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icang [17]
3 years ago
5

How to i solve for x in this situation? thank yous.

Mathematics
1 answer:
Lena [83]3 years ago
4 0

i did not understand the question can u explain again i can help u


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Which of the following cosine functions has a period of 3π?
skelet666 [1.2K]

Answer:

y = cos 2/3 x

Step-by-step explanation:

7 0
2 years ago
What is the measure of ∠EGF? __
Ray Of Light [21]

Answer:

∠EGF = 1/2( 180 - 50) = 1/2(130) = 65

∠CGF = 180 - 65 = 115.

Step-by-step explanation:

5 0
3 years ago
A corporate bond has a coupon rate of 5.5 percent, a $1,000 face value, and matures three years from today. The corporation is i
melomori [17]

Answer:

= \frac{\frac{75}{100}\times 1000 + \frac{25}{100} \times \frac{60}{100}\times 1000  }{(1+\frac{15}{100})^3 }

=\frac{0.75\times 1000 + 0.25\times 0.60 \times 1000}{(1+0.15)^3}

=\frac{750+0.25\times 0.60\times 1000}{1.15^3} \\\\=\frac{750+150}{1.520875} =\frac{900}{1.520875} \\\\=591.76

Step-by-step explanation:

= (probability of entire face value paid*face value+probability of entire face value not paid*percent of face value paid*face value)/(1+discount rate)^years to maturity

probability of entire face value paid = 75%

face value = 1000

probability of entire face value not paid = 25%

percent of face value paid= 60%

discount rate = 15%

years to maturity  = 3

= \frac{\frac{75}{100}\times 1000 + \frac{25}{100} \times \frac{60}{100}\times 1000  }{(1+\frac{15}{100})^3 }

=\frac{0.75\times 1000 + 0.25\times 0.60 \times 1000}{(1+0.15)^3}

=\frac{750+0.25\times 0.60\times 1000}{1.15^3} \\\\=\frac{750+150}{1.520875} =\frac{900}{1.520875} \\\\=591.76

6 0
3 years ago
Marie and her family are on a trip. So far, they have traveled 175 miles and are 25% of the way to their destination. How many t
kow [346]
To find the answer for this question you take 175 multiply it by 4 which gives you 700 which is the answer
7 0
3 years ago
Undefined Slope. Through (97 , 73)
Mnenie [13.5K]

Answer:

Equation of line through ((x_{1},y_{1})) having undefined slope is given by

 \frac{y-y_{1}}{x-x_{1}}=tan\alpha   where α =90°

Now ,it is given that line passes through (97, 93)

So , equation of line is

→ \frac{y-73}{x-97} = tan 90°

As, tan 90°=1/0

\frac{y-73}{x-97}=\frac{1}{0}\\

→  (x-97)×1= (y-73)×0

→ x-97 =0

→ x=97 , which is the required equation of line.

 

     




4 0
3 years ago
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