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Tju [1.3M]
2 years ago
6

A triangle has side lengths of

Mathematics
1 answer:
alexandr402 [8]2 years ago
6 0

Answer:

P = ( 4y + 6 inches) + ( 8y - B ) inches + ( 5y + 3 ) inches

Step-by-step explanation:

Formula of Perimeter of triangle is = Sum of all sides.

P = A + B + C

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Samantha spent 40 minutes on homework. Sixty percent of that time was spent on language arts. How many minutes were spent on lan
andrew-mc [135]

Answer: 24 minutes it the correct answer you would do 40(60) = 2,400

Deducte the 0's and you have 24

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What is the value of log <br><img src="https://tex.z-dn.net/?f=%20%7B16%7D%5E%7B4%7D%20" id="TexFormula1" title=" {16}^{4} " alt
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A particle is projected with a velocity of <img src="https://tex.z-dn.net/?f=40ms%5E-%5E1" id="TexFormula1" title="40ms^-^1" alt
Katena32 [7]

Answer:

2\sqrt{55}\text{ m/s or }\approx 14.8\text{m/s}

Step-by-step explanation:

The vertical component of the initial launch can be found using basic trigonometry. In any right triangle, the sine of an angle is equal to its opposite side divided by the hypotenuse. Let the vertical component at launch be y. The magnitude of 40\text{ m/s} will be the hypotenuse, and the relevant angle is the angle to the horizontal at launch. Since we're given that the angle of elevation is 60^{\circ}, we have:

\sin 60^{\circ}=\frac{y}{40},\\y=40\sin 60^{\circ},\\y=20\sqrt{3}(Recall that \sin 60^{\circ}=\frac{\sqrt{3}}{2})

Now that we've found the vertical component of the velocity and launch, we can use kinematics equation v_f^2=v_i^2+2a\Delta y to solve this problem, where v_f/v_i is final and initial velocity, respectively, a is acceleration, and \Delta y is distance travelled. The only acceleration is acceleration due to gravity, which is approximately 9.8\:\mathrm{m/s^2}. However, since the projectile is moving up and gravity is pulling down, acceleration should be negative, represent the direction of the acceleration.

What we know:

  • v_i=20\sqrt{3}\text{ m/s}
  • a=-9.8\:\mathrm{m/s^2}
  • \Delta y =50\text{ m}

Solving for v_f:

v_f^2=(20\sqrt{3})^2+2(-9.8)(50),\\v_f^2=1200-980,\\v_f^2=220,\\v_f=\sqrt{220}=\boxed{2\sqrt{55}\text{ m/s}}

3 0
3 years ago
WILL GIVE BRANLIEST!!!!!!!!!!!!!
Lunna [17]

Answer:

D is the answer

Step-by-step explanation:

6 0
2 years ago
Why is the value of q?
Rudiy27

Given:

QSR is a right triangle.

QT = 10

TR = 4

To find:

The value of q.

Solution:

Hypotenuse of QSR = QT + TR

                                 = 10 + 4

                                 = 14

Geometric mean of similar right triangle formula:

$\frac{\text { hypotenuse }}{\text { leg }}=\frac{\text { leg }}{\text { part }}$

$\Rightarrow \frac{14}{q} =\frac{q}{4}

Do cross multiplication.

$\Rightarrow14\times 4 =q\times q

$\Rightarrow 56 =q^2

Switch the sides.

$\Rightarrow q^2 =56

Taking square root on both sides.

\Rightarrow q =2 \sqrt{14}

The value of q is 2 \sqrt{14}.

8 0
3 years ago
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