by the fact that the sum of the measures of the angles of a triangle is 180 deg.

7.

We use the law of cosines to find C.

18^2 = 12^2 + 16^2 - 2(12)(16) \cos C

324 = 144 + 256 - 384 \cos C

-384 \cos C = -76

\cos C = 0.2

C = \cos^{-1} 0.2

C = 78.6^\circ

Now we use the law of sines to find angle A.

Law of Sines

\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}

We know c and C. We can solve for a.

\dfrac{a}{\sin A} = \dfrac{c}{\sin C}

\dfrac{12}{\sin A} = \dfrac{18}{\sin 78.6^\circ}

Cross multiply.

18 \sin A = 12 \sin 78.6^\circ

\sin A = \dfrac{12 \sin 78.6^\circ}{18}

\sin A = 0.6535

A = \sin^{-1} 0.6535

A = 40.8^\circ

To find B, we use

m<A + m<B + m<C = 180

40.8 + m<B + 78.6 = 180

m<B = 60.6 deg

8.

I'll use the law of cosines 3 times here to solve for all the angles.

Law of Cosines

a^2 = b^2 + c^2 - 2bc \cos A

b^2 = a^2 + c^2 - 2ac \cos B

c^2 = a^2 + b^2 - 2ab \cos C

Find angle A:

a^2 = b^2 + c^2 - 2bc \cos A

8^2 = 18^2 + 12^2 - 2(18)(12) \cos A

64 = 468 - 432 \cos A

\cos A = 0.9352

A = 20.7^\circ

Find angle B:

b^2 = a^2 + c^2 - 2ac \cos B

18^2 = 8^2 + 12^2 - 2(8)(12) \cos B

324 = 208 - 192 \cos A

\cos B = -0.6042

B = 127.2^\circ

Find angle C:

c^2 = a^2 + b^2 - 2ab \cos C

12^2 = 8^2 + 18^2 - 2(8)(18) \cos B

144 = 388 - 288 \cos A

\cos C = 0.8472

C = 32.1^\circ

8 0
3 years ago
6% of $33 is $______<br><br> someone please answer and thanks :) i'll give brainilist &lt;3
Zanzabum

Answer:

1.98

Step-by-step explanation:

7 0
3 years ago
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