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2 years ago

THERE IS 2 QUESTIONS PLEASE HELP ME!

Mathematics

1 answer:

lukranit [14]2 years ago

6 0

Answer:

Number 2 is 3:4.

Number 1 is d.

Step-by-step explanation:

Boxers and collies added together gives you 12:16. Divide them both by 4 and you get 3:4.

The number of student in class 4 (6) divided by the total number of kids in the entire class (18) gives you 0.33333333333. Multiply that by 100 and it gives you 33.3333333333. As a percentage it is 33 1/2 %

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500 divide 10/2 and divide 5 and times x divide 5

coldgirl [10]

(500 / (10/2) / 5 ) * x / 5 = 20 * x / 5 = (20 / 5) * x = 4 * x

3 0

3 years ago

Find the square. (1/4A + 1/4B)^2 a) 1/16A^2 + 1/8AB + 1/16B^2 b) 1/4A^2 + 1/8AB + 1/4B^2

alisha [4.7K]

$(a + b)^2 = a^2 + 2ab + b^2$

$(\dfrac{1}{4}A + \dfrac{1}{4}B)^2 =$

$= \dfrac{1}{16}A^2 + \dfrac{1}{8}AB + \dfrac{1}{16}B^2$

Answer: A.

Another way to solve by factoring 1/4.

$(\dfrac{1}{4}A + \dfrac{1}{4}B)^2 =$

$= [\dfrac{1}{4}(A + B)]^2$

$= (\dfrac{1}{4})^2(A + B)^2$

$= \dfrac{1}{16}(A^2 + 2AB + B^2)$

$= \dfrac{1}{16}A^2 + \dfrac{1}{16}2AB + \dfrac{1}{16}B)^2$

$= \dfrac{1}{16}A^2 + \dfrac{1}{8}AB + \dfrac{1}{16}B)^2$

6 0

3 years ago

Read 2 more answers

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.