2 years ago

Or is it d). 792 in^2

1 answer:

4 0

$A_1=8^2=64\ in.^2\\\\A_2=\dfrac{8\cdot11}{2}=\dfrac{88}{2}=44\ in.^2\\\\SA=A_1+4A_2\to AS=64+4\cdot44=64+176=240\ in.^2$

Answer: A) 240 in.²

You might be interested in

Please help Ill give brainiest!

Sliva [168]

Answer:

H. 4c <60

Step-by-step explanation:

4 0

2 years ago

Read 2 more answers

Jody, a statistics major, grows tomatoes in her spare time. She measures the diameters of each tomato. Assume the Normal model

d1i1m1o1n [39]

The 50th percentile is the median so the z score will be 0

3 0

3 years ago

What is the slope of the line on the graph?<br><br><br> Enter your answer in the box.

Andrews [41]

Answer:

slope = -1/3

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers

Any one know how to find the roots of an equation?

soldier1979 [14.2K]

So what you do is
make into ax^2+bx+c=0 form

add-9/2 to both sides
y^2-2y+9/2=0
now we use the quadratice formula which is
x= $\frac{-b+/- \sqrt{b^{2}-4ac} }{2a}$

ax^2+bx+c=0
1y^2-2y+9/2
a=1
b=-2
c=9/2
subsitute
x= $\frac{-(-2)+/- \sqrt{(-2)^{2}-4(1)(9/2)} }{2(1)}$
x= $\frac{2+/- \sqrt{4-(36/2)} }{2}$
x= $\frac{2+/- \sqrt{4-(18)} }{2}$
x= $\frac{2+/- \sqrt{-14} }{2}$
x= $\frac{2+/- \sqrt{-14} }{2}$
x= $1+/- \frac{\sqrt{-14} }{2}$
x= $1+/- \frac{\sqrt{14} \sqrt{-1}}{2}$
x= $1+/- \frac{\sqrt{14} i }{2}$
x= $1+/- \frac{ i \sqrt{14} }{2}$

answer is C

6 0

3 years ago

List the domain and range of the relation.<br> (5,- 9), (7,7), (0, -9), (7.1) (5,9)

nalin [4]

Answer:

Domain={ 5,7,0,}

Rang=={-9,7,1,9}

7 0

2 years ago