Answer:
117/50 or 2 17/50 or 2.34
Step-by-step explanation:
i don't know is just simple
You multiply 7 X5+4 is 39, 39/5 and then multiple
<span>As the given parent function is
f(x) = x^2
then updated one would be with teh k
f(x) = a(x-h)^2 + k
K allows us to shifts up the value
so
k=7 would shifts the graphs up 7 units.</span>
Answer:
a) 4 cm
b) 5 cm
c) 10 cm
Step-by-step explanation:
The side lengths of the reflected square are equal to the original, and the distance from the axis(2) also remains the same. From there, it is just addition.
Hope it helps <3
Answer:
.
Step-by-step explanation:
To find the exact value of the expression ![\sin(\tan^{-1}(\frac{8}{15} ))](https://tex.z-dn.net/?f=%5Csin%28%5Ctan%5E%7B-1%7D%28%5Cfrac%7B8%7D%7B15%7D%20%29%29)
First, we need to simplify the expression
.
Draw a triangle in the plane with vertices
,
, and the origin. Then
is the angle between the positive x-axis and the ray beginning at the origin and passing through
.
Therefore,
![\sin(\tan^{-1}(x))=\frac{x}{\sqrt{1+x^{2} } }](https://tex.z-dn.net/?f=%5Csin%28%5Ctan%5E%7B-1%7D%28x%29%29%3D%5Cfrac%7Bx%7D%7B%5Csqrt%7B1%2Bx%5E%7B2%7D%20%7D%20%7D)
![\mathrm{Multiply\:by\:the\:conjugate}\:\frac{\sqrt{1+x^2}}{\sqrt{1+x^2}}](https://tex.z-dn.net/?f=%5Cmathrm%7BMultiply%5C%3Aby%5C%3Athe%5C%3Aconjugate%7D%5C%3A%5Cfrac%7B%5Csqrt%7B1%2Bx%5E2%7D%7D%7B%5Csqrt%7B1%2Bx%5E2%7D%7D)
![\frac{x\sqrt{1+x^2}}{\sqrt{1+x^2}\sqrt{1+x^2}}](https://tex.z-dn.net/?f=%5Cfrac%7Bx%5Csqrt%7B1%2Bx%5E2%7D%7D%7B%5Csqrt%7B1%2Bx%5E2%7D%5Csqrt%7B1%2Bx%5E2%7D%7D)
![\sqrt{1+x^2}\sqrt{1+x^2}=1+x^2](https://tex.z-dn.net/?f=%5Csqrt%7B1%2Bx%5E2%7D%5Csqrt%7B1%2Bx%5E2%7D%3D1%2Bx%5E2)
![\sin(\tan^{-1}(x))=\frac{x\sqrt{1+x^2}}{1+x^2}](https://tex.z-dn.net/?f=%5Csin%28%5Ctan%5E%7B-1%7D%28x%29%29%3D%5Cfrac%7Bx%5Csqrt%7B1%2Bx%5E2%7D%7D%7B1%2Bx%5E2%7D)
Now, use the identity ![\sin(\tan^{-1}(x))=\frac{x\sqrt{1+x^2}}{1+x^2}](https://tex.z-dn.net/?f=%5Csin%28%5Ctan%5E%7B-1%7D%28x%29%29%3D%5Cfrac%7Bx%5Csqrt%7B1%2Bx%5E2%7D%7D%7B1%2Bx%5E2%7D)
![\sin(\tan^{-1}(\frac{8}{15} ))=\frac{\left(\frac{8}{15}\right)\sqrt{1+\left(\frac{8}{15}\right)^2}}{1+\left(\frac{8}{15}\right)^2}\\\\\frac{\frac{8}{15}\sqrt{\left(\frac{8}{15}\right)^2+1}}{1+\frac{8^2}{15^2}}\\\\\frac{\frac{136}{225}}{1+\frac{8^2}{15^2}}\\\\\frac{\frac{136}{225}}{1+\frac{64}{225}}\\\\\frac{136}{225\cdot \frac{289}{225}}\\\\\frac{136}{289}=\frac{8}{17}\\\\\sin \left(\tan^{-1} \left(\frac{8}{15}\right)\right)=\frac{8}{17}](https://tex.z-dn.net/?f=%5Csin%28%5Ctan%5E%7B-1%7D%28%5Cfrac%7B8%7D%7B15%7D%20%29%29%3D%5Cfrac%7B%5Cleft%28%5Cfrac%7B8%7D%7B15%7D%5Cright%29%5Csqrt%7B1%2B%5Cleft%28%5Cfrac%7B8%7D%7B15%7D%5Cright%29%5E2%7D%7D%7B1%2B%5Cleft%28%5Cfrac%7B8%7D%7B15%7D%5Cright%29%5E2%7D%5C%5C%5C%5C%5Cfrac%7B%5Cfrac%7B8%7D%7B15%7D%5Csqrt%7B%5Cleft%28%5Cfrac%7B8%7D%7B15%7D%5Cright%29%5E2%2B1%7D%7D%7B1%2B%5Cfrac%7B8%5E2%7D%7B15%5E2%7D%7D%5C%5C%5C%5C%5Cfrac%7B%5Cfrac%7B136%7D%7B225%7D%7D%7B1%2B%5Cfrac%7B8%5E2%7D%7B15%5E2%7D%7D%5C%5C%5C%5C%5Cfrac%7B%5Cfrac%7B136%7D%7B225%7D%7D%7B1%2B%5Cfrac%7B64%7D%7B225%7D%7D%5C%5C%5C%5C%5Cfrac%7B136%7D%7B225%5Ccdot%20%5Cfrac%7B289%7D%7B225%7D%7D%5C%5C%5C%5C%5Cfrac%7B136%7D%7B289%7D%3D%5Cfrac%7B8%7D%7B17%7D%5C%5C%5C%5C%5Csin%20%5Cleft%28%5Ctan%5E%7B-1%7D%20%5Cleft%28%5Cfrac%7B8%7D%7B15%7D%5Cright%29%5Cright%29%3D%5Cfrac%7B8%7D%7B17%7D)