Atotal of $9500 is invested, part at 10% simple interest and part at 9%. if the total annual return from the two investments is $914.00, how much is invested at each rate?

a total of $9500 is invested, part at 10% simple interest and part at 9%

if the total annual return from the two investments is $914.00

Amount invested = 9500

Total amount return means total yield = $914

Let x amount  is invested at 10% simple interest

Interest for first part =  10 % times x = 0.10x

Remaining part invested is 9500 -x

Interest for remaining amount = 9% (9500 -x) = 0.09(9500-x)

Total interest yield is 914. Now we frame an equation

0.10x + 0.09(9500-x) = 914

0.10x +855 - 0.09x = 914 (combine like terms)

0.01 x + 855 = 914 (subtract 855)

0.01x = 59 ( divide both sides by 0.01)

So x = 5900

$5900 is invested in 10% simple interest

9500 - 5900 = $3600 is invested in 9% simple interest

c.

step-by-step explanation:

substitution for (0,1)

y=x^2+4x+1

1=0^2+4(x)0+1

1=0+4(x)0+1

1=4(x)0+1

1=0+1

1=1

substitution for (-3,-2)

y=x+1

-2=-3+1

-2=-2

answer:   the paragraph proof is mentioned below.

step-by-step explanation:

given: in a δxwy where

such that, ∠wzx ≅ ∠yzx and zw≅zy

we have to prove that: segment zx is a perpendicular bisector of segment wy.

prove: since, zw≅zy ⇒ ∠wzy is a supplementary angle.

then, ∠wzy= 180°

⇒∠wzx + ∠yzx = 180° ( because xz divides ∠wzy into two parts)

but, ∠wzx ≅ ∠yzx ( given)

⇒∠wzy+∠wzy=180° ( by substitution)

⇒2∠wzy=180°

⇒∠wzy=90°

⇒∠wzy=∠yzx =90°

therefore, xz⊥wy where wz=zy

⇒zx is a perpendicular bisector of wy.