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2 years ago

5

Among all pairs of numbers whose difference is 16 find a pair whose product is as small as possible. What is the minimum produc

t?

1 answer:

avanturin [10]2 years ago

4 0

16,0
16-0=16
16*0=0
16,0 is the pair with the smallest product

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Alexandra and her children went into a grocery store and she bought $7.65 worth of

zalisa [80]

Answer:

$\large \boxed{\text{3 apples and 6 bananas}}$

Step-by-step explanation:

Let a = the number of apples

And b = the number of bananas. Then,

$\begin{array}{rcll}(1) \, 1.75a + 0.40 b & = & 7.65&\\(2)\, \, \, \, \quad \qquad a + b & = &9&\\(3) \, 0.40a + 0.40b & = &3.60&\text{Multiplied (2) by 0.40}\\1.35a & =&4.05\\(4) \, \, \quad \qquad \qquad a & =& \mathbf{3}&\text{Divided each side by 1.35}\\3 + b & = &9& \text{Substituted(4) into (2)}\\b &=& \mathbf{6}&\text{Subtracted 3 from each side}\\\end{array}\\\text{They bought $\large \boxed{\textbf{3 apples and 6 bananas}}$}$

Check:

$\begin{array}{cccl}3(1.75) + 6(0.40) = 7.65 & \qquad & 3 + 6 = 9\\5.25 + 2.40 = 7.65 & \qquad & 9 = 9\\7.65 = 7.65 & \qquad & \\\end{array}$

OK.

3 0

1 year ago

If a fish experiences an acceleration of 9.8 m/s^2 over a period of 1.4s (It must be falling), then what is its change in veloci

JulsSmile [24]

Answer:

Change In velocity= 13.72 ms^-1

Step-by-step explanation:

fish experiences an acceleration of

9.8 m/s^2 over a period of 1.4s

Change in velocity/time= acceleration

Time in seconds= 1.4 s

Acceleration in meter per second square= 9.8 m/s^2

Change in velocity/time= acceleration

Change In velocity= acceleration*time

Change In velocity= 9.8*1.4(m/s^2 *s)

Change In velocity= 13.72 ms^-1

6 0

2 years ago

Triangle PQR is dilated by a scale factor of 1.5 to form triangle P′Q′R′. This triangle is then dilated by a scale factor of 2 t

Ghella [55]

Answer: The correct option is (D) P″(9, -12) and Q″(15, -3).

Step-by-step explanation: Given that triangle PQR is dilated by a scale factor of 1.5 to form triangle P′Q′R′. This triangle is then dilated by a scale factor of 2 to form triangle P″Q″R″.

The co-ordinates of vertices P and Q are (3, -4) and (5, -1) respectively.

We are to find the co-ordinates of the vertices P″ and Q″.

<u>Case I :</u> ΔPQR dilated to ΔP'Q'R'

The co-ordinates of P' and Q' are given by

$P'(3\times 1.5, -4\times 1.5)=P'(4.5, -6),\\\\Q'(5\times 1.5, -1\times 1.5)=Q'(7.5, -1.5).$

<u>Case II :</u> ΔP'Q'R' dilated to ΔP''Q''R''

The co-ordinates of P'' and Q'' are given by

$P''(4.5\times 2, -6\times 2)=P'(9, -12),\\\\Q'(7.5\times 2, -1.5\times 2)=Q'(15, -3).$

Thus, the co-ordinates of the vertices P'' and Q'' are (9, -12) and (15, -3).

Option (D) is CORRECT.

<u></u>

4 0

2 years ago

Read 2 more answers

Ava wants to figure out the average speed she is driving. She starts checking her car’s clock at mile marker 0. It takes her 4 m

VLD [36.1K]

<h2>Answer:</h2>

The average speed of car is:

0.75 miles per minutes.

The equation of line is:

$n=\dfrac{3}{4}t$

<h2>Step-by-step solution:</h2>

The table that describes the time and number of miles marked is given by:

Time Miles

0 0

4 3

8 6

Clearly we could observe that with the increase in time by every 4 minutes the number of mile increases by 3.

i.e. the average speed is calculated as the ratio of change in miles to the change in time.

Hence,

$Average\ Speed=\dfrac{3-0}{4-0}\\\\\\i.e.\\\\\\Average\ Speed=\dfrac{3}{4}=0.75$

Hence, the average speed=0.75 miles per minutes.

Also we will find the equation of the lines that related the miles and time by using the two-point formula.

i.e. any line passing through (a,b) and (c,d) is calculated by using the formula:

$y-b=\dfrac{d-b}{c-a}\times (x-a)$

Here (a,b)=(0,0) and (c,d)=(4,3)

i.e. the equation is given by:

$n-0=\dfrac{3-0}{4-0}\times (t-0)\\\\i.e.\\\\n=\dfrac{3}{4}t$

6 0

1 year ago

Read 2 more answers

The weights of certain machine components are normally distributed with a mean of 8.01 g and a standard deviation of 0.06 g. Fin

Free_Kalibri [48]

Answer:

Option D) 7.90 g and 8.12 g

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 8.01 g

Standard Deviation, σ = 0.06 g

We are given that the distribution of weights is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

We have to find the value of x such that the probability is 0.03

P(X > x)

$P( X > x) = P( z > \displaystyle\frac{x - 8.01}{0.06})=0.03$

$= 1 -P( z \leq \displaystyle\frac{x - 8.01}{0.06})=0.03$

$=P( z \leq \displaystyle\frac{x - 8.01}{0.06})=0.97$

Calculation the value from standard normal z table, we have,

$\displaystyle\frac{x - 8.01}{0.06} = 1.881\\\\x = 8.12$

Thus, 8.17 g separates the top 3% of the weights.

P(X < x)

$P( X < x) = P( z < \displaystyle\frac{x - 8.01}{0.06})=0.03$

Calculation the value from standard normal z table, we have,

$\displaystyle\frac{x - 8.01}{0.06} = -1.881\\\\x = 7.90$

Thus, 7.90 separates the bottom 3% of the weights.

Thus, the correct answer is

Option D) 7.90 g and 8.12 g

7 0

2 years ago