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3 years ago

I need to factor 7r^2+6r-16

Mathematics

1 answer:

weeeeeb [17]3 years ago

6 0

I hope this helps you

7r^2+6r-16

7r -8

r +2

(7r-8)(r+2)

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Given n//p m<2=50 degrees what is m<7?

Bogdan [553]

Answer: m∠7=130°

Step-by-step explanation:

If two parallel lines are cut by another line (transversal), each one will have four angles surrounding the intersection.

By definition, the adjacent angles of each parallel line, are supplementary, this means that they add up 180°.

Then, if m∠2=50°, then m∠3 is:

m∠3 $=180-50=130$ °

By definition, the angles that are in the same relative position are know as corresponding angles and they are congruent.

Therefore:

m∠3=m∠7

m∠7=130°

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3 years ago

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The Hudson Bay tides vary between 3 feet and 9 feet. The tide is at its lowest point when time (t) is 0 and completes a full cyc

BaLLatris [955]

Answer:

$y(t)= 6-3cos(\dfrac{2\pi}{14}t )$

Step-by-step explanation:

The function that could model this periodic phenomenon will be of the form

$y(t) = y_0+Acos(wt)$

The tide varies between 3ft and 9ft, which means its amplitude $A$ is

$A =\dfrac{(9-3)ft}{2} \\\\\boxed{A = 3ft}$

and its midline $y_0$ is

$y_o=3+3 \\\\\boxed{y_o= 6ft}$ .

Furthermore, since at $t=0$ the tide is at its lowest ( 3 feet ), we know that the trigonometric function we must use is $-cos(\omega t)$ .

The period of the full cycle is 14 hours, which means

$\omega t =2\pi$

$\omega (t+14)= 4\pi$

giving us

$\boxed{\omega = \dfrac{2\pi}{14}.}$

With all of the values of the variables in place, the function modeling the situation now becomes

$\boxed{y(t)= 6-3cos(\dfrac{2\pi}{14}t ).}$

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4 years ago

A triangle has side lengths of 9 in, 13 in, and 20 in. What is the measurement of this triangle's largest angle?

Rama09 [41]

1. Given any triangle ABC with sides BC=a, AC=b and AB=c, the following are true :

i) the larger the angle, the larger the side in front of it, and the other way around as well. (Sine Law) Let a=20 in, then the largest angle is angle A.

ii) Given the measures of the sides of a triangle. Then the cosines of any of the angles can be found by the following formula:

a^{2}=b ^{2}+c ^{2}-2bc(cosA)

20^{2}=9 ^{2}+13 ^{2}-2*9*13(cosA)

400=81+169-234(cosA) 150=-234(cosA)

cosA=150/-234= -0.641

3. m(A) = Arccos(-0.641)≈130°,

4. Remark: We calculate Arccos with a scientific calculator or computer software unless it is one of the well known values, ex Arccos(0.5)=60°, Arccos(-0.5)=120° etc

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3 years ago

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Evaluate 1/6 + 2/3 PLEASEEEEEE HELP

mart [117]

2/3 = 4/6.

1/6 + 4/6 = 5/6.

The answer is 5/6.

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3 years ago

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Help please.........,