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3 years ago

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Heather's school is switching from block scheduling to a more traditional schedule. She is trying to determine how the parents a

nd students feel about this. Which question is the MOST appropriate for her to ask?

1 answer:

k0ka [10]3 years ago

8 0

Had to look for the options for this question and here is my answer.
Based on the given scenario above regarding Heather school's scheduling, I can say that the question that would be the most appropriate for her to ask is "How do you feel about your child having to take summer school in order to graduate based on this traditional schedule?" Hope this helps.

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Can someone help me with these two problems and show work please !!

irina1246 [14]

Work is attached to this post

5 0

2 years ago

Read 2 more answers

A golfer hits an errant tee shot that lands in the rough. A marker in the center of the fairway is 150 yards from the center of

Gwar [14]

$\bold{\huge{\underline{ Solution}}}$

<h3><u>Given </u><u>:</u><u>-</u></h3>

A marker in the center of the fairway is 150 yards away from the centre of the green
While standing on the marker and facing the green, the golfer turns 100° towards his ball
Then he peces off 30 yards to his ball

<h3><u>To </u><u>Find </u><u>:</u><u>-</u></h3>

<u>We </u><u>have </u><u>to </u><u>find </u><u>the </u><u>distance </u><u>between </u><u>the </u><u>golf </u><u>ball </u><u>and </u><u>the </u><u>center </u><u>of </u><u>the </u><u>green </u><u>.</u>

<h3><u>Let's </u><u> </u><u>Begin </u><u>:</u><u>-</u></h3>

Let assume that the distance between the golf ball and central of green is x

<u>Here</u><u>, </u>

Distance between marker and centre of green is 150 yards
<u>That </u><u>is</u><u>, </u>Height = 150 yards
For facing the green , The golfer turns 100° towards his ball
<u>That </u><u>is</u><u>, </u>Angle = 100°
The golfer peces off 30 yards to his ball
<u>That </u><u>is</u><u>, </u>Base = 30 yards

<u>According </u><u>to </u><u>the </u><u>law </u><u>of </u><u>cosine </u><u>:</u><u>-</u>

$\bold{\red{ a^{2} = b^{2} + c^{2} - 2ABcos}}{\bold{\red{\theta}}}$

Here, a = perpendicular height
b = base
c = hypotenuse
cos theta = Angle of cosine

<u>So</u><u>, </u><u> </u><u>For </u><u>Hypotenuse </u><u>law </u><u>of </u><u>cosine </u><u>will </u><u>be </u><u>:</u><u>-</u>

$\sf{ c^{2} = a^{2} + b^{2} - 2ABcos}{\sf{\theta}}$

<u>Subsitute </u><u>the </u><u>required </u><u>values</u><u>, </u>

$\sf{ x^{2} = (150)^{2} + (30)^{2} - 2(150)(30)cos}{\sf{100°}}$

$\sf{ x^{2} = 22500 + 900 - 900cos}{\sf{\times{\dfrac{5π}{9}}}}$

$\sf{ x^{2} = 22500 + 900 - 900( - 0.174)}$

$\sf{ x^{2} = 22500 + 900 + 156.6}$

$\sf{ x^{2} = 23556.6}$

$\bold{ x = 153.48\: yards }$

Hence, The distance between the ball and the center of green is 153.48 or 153.5 yards

5 0

2 years ago

Paladinen [302]

Answer:

d (w) = 0.1w² – w+ 5

Step-by-step explanation:

d(w) = c(w) - a(w)

= -0.3w²+2w+13 -(-0.4w²+3w+8)

= -0.3w²+0.4w²+2w-3w+13-8

= 0.1w²-w+5

4 0

3 years ago

Nicks iPhone cover is a rectangle that is 1 1/2 inches long, the width is 7/8 of the length. What is the area of his phone cover

Irina-Kira [14]

Sorry if it’s confusing
Answer is 1 3/4

7 0

3 years ago

The first three terms of an arithmetic series are 6p+2, 4p²-10 and 4p+3 respectively. Find the possible values of p. Calculate t

Doss [256]

Answer:

First Case:

$\displaystyle p=\frac{5}{2}\text{ and } d=-2$

Second Case:

$\displaystyle p=-\frac{5}{4}\text{ and } d=\frac{7}{4}$

Step-by-step explanation:

We know that the first three terms of an arithmetic series are:

$6p+2, 4p^2-10, \text{ and } 4p+3$

Since this is an arithmetic sequence, each subsequent term is <em>d</em> more than the previous term, where <em>d</em> is our common difference.

Therefore, we can write the second term as;

$4p^2-10=(6p+2)+d$

And, likewise, for the third term:

$4p+3=(6p+2)+2d$

Let's solve for <em>d</em> for each of the equations.

Subtracting in the first equation yields:

$d=4p^2-6p-12$

And for the second equation:

$2d=-2p+1$

To avoid fractions, let's multiply the first equation by 2. Hence:

$2d=8p^2-12p-24$

Therefore:

$8p^2-12p-24=-2p+1$

Simplifying yields:

$8p^2-10p-25=0$

Solve for <em>p</em>. We can factor:

$8p^2+10p-20p-25=0$

Factor:

$2p(4p+5)-5(4p+5)=0$

Grouping:

$(2p-5)(4p+5)=0$

Zero Product Property:

$\displaystyle p_1=\frac{5}{2} \text{ or } p_2=-\frac{5}{4}$

Then, we can use the second equation to solve for <em>d</em>. So:

$2d_1=-2p_1+1$

Substituting:

$\begin{aligned} 2d_1&=-2(\frac{5}{2})+1 \\ 2d_1&=-5+1 \\ 2d_1&=-4 \\ d_1&=-2\end{aligned}$

So, for the first case, <em>p</em> is 5/2 and <em>d</em> is -2.

Likewise, for the second case:

$\begin{aligned} 2d_2&=-2(-\frac{5}{4})+1 \\ 2d_2&=\frac{5}{2}+1 \\ 2d_2&=\frac{7}{2} \\ d_2&=\frac{7}{4}\end{aligned}$

So, for the second case, <em>p </em>is -5/4, and <em>d</em> is 7/4.

By using the values, we can determine our series.

For Case 1, we will have:

17, 15, 13.

For Case 2, we will have:

-11/2, -15/4, -2.

8 0

2 years ago