Can someone do 4, 5, 7 and 8 im failing math and avoiding a whooping

Tomas reparte 9 kg de mandarinas en mallas de tres cuartos de kilo cada una . ¿Cuantas mallas obtiene?

MrRissso [65]

Responder:

12

Explicación paso a paso:

Dado que:

Kilogramo total de mandarina = 9 kg

Tamaño de cada distribución = 3/4 de kilo

Número de mallas:

Kilogramo / tamaño total por distribución

9 kg ÷ 3/4 kg

9 * 4/3

= (9 * 4) / 3

= 36/3

= 12

7 0

2 years ago

3. Two cars are parked, 150m apart, and on opposite sides of a building. A camera on the top of the building rotates and can vie

allsm [11]

Answer:

<em>The building is 61.5 m tall</em>

Step-by-step explanation:

The image below is a diagram where all the given distances and angles are shown. We have additionally added some variables:

h = height of the building

a, b = internal angles of each triangle

x = base of each triangle

The angles a and b can be easily found by subtracting the given angles from 90° since they are complementary angles, thus:

a = 90° - 37° = 53°

b = 90° - 42° = 48°

Now we apply the tangent ratio on both triangles separately:

$\displaystyle \tan a=\frac{\text{opposite leg}}{\text{adjacent leg}}$

$\displaystyle \tan 53^\circ=\frac{150-x}{h}$

$\displaystyle \tan 48^\circ=\frac{x}{h}$

From the last equation:

$x=h.\tan 48^\circ$

Substituting into the first equation:

$\displaystyle \tan 53^\circ=\frac{150-h.\tan 48^\circ}{h}$

Operating on the right side:

$\displaystyle \tan 53^\circ=\frac{150}{h}-\tan 48^\circ$

Rearranging:

$\displaystyle \tan 53^\circ+\tan 48^\circ=\frac{150}{h}$

Solving for h:

$\displaystyle h=\frac{150}{\tan 53^\circ+\tan 48^\circ}$

Calculating:

h = 61.5 m

The building is 61.5 m tall

5 0

2 years ago

Solve for x.<br> Need help, please.

dlinn [17]

x=9

hope this helps you. :)

5 0

2 years ago

Major arc JL measures 300°. Which describes triangle JLM?

Fed [463]

Answer: equilateral

Step-by-step explanation:

8 0

2 years ago

Find the number b such that the line y = b divides the region bounded by the curves y = 36x2 and y = 25 into two regions with eq

Gemiola [76]

Answer:

b = 15.75

Step-by-step explanation:

Lets find the interception points of the curves

36 x² = 25

x² = 25/36 = 0.69444

|x| = √(25/36) = 5/6

thus the interception points are 5/6 and -5/6. By evaluating in 0, we can conclude that the curve y=25 is above the other curve and b should be between 0 and 25 (note that 0 is the smallest value of 36 x²).

The area of the bounded region is given by the integral

$\int\limits^{5/6}_{-5/6} {(25-36 \, x^2)} \, dx = (25x - 12 \, x^3)\, |_{x=-5/6}^{x=5/6} = 25*5/6 - 12*(5/6)^3 - (25*(-5/6) - 12*(-5/6)^3) = 250/9$

The whole region has an area of 250/9. We need b such as the area of the region below the curve y =b and above y=36x^2 is 125/9. The region would be bounded by the points z and -z, for certain z (this is for the symmetry). Also for the symmetry, this region can be splitted into 2 regions with equal area: between -z and 0, and between 0 and z. The area between 0 and z should be 125/18. Note that 36 z² = b, then z = √b/6.

$125/18 = \int\limits^{\sqrt{b}/6}_0 {(b - 36 \, x^2)} \, dx = (bx - 12 \, x^3)\, |_{x = 0}^{x=\sqrt{b}/6} = b^{1.5}/6 - b^{1.5}/18 = b^{1.5}/9$

125/18 = b^{1.5}/9

b = (62.5²)^{1/3} = 15.75

8 0

2 years ago