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3 years ago

3. For the equation-4y = &r, what is the constant of variation? (1 point) ООО 02

Mathematics

1 answer:

alexandr1967 [171]3 years ago

7 0

Answer:

3).

$- 4y = 8x \\ \frac{y}{x} = \frac{8}{ - 4} \\ \frac{y}{x} = - 2 \\ y = - 2x \\ hence \: constant \: is \: - 2$

4).

$y \alpha x \\ y = kx \\ where \: k \: is \: the \: constant \: of \: proportionality \\ when \: y = 24 \: and \: x = 8 \\ substitute \: in \: y = kx \\ 24 = 8k \\ k = \frac{24}{8} \\ k = 3 \\ value \: of \: constant = 3$

If it was by 10, k = 3 by 10

k = 30

5).

$general \: equation \: of \: line \\ y = mx + c \\ but \: m = \frac{2}{3} \: and \: c = 9 \\ substitute \: for \: m \: and \: c \: in \: y = mx + c \\ y = \frac{2}{3} x + 9 \\ alternatively \\ 3y = 2x + 27$

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Please answer Thank you please answer hurry PLease

Alik [6]

Answer:

$(80.5x + 69)

Step-by-step Explanation:

Recall, distributive property of multiplication can simply be expressed as follows, x(y + z) = xy + xz or x(y - z). Where x, y, and z could be any number.

Now, let's use the distributive property to show the total costs of plants Gloria decides to buy as an expression and also simplify it.

Thus, Gloria buys:

4 ferns = 4(0.5x - 3) = 4(0.5x) - 4(3) = $(2x - 12)

2 palms = 2(10x - 2.50) = 2(10x) - 2(2.50) = $(20x - 5.00)

10 lilies = 10(5 + 0.25x) = 10(5) + 10(0.25x) = $(50 + 2.5x)

8 shrubs = 8(4.50 + 7x) = 8(4.50) + 8(7x) = $(36 + 56x)

Total cost of the plants = (2x - 12) + (20x - 5) + (50 + 2.5x) + (36+ 56x)

Open the parentheses and then simplify:

Total cost = 2x - 12 + 20x - 5 + 50 + 2.5x + 36 + 56x

= 2x + 20x + 2.5x + 56x - 12 - 5 + 50 + 36

Total cost = $(80.5x + 69)

5 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.