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3 years ago

8

Perform the indicated operations and simplify: 8c + 5c - (5c+6)

1 answer:

sashaice [31]3 years ago

5 0

Answer:

Simplify the expression.

8c − 6

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Use the differential equation given by dy/dx=xy/3, y>0<br><br> complete the table of values

melisa1 [442]

The $\dfrac{\mathrm dy}{\mathrm dx}$ row is filled by just plugging in the corresponding values of $x$ and $y$ into the ODE.

For example, when $x=-1$ and $y=1$ , you have

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{xy}3\implies\dfrac{\mathrm dy}{\mathrm dx}\bigg|_{x=-1,y=1}=\dfrac{(-1)(1)}3=-\dfrac13$

and so on.

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3 years ago

Barbara found that 35 of the 60 people she surveyed had pets. About what percent of the students she surveyed had pets?

Sergio [31]

Answer:

35/60 = 0.583333333

= 0.58

0.58 * 100 = 58

58%

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3 years ago

Read 2 more answers

If x square plus 8x plus q =(x plus k) Square find the value of k and q

PIT_PIT [208]

Answer:

The value of k is 4 and q is 16.

Step-by-step explanation:

First, you have to expand the brackets by applying :

${(a + b)}^{2} = {a}^{2} + 2ab + {b}^{2}$

${x}^{2} + 8x + q = {(x + k)}^{2}$

${x}^{2} + 8x + q = {x}^{2} + 2kx + {k}^{2}$

Next, you have to do it by comparison :

${x}^{2} + 8x + q = {x}^{2} + 2kx + {k}^{2}$

$by \: comparison$

$8x = 2kx$

$8 = 2k$

$k = 4$

$q = {k}^{2}$

$q = {4}^{2}$

$q = 16$

7 0

3 years ago

What is h(-2)<br> H(-1.75)<br> H(1)

ruslelena [56]

$h(x)=\left\{\begin{array}{cccc}-2&if&-2\leq x$

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3 years ago

Find the value of x such that the area of a triangle whose vertices have coordinates (6, 5), (8, 2) and (x, 11) is 15 square uni

Serjik [45]

Answer:

$x =12\ or\ -8$

Step-by-step explanation:

Given

$(x_1,y_1) = (6,5)$

$(x_2,y_2) = (8,2)$

$(x_3,y_3) = (x,11)$

$Area = 15$

Required

Find x

The area is calculated as thus:

$Area = \frac{1}{2}|x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$

Substitute values

$15 = \frac{1}{2}|6*(2 - 11) + 8*(11 - 5) + x(5 - 2)|$

$15 = \frac{1}{2}|6*(-9) + 8*(6) + x(3)|$

$15 = \frac{1}{2}|-54 + 48 + 3x|$

$15 = \frac{1}{2}|-6 + 3x|$

Multiply through by 2

$2 * 15 = \frac{1}{2}|-6 + 3x| * 2$

$30 = |-6 + 3x|$

Remove absolute sign

$30 = -6 + 3x\ or\ -30 = -6 + 3x$

Add 6 to both sides

$36 = 3x\ or\ -24 = 3x$

Divide by 3

$12 = x\ or\ -8 = x$

$x =12\ or\ -8$

<em>So, the coordinates are (-8, 11) or (12,11)</em>

3 0

2 years ago