Please answer this as soon as possible.

When Sam gave Clarissa 105 chocolates, Clarissa had 5 times as many chocolates as Sam. They had 504 chocolates altogether. How m

IgorLugansk [536]

Answer:

315 chocolate

Step-by-step explanation:

Let Sam's chocolate be S

Let Clariss's chocolate be C

When Sam gave Clarissa 105 chocolates, Clarissa had 5 times as many chocolates as Sam.

This can be written as:

C = 5S

The sum of their chocolate is 504 i.e

S + C = 504

Now, let us determine the chocolate of Clarissa after receiving 105 chocolate from Sam. This can be obtained as follow:

S + C = 504

But: C = 5S

S + 5S = 504

6S = 504

Divide both side by 6

S = 504/6

S = 84.

C = 5S = 5 x 84 = 420

Therefore, Clarissa have 420 chocolate after receiving 105 chocolate from Sam.

Now, to know the amount of chocolate that Clarissa has at first, we simply subtract 105 from the present amount that Clarissa have. This is illustrated below:

Amount of chocolate that Clarissa has a first = 420 – 105 = 315

Therefore, Clarissa had 315 chocolate at first.

3 0

2 years ago

A mountain climber starts in a cave, then climbs up the mountain. After 1 hour, he is 6 m under ground. After 4 hours, he is 6 m

Nookie1986 [14]

Answer:

In Point Slope Form: y+6=4(x-1)

Step-by-step explanation:

First you have to find the points given which are:

(1, -6) and (4, 6)

Label the points x1, y1, x2, and y2

It already gives you the slope which is 4.

Plot the points in the equation of Point Slope Form:

y1-y2=m(x-x1)

y1-6=4(x-1)

Simplify: y1-6=4x-4

y=4x+2

It doesn't ask you to simplify but why not!

3 0

3 years ago

Can you show me exactly were to plot these?

sveticcg [70]

I need a photo to help you

5 0

3 years ago

Read 2 more answers

Simplify the expression please

Reika [66]

$\bf -2(3+2i)+2(4-5i)\implies 
\\\\\\
-6+8-10i-4i\implies \boxed{2-14i}$

3 0

3 years ago

If the sum of the even integers between 1 and k, inclusive, is equal to 2k, what is the value of k?

Alisiya [41]

If $k$ is odd, then

$\displaystyle\sum_{n=1}^{\lfloor k/2\rfloor}2n=2\dfrac{\left\lfloor\frac k2\right\rfloor\left(\left\lfloor\frac k2\right\rfloor+1\right)}2=\left\lfloor\dfrac k2\right\rfloor^2+\left\lfloor\dfrac k2\right\rfloor$

while if $k$ is even, then the sum would be

$\displaystyle\sum_{n=1}^{k/2}2n=2\dfrac{\frac k2\left(\frac k2+1\right)}2=\dfrac{k^2+2k}4$

The latter case is easier to solve:

$\dfrac{k^2+2k}4=2k\implies k^2-6k=k(k-6)=0$

which means $k=6$ .

In the odd case, instead of considering the above equation we can consider the partial sums. If $k$ is odd, then the sum of the even integers between 1 and $k$ would be

$S=2+4+6+\cdots+(k-5)+(k-3)+(k-1)$

Now consider the partial sum up to the second-to-last term,

$S^*=2+4+6+\cdots+(k-5)+(k-3)$

Subtracting this from the previous partial sum, we have

$S-S^*=k-1$

We're given that the sums must add to $2k$ , which means

$S=2k$
$S^*=2(k-2)$

But taking the differences now yields

$S-S^*=2k-2(k-2)=4$

and there is only one $k$ for which $k-1=4$ ; namely, $k=5$ . However, the sum of the even integers between 1 and 5 is $2+4=6$ , whereas $2k=10\neq6$ . So there are no solutions to this over the odd integers.

5 0

2 years ago