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katrin [286]
3 years ago
15

Which is equivalent to the following expression? (sp^2 + 5pq - q^2) + (p^2 + 3pq - 2q^2)

Mathematics
1 answer:
balandron [24]3 years ago
3 0
I think you've written s along with p^2 by mistake. if so, your answer is 2p^2+8pq-q^2.
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Plz help!!! marking brainiest!!
Ghella [55]
Answer: 3(x-12)(x+2)
5 0
3 years ago
what positive number satisfies the condition that twice the number minus three times the reciprocal of the number is equal to 1?
marshall27 [118]
Solve:

"<span>twice the number minus three times the reciprocal of the number is equal to 1."
                                                     3(1)
Let the number be n.  Then 2n - ------- = 1
                                                       n

Mult all 3 terms by n to elim. the fractions:

2n^2 - 3 = n.  Rearranging this, we get 2n^2 - n - 3 = 0.

We need to find the roots (zeros or solutions) of this quadratic equation.

Here a=2, b= -1 and c= -3.  Let's find the discriminant b^2-4ac first:

disc. = (-1)^2 - 4(2)(-3) = 1 + 24 = 25.

That's good, because 25 is a perfect square.
                -(-1) plus or minus 5         1 plus or minus 5
Then x = ------------------------------ = --------------------------
                            2(2)                                  4

x could be 6/4 = 3/2, or -5/4.

You must check both answers in the original equation.  If the equation is true for one or the other or for both, then you have found one or more solutions.</span>
8 0
4 years ago
Please help! 7/20, will give brainliest. I do not tolerate spam answers!
PtichkaEL [24]
(X+2)(3x+1) !!!!!!!!!!!!!!!!!!
3 0
3 years ago
Two ships leave a harbor together, traveling on courses that have an angle of 135°40' between them. If they each travel 402 mile
mario62 [17]

Answer:

Therefore they are 734.106 miles apart.

Step-by-step explanation:

Given that ,

Two ships have a harbor together. The angle between two ships  is  135°40'. Each of two ships travel 402 miles.

It forms a isosceles triangle whose two sides are 402 miles and one angle is 135°40'. Since it is isosceles triangle then other two angles of the triangle is equal.

Let ∠B= 135°40', and AB = 402 miles , BC =  402 miles

Then the distance between the ships = AC

We know

The sum of all angles = 180°

⇒∠A+∠B+∠C=180°

⇒∠A+135°40'+∠C=180°

⇒2∠A= 180°- 135°40'      [ since ∠A=∠C]

⇒2∠A=44°60'

⇒∠A= 22°30'

Again we know that,

\frac{AB}{sin\angle C}=\frac{BC}{sin \angle A}=\frac{AC}{sin \angle B}

Taking last two ratio,

\frac{BC}{sin \angle A}=\frac{AC}{sin \angle B}

Putting the value of BC , AC ,∠A,∠B

\frac{402}{sin 22^\circ30'}=\frac{AC}{sin 135^\circ40'}

\Rightarrow AC=\frac{402 \times sin135^\circ40'}{sin 22^\circ30'}

         ≈734.106 miles

Therefore they are 734.106 miles apart.

3 0
3 years ago
If f(x) = 2x - 6 and g(x) = 3x + 9, find (f + g)(x)
alex41 [277]

Answer:

5x+3

Step-by-step explanation:

(f + g)(x) = f(x) + g(x) \\  =2x - 6 + 3x + 9 \\  = 5x + 3

consider marking as Brainliest if this helped

5 0
3 years ago
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