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3 years ago

Help me plz!?!?! ASAP

Mathematics

2 answers:

Sauron [17]3 years ago

6 0

Limit the amount of vegetables he bought

Alona [7]3 years ago

5 0

Limit the amount of vegetables

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Helppp pleaseee this is due on 10 minutes with explanation pleasee thank you

Nat2105 [25]

Answer:

Jed's Market

Step-by-step explanation:

Find out the cost of one candy bar for each store:

Smith's:

9.00 ÷ 3 = $3.00

Green's:

10.00 ÷ 4 = $2.50

Jed's:

12.00 ÷ 5 = $2.40

Wan's:

It already tells you the price for each: $2.75

Finding the best deal:

Look at all our costs per 1 candy bar. Which one has the smallest price?
Because $2.40 is the smallest price, Jed's Market has the best buy.

I hope this helps!

8 0

2 years ago

Triangle P R Q is shown. Angle R P Q is 39 degrees. The length of P R is 6, the length of R Q is p, and the length of P Q is 8.

KATRIN_1 [288]

According to law of cosines the length of RQ can be written as $p^{2} =6^{2} +8^{2} -2*6*8cos(39)$ .

Given the length PR is 6 , the length of RQ is p, the length of PQ is 8 and the angle RPQ is 39 degrees.

A length of the triangle can be written as according to law of cosines if sides are given and one angle is $a^{2} =b^{2} +c^{2} -2bccos(A)$

We have to just put the values in the above equation.

as $p^{2} =6^{2} +8^{2} -2*6*8cos(39)$ .

p is the side opposite to angle given , the length of other sides are 6 and 8 and angle is 39 degrees.

Hence the side can be written as according to law of cosines if the angle is 39 degrees is as $p^{2} =6^{2} +8^{2} -2*6*8cos(39)$ .

Learn more about trigonometric functions at brainly.com/question/24349828

#SPJ10

7 0

1 year ago

In a circus performance, a monkey is strapped to a sled and both are given an initial speed of 3.0 m/s up a 22.0° inclined track

Aloiza [94]

Answer:

Approximately $0.31\; \rm m$ , assuming that $g = 9.81\; \rm N \cdot kg^{-1}$ .

Step-by-step explanation:

Initial kinetic energy of the sled and its passenger:

$\begin{aligned}\text{KE} &= \frac{1}{2}\, m \cdot v^{2} \\ &= \frac{1}{2} \times 14\; \rm kg \times (3.0\; \rm m\cdot s^{-1})^{2} \\ &= 63\; \rm J\end{aligned}$ .

Weight of the slide:

$\begin{aligned}W &= m \cdot g \\ &= 14\; \rm kg \times 9.81\; \rm N \cdot kg^{-1} \\ &\approx 137\; \rm N\end{aligned}$ .

Normal force between the sled and the slope:

$\begin{aligned}F_{\rm N} &= W\cdot \cos(22^{\circ}) \\ &\approx 137\; \rm N \times \cos(22^{\circ}) \\ &\approx 127\; \rm N\end{aligned}$ .

Calculate the kinetic friction between the sled and the slope:

$\begin{aligned} f &= \mu_{k} \cdot F_{\rm N} \\ &\approx 0.20\times 127\; \rm N \\ &\approx 25.5\; \rm N\end{aligned}$ .

Assume that the sled and its passenger has reached a height of $h$ meters relative to the base of the slope.

Gain in gravitational potential energy:

$\begin{aligned}\text{GPE} &= m \cdot g \cdot (h\; {\rm m}) \\ &\approx 14\; {\rm kg} \times 9.81\; {\rm N \cdot kg^{-1}} \times h\; {\rm m} \\ & \approx (137\, h)\; {\rm J} \end{aligned}$ .

Distance travelled along the slope:

$\begin{aligned}x &= \frac{h}{\sin(22^{\circ})} \\ &\approx \frac{h\; \rm m}{0.375}\end{aligned}$ .

The energy lost to friction (same as the opposite of the amount of work that friction did on this sled) would be:

$\begin{aligned} & - (-x)\, f \\ = \; & x \cdot f \\ \approx \; & \frac{h\; {\rm m}}{0.375}\times 25.5\; {\rm N} \\ \approx\; & (68.1\, h)\; {\rm J}\end{aligned}$ .

In other words, the sled and its passenger would have lost (approximately) $((137 + 68.1)\, h)\; {\rm J}$ of energy when it is at a height of $h\; {\rm m}$ .

The initial amount of energy that the sled and its passenger possessed was $\text{KE} = 63\; {\rm J}$ . All that much energy would have been converted when the sled is at its maximum height. Therefore, when $h\; {\rm m}$ is the maximum height of the sled, the following equation would hold.

$((137 + 68.1)\, h)\; {\rm J} = 63\; {\rm J}$ .

Solve for $h$ :

$(137 + 68.1)\, h = 63$ .

$\begin{aligned} h &= \frac{63}{137 + 68.1} \approx 0.31\; \rm m\end{aligned}$ .

Therefore, the maximum height that this sled would reach would be approximately $0.31\; \rm m$ .

7 0

2 years ago

An airline claims that the no-show rate for passengers is less than 5%. In a sample of 420 randomly selected reservations, 19 we

aleksandr82 [10.1K]

Answer:

An airline claims that the no-show rate for passengers is less than 5%. In a sample of 420 randomly selected reservations, 19 were no-shows. At α=0.01, test the airline's claim. State the sample percentage and round it to three decimal places.

State the hypotheses.

State the critical value(s).

State the test statistics.

State the decision

State the conclusion.

3 0

2 years ago

the sum of three numbers is 131. the third number is 4 times the first. the second number is 5 more than the first. what are the

Sphinxa [80]

Step-by-step explanation:

let the numbers are:

a, b, and c

the equation would be:

a+b+c = 131

c = 4a

b = a+5

a + a+5 +4a = 131

6a +5 = 131

6a = 131-5

6a = 126

a= 126/6

a = 21

b = a+5 = 21+5

b = 26

c = 4a = 4(21)

c = 84

4 0

2 years ago