Answer:
X = 20 and Y = 5
Step-by-step explanation:
you're supposed to substitute the numbers it so the problem x * 2 also means 10 * 2 because x = 10 and y * 1 also means 5 * 1 because y = 5
7+14+15+5+8 + 9 =58 yard
answer 58 yd
We know that<span>
<span>Figures can be proven similar if one, or more,
similarity transformations (reflections, translations, rotations, dilations)
can be found that map one figure onto another.
In this problem to prove circle 1 and circle 2 are similar, a
translation and a scale factor (from a dilation) will be found to map one
circle onto another.
we have that</span>
<span> Circle 1 is centered at (5,8) and has a
radius of 8 centimeters
Circle 2 is centered at (1,-2) and has a radius of 4 centimeters
</span>
step 1
<span>Move the center of the circle 1 onto the
center of the circle 2
the transformation has the following rule</span>
(x,y)--------> (x-4,y-10)
so
(5,8)------> (5-4,8-10)-----> (1,-2)
so
center circle 1 is now equal to center circle 2
<span>The circles are now concentric (they have the
same center)
</span>
step 2
<span>A dilation is needed to decrease the size of
circle 1 to coincide with circle 2
</span>
scale factor=radius circle 2/radius circle
1-----> 4/8----> 0.5
radius circle 1 will be=8*scale factor-----> 8*0.5-----> 4 cm
radius circle 1 is now equal
to radius circle 2
<span>A
translation, followed by a dilation will map one circle onto the other,
thus proving that the circles are similar
the answer is
</span></span>The circles are similar because you can translate Circle 1 using the transformation rule (x-4,y-10) and then dilate it using a scale factor of (0.5)
Answer:
144 cubic inches
Step-by-step explanation:
volume = l x w x h. This = 8 x 6 x 3.
Hope this helps :)
Answer:
Step-by-step explanation:
Find two linear functions p(x) and q(x) such that (p (f(q(x)))) (x) = x^2 for any x is a member of R?
Let p(x)=kpx+dp and q(x)=kqx+dq than
f(q(x))=−2(kqx+dq)2+3(kqx+dq)−7=−2(kqx)2−4kqx−2d2q+3kqx+3dq−7=−2(kqx)2−kqx−2d2q+3dq−7
p(f(q(x))=−2kp(kqx)2−kpkqx−2kpd2p+3kpdq−7
(p(f(q(x)))(x)=−2kpk2qx3−kpkqx2−x(2kpd2p−3kpdq+7)
So you want:
−2kpk2q=0
and
kpkq=−1
and
2kpd2p−3kpdq+7=0
Now I amfraid this doesn’t work as −2kpk2q=0 that either kp or kq is zero but than their product can’t be anything but 0 not −1 .
Answer: there are no such linear functions.
