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krek1111 [17]
3 years ago
14

HI CAN ANYONE PLS ANSWER DIS MATH PROBLEM!!!!!!!!!!!

Mathematics
2 answers:
Gekata [30.6K]3 years ago
7 0
I’d go with B. If that’s not it try C.
maks197457 [2]3 years ago
5 0

C

Because it is not a constant change

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Solving Expressions Analytically Consider the following equation: x3+4x2+4x+16=0. Solve for x. Your answer should be stored in a
Shalnov [3]

Answer:

List is {-2i, 2i, -4)

Step-by-step explanation:

This is a third order polynomial, so we expect three roots (solutions).

Note that we can factor this expression by grouping:

(x²)(x + 4) + (4)(x + 4) = 0, or

(x + 4)(x² + 4) = 0

The factor (x + 4) corresponds to the root/solution x = -4, and:

The other factor (x² + 4) corresponds to the roots/solutions x = 2i and x = -2i.

List is {-2i, 2i, -4)

5 0
4 years ago
Which equation has an a-value of 1, a b value of -3 and a c- value of -5?
Vitek1552 [10]
If the answers are in the standard form of a line then it would be -x + (-3x) = -5y. This is because the standard form of a line is written as Ax + Bx = Cy
9 0
3 years ago
Answer a and b ASAP
melomori [17]

Answer:

The area of the rectangle is 48cm2

6 0
3 years ago
Simplify each expression to a single trig function or number cosx(secx-cosx)
AlladinOne [14]
I did this test  b4, yours is answer #number 12

Convert things to their basic forms. 
<span>Remember a few identities </span>
<span>sin^2 + cos^2 = 1 so </span>
<span>sin^2 = 1 - cos^2 and </span>
<span>cos^2 = 1 - sin^2 </span>

<span>I'm going to skip typing the theta symbol, just to make things faster. Just assume it is there and fill it in as you work the problems. </span>

<span>Follow along to see how each problem was worked out. You'll catch on to the general technique. </span>

<span>====== </span>
<span>1. sec θ sin θ </span>
<span>1/cos * sin = sin/cos = tan </span>

<span>2. cos θ tan θ </span>
<span>cos * sin/cos = sin </span>

<span>3. tan^2 θ- sec^2 θ </span>
<span>sin^2 / cos^2 - 1/cos^2 </span>
<span>(sin^2 - 1)/cos^2 </span>
<span>-(1-sin^2)/cos^2 </span>
<span>-cos^2/cos^2 </span>
<span>-1 </span>

<span>4. 1- cos^2θ </span>
<span>sin^2 </span>

<span>5. (1-cosθ)(1+cosθ) </span>
<span>Remember (a+b)(a--b) = a^2 - b^2 </span>
<span>1-cos^2 = sin^2 </span>

<span>6. (secx-1) (secx+1) </span>
<span>sec^2 -1 </span>
<span>1/cos^2 - 1 </span>
<span>1/cos^2 - cos^2/cos^2 </span>
<span>(1-cos^2)/cos^2 </span>
<span>sin^2/cos62 </span>
<span>tan^2 </span>

<span>7. (1/sin^2A)-(1/tan^2A) </span>
<span>1/sin^2 - 1/(sin^2/cos^2) </span>
<span>1/sin^2 - cos^2/sin^2 </span>
<span>(1-cos^2)/sin^2 </span>
<span>sin^2/sin^2 </span>
<span>1 </span>

<span>8. 1- (sin^2θ/tan^2θ) </span>
<span>1-sin^2/(sin^2/cos^2) </span>
<span>1 - sin^2*cos^2/sin^2 </span>
<span>1-cos^2 </span>
<span>sin^2 </span>


<span>9. (1/cos^2θ)-(1/cot^2θ) </span>
<span>1/cos^2 - 1/(cos^2/sin^2) </span>
<span>1/cos^2 - sin^2/cos^2 </span>
<span>(1-sin^2)/cos^2 </span>
<span>cos^2/cos^2 </span>
<span>1 </span>

<span>10. cosθ (secθ-cosθ) </span>
<span>cos *(1/cos - cos) </span>
<span>1-cos^2 </span>
<span>sin^2 </span>

<span>11. cos^2A (sec^2A-1) </span>
<span>cos^2 * (1/cos^2 - 1) </span>
<span>1 - cos^2 </span>
<span>sin^2 </span>


<span>12. (1-cosx)(1+secx)(cosx) </span>
<span>(1-cos)(1+1/cos)cos </span>
<span>(1-cos)(cos + 1) </span>
<span>-(cos-1)(cos+1) </span>
<span>-(cos^2 - 1) </span>
<span>-(-sin^2) </span>
<span>sin^2 </span>

<span>13. (sinxcosx)/(1-cos^2x) </span>
<span>sin*cos/sin^2 </span>
<span>cos/sin </span>
<span>cot </span>

<span>14. (tan^2θ/secθ+1) +1 </span>
<span>(sin^2/cos^2)/(1/cos) + 2 </span>
<span>sin^2/cos + 2 </span>
<span>sin*tan + 2 </span>
5 0
3 years ago
How to find r in this equation using combination formula C(8,r)=28<br>​
slavikrds [6]

Answer:

r = 2

Step-by-step explanation:

We have the formula of ^nC_r = \frac{n!}{r! (n-r)!}

Now, it is given that ^8C_r = \frac{8!}{r! (8-r)!} = 28 ........ (1)

And we have to find the value of r which satisfy the above equation.

So, r! (8-r)! = \frac{8!}{28} = \frac{40320}{28} = 1440

Now, we have to use the trial method to find the value of r.

For r = 1, 1! (8-1)! = 7! = 5040 \neq 1440

Hence, r can not be 1.

Now, put r = 2, 2! (8-2)! = 2 \times 6! = 1440

Therefore, r = 2 (Answer)

5 0
3 years ago
Read 2 more answers
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