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Mashcka [7]
3 years ago
7

Please help me I don’t understand how to get the answers

Mathematics
1 answer:
Lunna [17]3 years ago
3 0

Answer:

Step-by-step explanation:

1.

What is the slope of the line? The graph is hard to see, so can't really calculate that, please reupload

What does the slope represent? Increase in costs of Hot Yoga

2. (0,25)

Not sure if that number is 25, but the most bottom dot & on the black line

Y int represents where they started at, minimum cost

Unsure about last 2

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Solve this inequality for x. 81-1 1/5 x <55
katen-ka-za [31]

Answer:

A. x > 21 2/3

Step-by-step explanation:

x is greater than 21 and two thirds

i did the math :]

4 0
3 years ago
A cube has a surface area of 486 cm2. What is the side length ?
astraxan [27]
A Cube, has all equal sides, namely, the length, width and height
are all equal to each other

so..  notice the picture added here
you have really, 6 squares, stacked up to each other at the edges

so...what is the Area of one of those squares?
well, if the sides are equal, let's say the side is "x" long, then
the Area is x\cdot x \implies x^2

well, you have 6 of those squares, thus \bf \textit{surface area of a cube}=
\begin{cases}
(x\cdot x)+\\
(x\cdot x)+\\
(x\cdot x)+\\
(x\cdot x)+\\
(x\cdot x)+\\
(x\cdot x)\\
 --------------\\
 x^2+x^2+x^2+x^2+x^2\\
6x^2
\end{cases}
\\\\
\textit{we know the Area is }486\ cm^2\qquad thus
\\\\
\textit{surface area}=6x^2\implies 486=6x^2

solve for "x", to get one side's length

3 0
3 years ago
Which of these values are farthest from zero 0.75 2/3 and 0.25
leva [86]
.75 is farthest because 2/3 is .66 (infinitely) and .25 is smaller than both
8 0
3 years ago
Can somebody please help me
zubka84 [21]

Answer:

B

Step-by-step explanation:

6 0
3 years ago
Let T:P3→P3 be the linear transformation such that T(â’2x2)=3x2+4x, T(0.5x+3)=â’3x2+3xâ’4, and T(2x2â’1)=â’3x+4. Find T(1), T(
balu736 [363]

It looks like we're told that

T(-2x^2)=3x^2+4x

T\left(\dfrac12x+3\right)=-3x^2+3x-4

T(2x^2-1)=-3x+4

We use the fact that T is linear to find T(1),T(x),T(x^2). First, we notice that

T(-2x^2)+T(2x^2-1)=T(-2x^2+2x^2-1)=T(-1)=-T(1)

We also have

T\left(\dfrac12x+3\right)=T\left(\dfrac12x\right)+T(3)=\dfrac12T(x)+3T(1)

T(2x^2-1)=T(2x^2)-T(1)=2T(x^2)-T(1)

So once we find T(1), we can determine T(x) and T(x^2). We have

T(1)=-\left(T(-2x^2)+T(2x^2-1)\right)\implies T(1)=-3x^2-x-4

and using this we find

T(x)=12x^2+12x+16

T(x^2)=-\dfrac32x^2-2x

Then

T(ax^2+bx+c)=T(ax^2)+T(bx)+T(c)=aT(x^2)+bT(x)+cT(1)

T(ax^2+bx+c)=-\dfrac32\left(a-8b+2c\right)x^2-(2a-12b+c)x+16b-4c

8 0
3 years ago
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