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2 years ago

I need help! I've been sitting here for hours. Choose the correct simplification of 4X^3 + 3X^2 - 6X -10X^3 + 3X^2

Mathematics

2 answers:

Serhud [2]2 years ago

4 0

You've chosen the correct answer ~

Nookie1986 [14]2 years ago

3 0

(4x^3 + 3x^2 - 6x) - (10x^3 + 3x^2)
4x^3 + 3x^2 -6x - 10x^3 - 3x^2
+3x^2 and - 3x^2 // cancel out
-6x^3 - 6x <- Answer

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Evaluate 4(x - 3) + 5x - x^2 for x = 3 <br>A. 2 <br>B. -6 <br>C. 6<br>D. -2

ipn [44]

Answer:

C) x=6...

Step-by-step explanation:

hope this helps...

3 0

2 years ago

Help me please find x for this triangle!!!!

Ne4ueva [31]

Answer:

x=30

Step-by-step explanation:

Since the measure of an exterior angle is equal to the sum of any two angles in a triangle, the equation would be 4x-18=2x+6+36. Solve this, and we get x=30.

Hope this helped!

5 0

3 years ago

Find the value of x.

koban [17]

Step-by-step explanation:

360 - 126 =234

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2 years ago

Suppose that θ is an acute angle of a right triangle and that sec(θ)=52. Find cos(θ) and csc(θ).

insens350 [35]

Answer:

$\cos{\theta} = \dfrac{1}{52}$

$\csc{\theta} = \dfrac{52}{\sqrt{2703}}$

Step-by-step explanation:

To solve this question we're going to use trigonometric identities and good ol' Pythagoras theorem.

a) Firstly, sec(θ)=52. we're gonna convert this to cos(θ) using:

$\sec{\theta} = \dfrac{1}{\cos{\theta}}$

we can substitute the value of sec(θ) in this equation:

$52 = \dfrac{1}{\cos{\theta}}$

and solve for for cos(θ)

$\cos{\theta} = \dfrac{1}{52}$

side note: just to confirm we can find the value of θ and verify that is indeed an acute angle by $\theta = \arccos{\left(\dfrac{1}{52}\right)} = 88.8^\circ$

b) since right triangle is mentioned in the question. We can use:

$\cos{\theta} = \dfrac{\text{adj}}{\text{hyp}}$

we know the value of cos(θ)=1\52. and by comparing the two. we can say that:

length of the adjacent side = 1
length of the hypotenuse = 52

we can find the third side using the Pythagoras theorem.

$(\text{hyp})^2=(\text{adj})^2+(\text{opp})^2$

$(52)^2=(1)^2+(\text{opp})^2$

$\text{opp}=\sqrt{(52)^2-1}$

$\text{opp}=\sqrt{2703}$

length of the opposite side = √(2703) ≈ 51.9904

we can find the sin(θ) using this side:

$\sin{\theta} = \dfrac{\text{opp}}{\text{hyp}}$

$\sin{\theta} = \dfrac{\sqrt{2703}}{52}}$

and since $\csc{\theta} = \dfrac{1}{\sin{\theta}}$

$\csc{\theta} = \dfrac{52}{\sqrt{2703}}$

4 0

3 years ago

Let f(x)=g(x)h(x), g(10)=-4, h(10)=560, g'(10)=0, and h'(10)=35. find f'(10)

Ulleksa [173]

This question has extraneous info to trick you.
f(x) = g(x)h(x) ⇒ f'(x) = g'(x)h'(x) . Letting x = 10, we get f'(10) = g'(10)h'(10). then just plug in the values provided. g(10) and h(10) are there to throw you off, just use g'(10) and h'(10).
So f'(10), pronounced "eff prime of ten", = 0 * 35 = 0.

If the question were asking for f(10) instead of f'(10) then you would use g(10) and h(10), ⇒-4*560=90.

7 0

3 years ago