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lukranit [14]
2 years ago
5

Can you solve this??​

Mathematics
1 answer:
gavmur [86]2 years ago
3 0

Answer:

The answer is c

Step-by-step explanation:

The middle problem is adding, the others are taking away, so C is the correct answer

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PLs help
Pavel [41]

A) The hypotenuse is the longest side of the triangle .In the figure EG is the hypotenuse.The given angle is <G=40 degrees.The side adjacent is FG= 10 units.

The question is asking us to find the trigonometry ratio which will  help us find EG.We can use cos  as cos is ratio of adjacent and hypotenuse.

cos 40=\frac{10}{Hypotenuse EG.}

B) Cos  40 =0.766.

Substituting tan40 value we have:

0.766=\frac{10}{EG}

EG=13.05.

C)Sum of angles in  triangle is 180.

<E+<F+<G=180

<E+90+40=180

<E+130=180

<E=50 degrees.

D) Tan  40=\frac{EF}{10}

EF=8.39

E) Area of triangle EFG= \frac{1}{2}x base x height.

A rea of the triangle = \frac{1}{2}x8.39x13.05

Area= 54.75 square units.





5 0
3 years ago
What is a difference of squares that has a factor of x + 8?
Goshia [24]
Hello,

x²-64=(x+8)(x-8)
============
8 0
3 years ago
N<br> Which expression has twice the value of 2-2.2-2.2?
Free_Kalibri [48]

Answer:

2^5

explanation:

I did this already man this is easy

7 0
3 years ago
A quantity P is an exponential function of time I, such that P = 160 when t = 6 and P = 150 when I = 4. Use the given informatio
Klio2033 [76]

Answer:

  • k = 0.032
  • P0 = 131.836

Step-by-step explanation:

Perhaps you want to use the points (t, P) = (4, 150) and (6, 160) to find the parameters P0 and k in the equation ...

  P(t)=P_0\cdot e^{kt}

We know from the given points that we can write the equation as ...

  P(t)=150\left(\dfrac{160}{150}\right)^{(t-4)/(6-4)}=150\left(\dfrac{16}{15}\right)^{\frac{t}{2}-2}\\\\=150\left(\dfrac{16}{15}\right)^{-2}\times\left(\left(\dfrac{16}{15}\right)^{\frac{1}{2}}\right)^t

Comparing this to the desired form, we see that ...

  P_0=150\left(\dfrac{16}{15}\right)^{-2}\approx 131.836\\\\e^{k}=\left(\dfrac{16}{15}\right)^{1/2}\rightarrow k=\dfrac{1}{2}(\ln{16}-\ln{15})\approx 0.0322693

So, the approximate equation for P is ...

  P(t)=131.836\cdote^{0.032t}

And the parameters of interest are ...

  • k = 0.032
  • P0 = 131.836
4 0
3 years ago
What is the solution to the equation 2^x+4 -12=20?<br> A. x=0<br> B. x=1<br> C. x=2<br> D. x=9
seropon [69]
<span>2^x+4 -12=20
</span><span>2^x+4 = 20+12
</span><span>2^x+4 = 32
</span><span>2^x+4 = 2^5
x+4 = 5
x = 1

answer is </span><span>B. x=1</span>
8 0
3 years ago
Read 2 more answers
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