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stira [4]
3 years ago
9

Sarah can take no more than 22 pounds and I like the trap her suitcase weighs 112 ounces how many more pounds can she park witho

ut going over the limit
Mathematics
1 answer:
Leya [2.2K]3 years ago
3 0

Answer:

15 pounds

Step-by-step explanation:

there are 16 ounces in one pound

so...

first find the limit in ounces

22  =  \frac{x}{16}

because there are 16 ounces in every one pound you need to multiply 22 by 16

which is 352

now subtract what you already used

352 - 112

which is 240

now convert to pounds

x =  \frac{240}{16}

which is 15

so 15 more pounds

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vodka [1.7K]

The answer is 270 found out by google calculator.

6 0
3 years ago
9(y-4) -3y = 2(3y-2 need solved as fast as possible
Vlad [161]

Step-by-step explanation:

did you give us all the information ? the equating looks cut off.

so, from what I can see we have

9(y-4) - 3y = 2(3y-2)

let's try :

9y - 36 - 3y = 6y - 4

0 = 32

so, no, this can't be it. this does not have a solution. you must have cut off some additional information.

7 0
3 years ago
What is the area of rectangle ABCD?
sashaice [31]

Answer:

20

Step-by-step explanation:

area=width×height

=4×5 (because of how many squares there are)

=20

or you could count all the squares but timesing is eaier

4 0
3 years ago
Read 2 more answers
Question 9 of 10
sleet_krkn [62]

Answer:

y = 51.32º

Step-by-step explanation:

z = 180 - 47 - 90

z = 43º

sin43 = opp/hyp

sin43 = 35/y

y = 35/sin43

y = 51.32º

8 0
3 years ago
I really am bad with math and just have given up
Slav-nsk [51]
Well hmmm does the graph open upwards or downwards?  well
is a quadratic, if the leading term's coefficient is negative, Down, if positive, Up

a)
now, let's see the leading term, -x², what's its coefficient? is -1 * x² or -1x², is -1, so is negative, thus is opening downwards

c)
x-intercepts occur, when y = 0, namely y = -x²-4x-3, so setting y to 0
0 = -x² -4x -3 take a common factor of -1, thus

0 = -1 (x²+4x+3)  <--- now, if we factor that out, notice, surely you've done many of these by now, so we end up with 

\bf \begin{array}{lcclll}&#10;0=x^2&+4x&+3\\&#10;&\uparrow &\uparrow \\&#10;&3+1&3\cdot 1&#10;\end{array}\\\\&#10;-----------------------------\\\\&#10;0=(x+3)(x+1)\implies &#10;\begin{cases}&#10;0=x+3\implies &-3=x\\&#10;0=x+1\implies &-1=x&#10;\end{cases}

so, the x-intercepts are at -3 and -1

d)

now, the y-intercepts, just set x = 0
y = -x²-4x-3, settting x to 0 y = -0²-4(0)-3, which is y = -3

so the sole y-intercept is at y = -3


now, let's get on to b)

b)   \bf \textit{vertex of a parabola}\\\\&#10;&#10;\begin{array}{lccclll}&#10;f(x)=&-1x^2&-4x&-3\\&#10;&\uparrow &\uparrow &\uparrow \\&#10;&a&b&c&#10;\end{array}\qquad &#10;\left(-\cfrac{{{ b}}}{2{{ a}}}\quad ,\quad  {{ c}}-\cfrac{{{ b}}^2}{4{{ a}}}\right)

and those are the coordinates, notice a = -1, b = -4 and c = -3

now, for

e)

well, all you have to do is, once you have the vertex, pick an x-value on the left-hand-side of the vertex, get they value for "y", or OUTPUT,

then pick another x-value, on the right-hand-side of the vertex, get the "y" value again, and plot away
7 0
3 years ago
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