HELP ASAP PLEASE FOR A TEST!!!!!

PLEASE HELP QUICK!!! A bowl has 85 pieces of candy. Nineteen children empty the bowl of candy. Some children take 3 pieces, some

Mekhanik [1.2K]

Answer: 3 kids take 3 pieces of candy each

Step-by-step explanation:

Let the number of children that took 3 pieces is x ( total take 3*x pieces of candy)

Number of children that took 5 pieces is y ( total take 5*y pieces of candy)

1 child took 1 piece that actually means that x+y=18 and 3*x+5*y=84.

( Because total number of all kids is 19. We just deduct one kid (Let his name is John) who took only 1 candy. So we have 19-1 =18 kids without John. The similar is with the candies. Total number is 85. We deduct 1 piece which John has taken. )

So we have 2 equations or the system of 2 equations:

1). x+y=18

2). 3*x +5*y=84

Multuply both sides of equation 1) by 3

We have 3*x+3*y=18*3

Deduct 3*y from both sides of this equation

3*x+3*y-3*y=54-3*y

3*x=54-3*y

Substitute 3*x in equation 2). by 54-3*y

2) 54-3*y+5*y=84

2*y=30

y=15 ( kids take 5 pieces of candy each)

Using equation 1) find x

x+15=18

x=3 (kids take 3 pieces of candy each)

5 0

3 years ago

Which equation has no solution?

EastWind [94]

Answer: It should be the third one

Step-by-step explanation:

7 0

2 years ago

Find a such that the solution of y'' + y' − 2y = 0, y(0) = a, y' (0) = 1 tends to zero as t → +∞.

Kay [80]

Answer:

Step-by-step explanation:

$y'' + y' − 2y = 0, y(0) = a, y' (0) = 1$

Auxialary equation is

$m^2+m-2=0\\m=-2,1$

General solution is

$y=Ae^{-2x} +Be^x$

$y(0) = A+B =a$

$y'(0) = -2Ae^{2x} +Be^x = -2A+B = 1$

Eliminate B to get

3A =a-1

We know that y tends to 0 when x tends to infinity for any finite A

i.e. a should be a finite real number.

8 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 $\rm \triangle OBA$ :

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