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strojnjashka [21]
3 years ago
7

Why is Josh incorrect? Explain his error.

Mathematics
1 answer:
schepotkina [342]3 years ago
4 0

Answer: the reason josh is incorrect is because he spammed the same 2 questions 17 times, and never gave us the actual thing josh had done that is incorrect. (add the problem that josh had done incorrectly)

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Write a rule for the nth term of the arithmetic sequence. Then graph the first six terms of the sequence. a6=-12 a12=-36
denis-greek [22]

Answer:

We need to find the first term.  We can use the formula

an = a1 + d(n - 1)

to solve for a1.  We already know the 12th term, a12, common difference, d, and nth sequence, n.

a12 = -36

d = -4

n = 12

-36 = a1 - 4(12 - 1)

-36 = a1 - 44

8 = a1

The first term is 8.  Therefore, your formula is

an = 8 - 4(n - 1)

an = -4n + 12

Then use this formula to graph.

n is the independent variable.

an is the dependent variable.

Your graph will be a line.

n    |       an

___________

1          8

2          4

3          0

4         -4

5         -8

6         -12

Step-by-step explanation:

give me brainliest.

3 0
3 years ago
The legs of the right triangle have Links of 28 metres and 21 metres
GarryVolchara [31]
If we're trying to find the hypotenuse we'll take the \sqrt{28^{2}+ 21^{2}  }
Getting 35.

35
4 0
3 years ago
Can someone please teach me how to do this? You can please do like 2 questions and please explain exactly how you got the answer
spayn [35]

All of these questions require one thing: trigonometric functions.

There are 3 main trigonometric functions, which can only be used on right triangles: sine, cosine, and tangent.

Sine = opposite / hypotenuse

Cosine = adjacent / hypotenuse

Tangent = opposite / adjacent

When trying to figure out what function to use, we always start by looking from the angle. Take problem a, for example. Looking from angle E, of which the value is not given, we have the side opposite and the side adjacent. Therefore, we should use the tangent function.

---The hypotenuse is always the longest side of the triangle. It is never considered the opposite or adjacent side.

Let's set up our function with the given information from problem a.

tan(x) = 9.7 / 5.2

---The tangent of an unknown angle is equal to the quotient of the opposite side and the adjacent side.

Now, solving for the value of x will require a calculator. We'll need to use what's called an inverse trigonometric function. Most calculators have these directly above the regular trigonometric functions, and the inverse functions are accessed using a "second" key.

---Ensure that your calculator is in degrees, not radians!

x = tan^-1(9.7 / 5.2)

x = 61.805 = 62 degrees

Next, let's take a look at problem b. This time, we're solving for a side length instead of an angle. But, we're still going to start by looking from our angle.


Looking from the 38 degree angle, we are given the adjacent side and an unknown hypotenuse. Therefore, we should use the cosine function.

cosine(38) = 53.1 / r

---The cosine of a 38 degree angle is equal to the quotient of 53.1 and an unknown hypotenuse, r.

Use your algebra skills to isolate the variable r.

r * cosine(38) = 53.1

r = 53.1 / cosine(38)

---From here, all you need to do is plug this into your calculator. Since we are solving for a side length (and given an angle), we are just using the regular trigonometric function buttons on the calculator.

r = 67.385 = 67.4 units

Hope this helps!

4 0
2 years ago
Solve the simultaneous equations<br> x + 6y = 20<br> x + 3y = 14<br> x =<br> Y=
xxTIMURxx [149]

Answer:

x = 8, y = 2

Step-by-step explanation:

Multiply the second equation by -2:

x + 6y = 20

-2x - 6y = -28

Add the equations and simplify:

-x = -8

x = 8

Plug x = 8 back into the first equation and solve for y:

8 + 6y = 20

6y = 12

y = 2

6 0
3 years ago
Read 2 more answers
Solve for d <br> d + 8 &lt; 35
9966 [12]

Answer:

d    <   27

Step-by-step explanation:

Solve for d by subtracting 8 from both sides.  This isolates d:

d + 8 < 35

   -8      -8

-------    -------

  d    <   27

5 0
4 years ago
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