Not entirely sure, but it seem the answer could be B.) It will not be spread out vertically across the entire coordinate plane because in step 5, Nancy selected an incorrect scale on the y-axis.
Answer:
B
Step-by-step explanation:
Put x as 1 and the output will be 9.
y = (3x)²
y = (3(1))²
y = 3²
y = 9
True
7/10 = 0.7
198/100 = 1.98
85/100 = 0.85
45/100 = 0.45
1/4 = 0.25
3/2 = 1.5
3/8 = 0.375
1/9 = 0.11111111 ( goes on forever) ( sigue para siempre )
Hope it Helps!
¡Espero eso ayude!
Answer:
10. Let m<Y = y.
Then,
m<Y + m<X +m<Z = 180 [ Sum of angles of triangle is 180 degrees]
or, y + (6x-23) + (4x + 9) = 180
or, y + 10x -14 = 180
or, y = 194 - 10x
or, m<Y = 194 - 10x
12. Solution,
a. m<1 = 60 degrees [Each angles of equilateral triangle is equal to 60 degrees]
b. In triangle WYZ,
m<Z + m<ZWY + m<ZYW =180 [Sum of angles of triangle is 180]
or, 138 + m<ZWY +m<ZWY =180 [Base angles of isosceles triangle are equal, i.e. m<ZYW = m <ZWY]
or, 2 (m<ZWY) = 180 -138 = 42
or, m<ZWY = 21 = m<ZYW
or, m<3 = 21 = m<5
c. Solution,
m<XWY = m<2 + m<3 [Addition axiom]
or, 60 = m<2 + 21 [Each angle of equilateral triangle is 60]
or, m<2 = 39
d. Solution,
m<4 + m<5 =60 [<XYW = 60 ]
or, m<4 + 21 = 60
or, m<4 = 39
Answer:
<em>A) (-5,7)</em>
Step-by-step explanation:
<u>Functions and Relations</u>
A set of values A can have a relation with another set B as long as at least one element of A has at least one image in B. Functions are special relations where each element of A (the domain of the function) has one and only one image on B (the range of the function).
By looking at the options, we can see that x=9, x=-8, and x=-1 already have defined values in Y, so if we define another value for any of them the relation will stop being a function. The only possible choice to preserve the function is the option
