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PIT_PIT [208]
4 years ago
14

Can you please check my answer for me ?(:

Mathematics
2 answers:
Alexandra [31]4 years ago
8 0
Dont doubt yourself you did good!
Vinvika [58]4 years ago
7 0
I don't believe you would shade anywhere, so your work is correct :D 
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What is the value of 3x 2 squared + 5x when x=3
Serhud [2]
3•2=6. When you square 6, you get 36. Add 5x to that. 5x=15. 36+15=51

Cassie read 155 pages one day, and we need to find the unknown value of “x”. All you have to do is add 155+x=325.

6 0
4 years ago
What’s the equation of the blue line??
Sholpan [36]

Answer:

y = -1/2 + 2 (Slope intercept)

Step-by-step explanation:

Y-intercept is when the x-value is 0.

So the y-intercept is (0, 2)

To find slope, find two random points:

(0, 2) , (2, 1)

Slope (m) =ΔY/ΔX  = -1/2 =  -0.5

Equation of the line:

y = -0.5x + 2

or

y = -1/2 + 2

When x=0, y = 2

When y=0, x = 4

4 0
2 years ago
So I'm still trying to figure out but tell me. What are two demicals rounded to 1.5
Andrej [43]
451.76 and <span>451.7576 these two</span>
7 0
3 years ago
I don’t get it. <br> Use the net to find the lateral area of the prism.
sesenic [268]

the net is pretty much the net of a long box, kinda like the one in the example in the picture below.  Due to that, we can pretty much assume the two sides sticking up and down, are just two small 6x3 rectangles, namely, they have a height of 3, reason why we assume that, if that if we fold the other sides to make out the box, those two sides sticking out, must be 3m to neatly snugfit.


now, if we close the box as it stands, the sides(laterals) will be on the left-right sides two 3x15 rectangles, and on the front-back sides, two 6x15 rectangles.


we're excluding the top and bottom sides, because those are not "laterals", or sides of the box.


\bf \stackrel{\textit{two rectangles, left and right}}{2(3\cdot 15)}+\stackrel{\textit{two rectangles, front and back}}{2(6\cdot 15)}&#10;\\\\\\&#10;90+180\implies 270

6 0
4 years ago
Simplify the monomials
Zielflug [23.3K]

Answer:

(4a)^{-3}*^{-4}

x^m*x^n=x^{m+n}

(4a)^{-3}*a^{-4}

(-4)^{-3}\: a^{-3}*a^{-4}

4^{-3}\:a^{-3-4}

\cfrac{1}{4^3} \:a^{-7}

\boxed{\frac{1}{64a^{7}} }

~

(3xy)^2*(-4x^3y^2)^3

9x^2y^2*(64x^9y^6)

\boxed{-576x^{11}y^8}

~

(4a^{-1}b^5c^{-3})^3

(4)^3(a^{-1})^3(b^5)^3(c^{-3})^3

(4*4*4)\:a^{-1*3}\:b^{5*3}\:c^{-3*3}

64\:a^{-3}b^{15}c^{-9}

\boxed{\frac{64b^{15}}{a^3c^9}}

7 0
2 years ago
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