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cestrela7 [59]
3 years ago
14

Christina and Sarah went on a fishing trip and caught 35 fish. Christina

Mathematics
1 answer:
Dmitriy789 [7]3 years ago
4 0

Answer:

f=35x-1 .................

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14a+2=<br> What is this answer??
Bogdan [553]
16 a hope this helped
8 0
3 years ago
Determine the 20th term of the sequence: -14,-8, -2, 4, 10,
Pie

Answer: The 20th term is 100

Step-by-step explanation: It adds by 6 each time

-14, -8, -2, 4, 10, 16, 22, 28, 34, 40, 46, 52, 58, 64, 70, 76, 82, 88, 94, 100

3 0
3 years ago
an isosceles triangle has congruent sides of 20cm. the base is 10cm. find the height of the triangle.​
dolphi86 [110]

Answer:

\large\boxed{A_\triangle=25\sqrt{15}\ cm^2}

Step-by-step explanation:

Look at the picture.

The formula of an area of a triangle:

A_\triangle=\dfrac{bh}{2}

<em>b</em><em> - base</em>

<em>h</em><em> - height</em>

<em />

We need a length of a height.

Use the Pythagorean theorem:

leg^2+leg^2=hypotenuse^2

We have:

leg=5,\ leg=h,\ hypotenuse=20

Substitute:

5^2+h^2=20^2

25+h^2=400             <em>subtract 25 from both sides</em>

h^2=375\to h=\sqrt{375}\\\\h=\sqrt{(25)(15)}\\\\h=\sqrt{25}\cdot\sqrt{15}\\\\h=5\sqrt{15}\ cm

Calculate the area:

A_\triangle=\dfrac{(10)(5\sqrt{15})}{2}=\dfrac{50\sqrt{15}}{2}=25\sqrt{15}\ cm^2

8 0
3 years ago
Solve for x to the nearest 10th.<br> 410 = 1080(0.915)^x
Hunter-Best [27]

Answer:

410/1080=(0.915)^x

0.379=(0.915)^x

ln0. 379/ln0.915=x

X=10.903

4 0
3 years ago
State one form of the Law of Cosines and provide a trick for writing the other two forms and explain when Law of Cosines should
Yuki888 [10]

Solving for <em>Angles</em>

\displaystyle \frac{a^2 + b^2 - c^2}{2ab} = cos∠C \\ \frac{a^2 - b^2 + c^2}{2ac} = cos∠B \\ \frac{-a^2 + b^2 + c^2}{2bc} = cos∠A

* Do not forget to use the <em>inverse</em> function towards the end, or elce you will throw your answer off!

Solving for <em>Edges</em>

\displaystyle b^2 + a^2 - 2ba\:cos∠C = c^2 \\ c^2 + a^2 - 2ca\:cos∠B = b^2 \\ c^2 + b^2 - 2cb\:cos∠A = a^2

You would use this law under <em>two</em> conditions:

  • One angle and two edges defined, while trying to solve for the <em>third edge</em>
  • ALL three edges defined

* Just make sure to use the <em>inverse</em> function towards the end, or elce you will throw your answer off!

_____________________________________________

Now, JUST IN CASE, you would use the Law of Sines under <em>three</em> conditions:

  • Two angles and one edge defined, while trying to solve for the <em>second edge</em>
  • One angle and two edges defined, while trying to solve for the <em>second angle</em>
  • ALL three angles defined [<em>of which does not occur very often, but it all refers back to the first bullet</em>]

* I HIGHLY suggest you keep note of all of this significant information. You will need it going into the future.

I am delighted to assist you at any time.

7 0
2 years ago
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