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mihalych1998 [28]

4 years ago

9

The volume of a box(V) varies directly with its length(l). If one of the boxes has a volume of 288 cubic inches and a length of

12 inches, what is the constant of proportionality for the group of boxes?

1 answer:

ziro4ka [17]4 years ago

8 0

Answer:

24

Step-by-step explanation:

Varies directly means that something constant times l equals V or V=kl where k is that constant.

Plug in point given 288=k(12) and solve the constant of proportionality by dividing both sides by 12.

k=288/12=24

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Steve is buying a sandwich for lunch and some fresh fruit juice for his friends. The sandwich costs $3.92 and the fruit juice co

Jobisdone [24]

Answer:

Steve is buying a sandwich for lunch and some fresh fruit juice for his friends. The sandwich costs $3.92 and the fruit juice costs $0.80 per ounce. He can spend at most $23.52 on all the food items. How many ounces of juice, x, can he purchase and stay within his budget? Select the inequality that includes the largest amount of ounces of juice he can purchase and still stay under his budget

Step-by-step explanation: Steve is buying a sandwich for lunch and some fresh fruit juice for his friends. The sandwich costs $3.92 and the fruit juice costs $0.80 per ounce. He can spend at most $23.52 on all the food items. How many ounces of juice, x, can he purchase and stay within his budget? Select the inequality that includes the largest amount of ounces of juice he can purchase and still stay under his budget

3 0

4 years ago

The service elevator in a high-rise building travels between the lowest underground parking level and the seventeenth story of t

Vitek1552 [10]

15 + 51 = 66 Meters total

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6 0

4 years ago

Read 2 more answers

The n term of a geometric sequence is denoted by Tn and the sum of the first n terms is denoted by Sn.Given T6-T4=5/2 and S5-S3=

Leno4ka [110]

1 step: $S_{5}=T_{1}+T_{2}+T_{3}+T_{4}+T_{5}$ , $S_{3}=T_{1}+T_{2}+T_{3}$ , then
$S_{5}-S_{3}=T_{4}+T_{5}=5$ .

2 step: $T_{n}=T_{1}*q^{n-1}$ , then
$T_{6}=T_{1}*q^{5}$
$T_{5}=T_{1}*q^{4}$
$T_{4}=T_{1}*q^{3}$
$T_{3}=T_{1}*q^{2}$
and $\left \{ {{T_{6}-T_{4}= \frac{5}{2} } \atop {T_{5}+T_{4}=5}} \right.$ will have form $\left \{ {{T_1*q^{5}-T_{1}*q^{3}= \frac{5}{2} } \atop {T_{1}*q^{4}+T_{1}*q^{3}=5} \right.$ .

3 step: Solve this system $\left \{ {{T_1*q^{3}*(q^{2}-1)= \frac{5}{2} } \atop {T_{1}*q^{3}*(q+1)=5} \right.$ and dividing first equation on second we obtain $\frac{q^{2}-1}{q+1}= \frac{ \frac{5}{2} }{5}$ . So, $\frac{(q-1)(q+1)}{q+1} = \frac{1}{2}$ and $q-1= \frac{1}{2}$ , $q= \frac{3}{2}$ - the common ratio.

4 step: Insert $q= \frac{3}{2}$ into equation $T_{1}*q^{3}*(q+1)=5$ and obtain $T_{1}* \frac{27}{8}*( \frac{3}{2}+1 ) =5$ , from where $T_{1}= \frac{16}{27}$ .

5 0

3 years ago

NEED HELP!! Trigonometry questions

anzhelika [568]

Answer:

Step-by-step explanation:

Question 1: Assumption: This is a 30-60-90 triangle.

Remember that the sides of a 30-60-90 triangle are in the ratio 1:√3:2

The side opposite the 90° angle is 16, so the side opposite the 30° angle is 16/2 = 8

x = 8 units.

:::::

Question 2: Assumption: This is an isosceles triangle.

Draw the altitude to the vertex angle and you get a 30-60-90 triangle.

The side opposite the 90° angle has length 22, so the side opposite the 30° angle has length 11.

x/2 = 11

x = 22 units

:::::

Question 3: Assumption: This is a 45-45-90 triangle.

The sides of a 45-45-90 triangle are in the ratio 1:1:√2

The sides opposite the 45° angles are 19 and x.

x = 19

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Question 4: Assumption: This is an isosceles triangle.

x = 13 units

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Question 5: Assumption: This is a right triangle.

sin(54°) = x/45

x = 45sin(54°) ≅ 36.4 units

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Question 6: Assumption: This is a right triangle.

sin(35°) = z/23

z = 23sin(35°) ≅ 13.2 units

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3 years ago

Express the ratio of A's to N's in the word BANANA, in simplest form.

Ilya [14]

Answer:

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3 years ago