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3 years ago

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there are 158 4th grade students and 139 fifth grade students going on a field trip .there needs to be at least 1 adult chaperon

e for evrey 8 students .

1 answer:

strojnjashka [21]3 years ago

3 0

Answer:

(D) is the best explanation

Step-by-step explanation:

The total number of students is 158 + 139 = 297. This number can be rewritten as ...

297 = 37 × 8 + 1

That is, if there were 37 chaperones, 1 student would remain without a chaperone. That means the number of chaperones must be increased by 1 to 38. This is best explained by choice D.

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Please help asap I need it

hjlf

When you plot the points and use the vertical line test, you'll see that none of the vertical lines go through two points. Therefore verifying that this set of points represents a function.

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3 years ago

Hey hey hey. help pls

Firdavs [7]

Answer:

Emma spend 8 hours in homework in a week and Jared spends 16 hours in homework in 2 weeks.

Step-by-step explanation:

Two quantities P and Q are said to be in PROPORTION if and only if for all values of P and q, we get:

$\frac{P}{Q} = k$

Here, k : PROPORTIONALITY CONSTANT

Case: 1

5 hour piano practice in 1 week.

So, the piano practice in 2 weeks( 1 x 2) = 2 x 5 = 10 hours

But here, it is given to be 12 hours in 2 weeks.

Hence, NOT PROPORTIONAL.

Case 2:

70 words in 60 seconds.

So, the words typed in 210 seconds ( 60 x 3.5) = 70 x 3.5 = 240 words

But here, it is given to be 280 words in 210 seconds.

Hence, NOT PROPORTIONAL.

Case: 3

8 hour homework in 1 week.

So, the homework done in 2 weeks( 1 x 2) = 2 x 8 = 16 hours

It is given to be 16 hours in 2 weeks.

Hence, PROPORTIONAL.

Case: 4

15 miles in 78 minutes.

So, the distance covered in 102 minutes (78 x 1.3) = 15 x 1.3 = 19.6 miles

But here, it is given to be 17 miles in 102 minutes.

Hence, NOT PROPORTIONAL.

Case: 5

18 miles in 45 minutes.

So, the distance covered in 55 minutes (45 x 1.2) = 18 x 1.2 = 21.6 miles

But here, it is given to be 22 miles in 55 minutes.

Hence, NOT PROPORTIONAL.

Hence, the only proportional case is Emma spend 8 hours in homework in a week and Jared spends 16 hours in homework in 2 weeks.

6 0

3 years ago

Read 2 more answers

How do I do this greatest common factor

zaharov [31]

The greatest common factor is 5

8 0

3 years ago

What is the slope of the line with equation 6x+3y=12?

max2010maxim [7]

Answer:

<h2>The slope m = -2</h2>

Step-by-step explanation:

The slope-intercept form of an equation of a line:

$y=mx+b$

<em>m</em><em> - slope</em>

<em>b</em><em> - y-intercept</em>

<em />

We have the equation in the standard form:

$6x+3y=12$

Convert to the slope intercept form:

$6x+3y=12$ <em>subtract 6x from both sides</em>

$3y=-6x+12$ <em>divide both sides by 3</em>

$y=-2x+4$

slope: <em>m = -2</em>

y-intercept: <em>b = 4</em>

6 0

3 years ago

The angle of elevation from me to the top of a hill is 51 degrees. The angle of elevation from me to the top of a tree is 57 deg

julia-pushkina [17]

Answer:

Approximately $101\; \rm ft$ (assuming that the height of the base of the hill is the same as that of the observer.)

Step-by-step explanation:

Refer to the diagram attached.

Let $\rm O$ denote the observer.
Let $\rm A$ denote the top of the tree.
Let $\rm R$ denote the base of the tree.
Let $\rm B$ denote the point where line $\rm AR$ (a vertical line) and the horizontal line going through $\rm O$ meets. $\angle \rm B\hat{A}R = 90^\circ$ .

Angles:

Angle of elevation of the base of the tree as it appears to the observer: $\angle \rm B\hat{O}R = 51^\circ$ .
Angle of elevation of the top of the tree as it appears to the observer: $\angle \rm B\hat{O}A = 57^\circ$ .

Let the length of segment $\rm BR$ (vertical distance between the base of the tree and the base of the hill) be $x\; \rm ft$ .

The question is asking for the length of segment $\rm AB$ . Notice that the length of this segment is $\mathrm{AB} = (x + 20)\; \rm ft$ .

The length of segment $\rm OB$ could be represented in two ways:

In right triangle $\rm \triangle OBR$ as the side adjacent to $\angle \rm B\hat{O}R = 51^\circ$ .
In right triangle $\rm \triangle OBA$ as the side adjacent to $\angle \rm B\hat{O}A = 57^\circ$ .

For example, in right triangle $\rm \triangle OBR$ , the length of the side opposite to $\angle \rm B\hat{O}R = 51^\circ$ is segment $\rm BR$ . The length of that segment is $x\; \rm ft$ .

$\begin{aligned}\tan{\left(\angle\mathrm{B\hat{O}R}\right)} = \frac{\,\rm {BR}\,}{\,\rm {OB}\,} \; \genfrac{}{}{0em}{}{\leftarrow \text{opposite}}{\leftarrow \text{adjacent}}\end{aligned}$ .

Rearrange to find an expression for the length of $\rm OB$ (in $\rm ft$ ) in terms of $x$ :

$\begin{aligned}\mathrm{OB} &= \frac{\mathrm{BR}}{\tan{\left(\angle\mathrm{B\hat{O}R}\right)}} \\ &= \frac{x}{\tan\left(51^\circ\right)}\approx 0.810\, x\end{aligned}$ .

Similarly, in right triangle $\rm \triangle OBA$ :

$\begin{aligned}\mathrm{OB} &= \frac{\mathrm{AB}}{\tan{\left(\angle\mathrm{B\hat{O}A}\right)}} \\ &= \frac{x + 20}{\tan\left(57^\circ\right)}\approx 0.649\, (x + 20)\end{aligned}$ .

Equate the right-hand side of these two equations:

$0.810\, x \approx 0.649\, (x + 20)$ .

Solve for $x$ :

$x \approx 81\; \rm ft$ .

Hence, the height of the top of this tree relative to the base of the hill would be $(x + 20)\; {\rm ft}\approx 101\; \rm ft$ .

6 0

3 years ago