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2 years ago

With similar triangles, the corresponding angles are equal. True or False?

Mathematics

2 answers:

Svetradugi [14.3K]2 years ago

6 0

Answer: its true

i did this on apex too so i got you

Stels [109]2 years ago

5 0

Its true because of the rule (cpct)corresponding parts of congruence triangle.

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Jerry calculates that if he makes a deposit of $6 each month at an APR of 4.8%, then at the end of two years he'll have $100. Be

Nimfa-mama [501]

Assuming Jerry calculates that if he makes a deposit of $6 each month at an APR of 4.8%, then at the end of two years the correct balance will be: $158.5

First step is to determine Jerry total deposit

over the two years

Total deposit = 24×$5

Total deposit= $144

Now let determine what the correct balance will be at end of two years

Using this formula

Maximum Amount=Principal (1+r)^t

<em>Where</em>:

Principal=$144

r=4.8%/12 = 0.4% or 0.004

t=24 months

Let plug in the formula

Maximum Balance = $144 (1.004)^24

Maximum Balance = $158.5

Based on the above calculation both Jerry $100 and Benny $163 balance are eliminated or rule out because the correct balance after two years is $158.5

Inconclusion Assuming Jerry calculates that if he makes a deposit of $6 each month at an APR of 4.8%, then at the end of two years the correct balance will be: $158.5

Learn more here:

brainly.com/question/3658861

4 0

2 years ago

Simplify each expression completely. please show work. thanks!

yulyashka [42]

I know the answers but I hate simplifying
40) 9y-12
42) -3x+3y-5

8 0

2 years ago

Read 2 more answers

The quadratic function g(x) = a.ca + bx+c has the

Mumz [18]

<em>The value of b is 14 and the value of c is 65</em>

<h2>Explanation:</h2>

The quadratic function is a function of the form:

$f(x)=ax^2+bx+c$

Here we know that the leading coefficient $a=1$ so we reduce our equation to:

$g(x)=x^2+bx+c$

The roots are those values at which $g(x)=0$

So:

$x^2+bx+c=0 \\ \\ First \ root: \\ \\ (-7+4i)^2+b(-7+4i)+c=0 \\ \\ (-7)^2-2(7)(4i)+(4i)^2-7b+4bi+c=0 \\ \\ 49-56i+16i^2-7b+4bi+c=0 \\ \\ \\ Simplifying: \\ \\ 49-56i+16(-1)-7b+4bi+c=0 \\ \\ 49-56i-16-7b+4bi+c=0 \\ \\ 33-56i-7b+4bi+c=0 \\ \\ \\$

$Second \ root: \\ \\ (-7-4i)^2+b(-7-4i)+c=0 \\ \\ (-1)^2(7+4i)^2+b(-7-4i)+c=0 \\ \\ (7)^2+2(7)(4i)+(4i)^2-7b-4bi+c=0 \\ \\ 49+56i+16i^2-7b-4bi+c=0 \\ \\ \\ Simplifying: \\ \\ 49+56i+16(-1)-7b-4bi+c=0 \\ \\ 49+56i-16-7b-4bi+c=0 \\ \\ 33+56i-7b-4bi+c=0$

So we have:

$(1) \ 33-56i-7b+4bi+c=0 \\ \\ (2) \ 33+56i-7b-4bi+c=0 \\ \\ \\ Subtract \ 2 \ from: \\ \\ 33-56i-7b+4bi+c-(33+56i-7b-4bi+c)=0 \\ \\ 33-56i-7b+4bi+c-33-56i+7b+4bi-c=0 \\ \\ \\ Combine \ like \ terms: \\ \\ 33-33-56i-56i-7b+7b+4bi+4bi+c-c=0 \\ \\ -112i+8bi=0 \\ \\ Isolating \ b: \\ \\ b=\frac{112i}{8i} \\ \\ \boxed{b=14}$