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2 years ago

11

An ice company has y lbs of ice in stock. The company sells 500 lbs of ice and then receives a new shipment of 5x lbs of ice. Wh

ich expression represents the
weight of ice the ice company has now?

1 answer:

julsineya [31]2 years ago

8 0

Answer:

yes

Step-by-step explanation:

no sabo

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If f(x) = 1-x^2+x^3, then f(-1)=

Yuri [45]

Answer: -1

Step-by-step explanation:

1-(-1^2)+(-1^3) = 1-1+(-1)= -1

7 0

2 years ago

pls help i need this now subtract -9 1/3 - ( - 1 1/3 enter your answer as a simplified fraction in the box

Neko [114]

Answer:

-8/1 or -8

Step-by-step explanation:

when you subtract a negative, it means add. so first lets add the one. we get -8 1/3. then we can add the 1/3. our final answer is -8

4 0

2 years ago

The length of a rectangle is 8 inches more than twice its width. The area of the rectangle is 120 square inches. How many inches

Irina-Kira [14]

Answer:

L = 20 inches

Step-by-step explanation:

w = width

L = length

area = 120

W x L = 120

L - 8 = 2W then L = 2W + 8

substitute for L:

W x (2W + 8) = 120

2W² + 8W -120 = 0

(2W - 12)(W + 10) = 0

2W-12 = 0

2W = 12

W = 6

L = 20

7 0

3 years ago

If RTBA is a rhombus, solve for y and find the m<RAB

navik [9.2K]

Have a nice day!!!!

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2 years ago

Find the two intersection points

bogdanovich [222]

Answer:

Our two intersection points are:

$\displaystyle (3, -2) \text{ and } \left(-\frac{53}{25}, \frac{46}{25}\right)$

Step-by-step explanation:

We want to find where the two graphs given by the equations:

$\displaystyle (x+1)^2+(y+2)^2 = 16\text{ and } 3x+4y=1$

Intersect.

When they intersect, their <em>x-</em> and <em>y-</em>values are equivalent. So, we can solve one equation for <em>y</em> and substitute it into the other and solve for <em>x</em>.

Since the linear equation is easier to solve, solve it for <em>y: </em>

<em /> $\displaystyle y = -\frac{3}{4} x + \frac{1}{4}$ <em />

<em />

Substitute this into the first equation:

$\displaystyle (x+1)^2 + \left(\left(-\frac{3}{4}x + \frac{1}{4}\right) +2\right)^2 = 16$

Simplify:

$\displaystyle (x+1)^2 + \left(-\frac{3}{4} x + \frac{9}{4}\right)^2 = 16$

Square. We can use the perfect square trinomial pattern:

$\displaystyle \underbrace{(x^2 + 2x+1)}_{(a+b)^2=a^2+2ab+b^2} + \underbrace{\left(\frac{9}{16}x^2-\frac{27}{8}x+\frac{81}{16}\right)}_{(a+b)^2=a^2+2ab+b^2} = 16$

Multiply both sides by 16:

$(16x^2+32x+16)+(9x^2-54x+81) = 256$

Combine like terms:

$25x^2+-22x+97=256$

Isolate the equation:

$\displaystyle 25x^2 - 22x -159=0$

We can use the quadratic formula:

$\displaystyle x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}$

In this case, <em>a</em> = 25, <em>b</em> = -22, and <em>c</em> = -159. Substitute:

$\displaystyle x = \frac{-(-22)\pm\sqrt{(-22)^2-4(25)(-159)}}{2(25)}$

Evaluate:

$\displaystyle \begin{aligned} x &= \frac{22\pm\sqrt{16384}}{50} \\ \\ &= \frac{22\pm 128}{50}\\ \\ &=\frac{11\pm 64}{25}\end{aligned}$

Hence, our two solutions are:

$\displaystyle x_1 = \frac{11+64}{25} = 3\text{ and } x_2 = \frac{11-64}{25} =-\frac{53}{25}$

We have our two <em>x-</em>coordinates.

To find the <em>y-</em>coordinates, we can simply substitute it into the linear equation and evaluate. Thus:

$\displaystyle y_1 = -\frac{3}{4}(3)+\frac{1}{4} = -2$

And:

$\displaystyle y _2 = -\frac{3}{4}\left(-\frac{53}{25}\right) +\frac{1}{4} = \frac{46}{25}$

Thus, our two intersection points are:

$\displaystyle (3, -2) \text{ and } \left(-\frac{53}{25}, \frac{46}{25}\right)$

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2 years ago