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3 years ago

Which expression is equal to 3(5 + 6)?

Mathematics

1 answer:

Alina [70]3 years ago

6 0

D. 15+18 Hope that helps!

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HELP PLS!!! A rectangular garden has a walkway around it. The area of the garden is 4(5.5x + 2.5) The combined area of the garde

pishuonlain [190]

Answer:

4x5+2x5= 30

Step-by-step explanation:

4 0

2 years ago

We are standing on the top of a 1680 ft tall building and throw a small object upwards. At every second, we measure the distance

MAVERICK [17]

Answer:

a) The height of the small object 3 seconds after being launched is 2304 feet.

b) The small object ascends 128 feet between 5 seconds and 7 seconds.

c) The object will take 6 and 10 seconds after launch to reach a height of 2640 feet.

d) The object will take 21 seconds to hit the ground.

Step-by-step explanation:

The correct formula for the height of the small object is:

$h(t) = -16\cdot t^{2}+256\cdot t+1680$ (1)

Where:

$h$ - Height above the ground, measured in feet.

$t$ - Time, measured in seconds.

a) The height of the small object at given time is found by evaluating the function:

$h(3\,s)= -16\cdot (3\,s)^{2}+256\cdot (3\,s)+1680$

$h(3\,s) = 2304\,ft$

The height of the small object 3 seconds after being launched is 2304 feet.

b) First, we evaluate the function at $t = 5\,s$ and $t = 7\,s$ :

$h(5\,s)= -16\cdot (5\,s)^{2}+256\cdot (5\,s)+1680$

$h(5\,s) = 2560\,s$

$h(7\,s)= -16\cdot (7\,s)^{2}+256\cdot (7\,s)+1680$

$h(7\,s) = 2688\,s$

We notice that the small object ascends in the given interval.

$\Delta h = h(7\,s)-h(5\,s)$

$\Delta h = 128\,ft$

The small object ascends 128 feet between 5 seconds and 7 seconds.

c) If we know that $h = 2640\,ft$ , then (1) is reduced into this second-order polynomial:

$-16\cdot t^{2}+256\cdot t-960=0$ (2)

All roots of the resulting equation come from the Quadratic Formula:

$t_{1} = 10\,s$ and $t_{2}= 6\,s$

The object will take 6 and 10 seconds after launch to reach a height of 2640 feet.

d) If we know that $h = 0\,ft$ , then (1) is reduced into this second-order polynomial:

$-16\cdot t^{2}+256\cdot t +1680 = 0$ (3)

All roots of the resulting equation come from the Quadratic Formula:

$t_{1} = 21\,s$ and $t_{2} = -5\,s$

Just the first root offers a solution that is physically reasonable.

The object will take 21 seconds to hit the ground.

8 0

3 years ago

Which Value From The Number Set 1,3,5,7 Makes The Inequality X+5>10 True

Ne4ueva [31]

Put all number choices in for x. 1+5=6, 3+5=8, 5+5=10, 7+5=12. The equation says that the answer has to be greater than ten. 6 and 8 are less than ten, and ten is equal to ten, so the only answer is 7, because 7+5=12 :)

6 0

3 years ago

Please help if you can. I will give Brainiest to best answer. This question is a Calculus/Trigonometry question. I ask that you

Ierofanga [76]

7 is incorrect. The answer should be -4. Here's how I'd derive it:

$\tan^2\dfrac x2=\dfrac{\sin^2\frac x2}{\cos^2\frac x2}=\dfrac{\frac{1-\cos x}2}{\frac{1+\cos x}2}=\dfrac{1-\cos x}{1+\cos x}$

With $\pi$ , we should expect $\cos x$ . If $\tan x=\dfrac8{15}$ , then

$\sec x=-\sqrt{1+\tan^2x}=-\dfrac{17}{15}\implies\cos x=-\dfrac{15}{17}$

Also,

$\pi$

so we should expect $\tan\dfrac x2$ and

$\tan\dfrac x2=-\sqrt{\dfrac{1-\cos x}{1+\cos x}}=-4$

You seem to be taking

$\tan\dfrac x2=\dfrac{1+\cos x}{-\sin x}$

but this is not an identity.

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8 is incorrect. Just a silly mistake, you swapped the order of the terms in the numerator. It should be

$\dfrac{\sqrt6-\sqrt2}4$

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9. Use the identity from (7). $\dfrac{7\pi}{12}$ lies in the second quadrant, so

$\tan\dfrac{7\pi}{12}=-\sqrt{\dfrac{1-\cos\frac{7\pi}6}{1+\cos\frac{7\pi}6}}=-\sqrt{7+4\sqrt3}=-2-\sqrt3$

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12.

$\dfrac{\cot x-1}{1-\tan x}=\dfrac{\frac{\cos x}{\sin x}-1}{1-\frac{\sin x}{\cos x}}$

$=\dfrac{\cos x(\cos x-\sin x)}{\sin x(\cos x-\sin x)}$

$=\dfrac{\cos x}{\sin x}$

$=\dfrac{\csc x}{\sec x}$

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13.

$\dfrac{1+\tan x}{\sin x+\cos x}=\dfrac{1+\frac{\sin x}{\cos x}}{\sin x+\cos x}$

$=\dfrac{\cos x+\sin x}{\cos x(\sin x+\cos x)}$

$=\dfrac1{\cos x}$

$=\sec x$

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14.

$\sin2x(\cot x+\tan x)=2\sin x\cos x\left(\dfrac{\cos x}{\sin x}+\dfrac{\sin x}{\cos x}\right)$

$=2\sin x\cos x\dfrac{\cos^2x+\sin^2x}{\sin x\cos x}$

$=2$

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15.

$\dfrac{1-\tan^2\theta}{1+\tan^2\theta}=\dfrac{1-\frac{\sin^2\theta}{\cos^2\theta}}{1+\frac{\sin^2\theta}{\cos^2\theta}}$

$=\dfrac{\cos^2\theta-\sin^2\theta}{\cos^2\theta+\sin^2\theta}$

$=\cos2\theta$

3 0

3 years ago

PLZ ASAP, I NEED HELP ASAP

loris [4]

I won't give you the answer because I am sure you are capable of finding it yourself once given a push, but what I would do is simplify all of the numbers into a mixed fraction (a b/c) and then go from there. Don't subtract the two numbers with the denominator of 25 first because they goes against PEMDAS. Change all of the fractions so they have like denominators, and once you solve it, just write the fraction part of the answer (b/c). If you have any other questions, just ask.

8 0

3 years ago