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3 years ago

A scared tarantula was able to go a distance of 20 meters in 6 seconds. What was its

Mathematics

1 answer:

GuDViN [60]3 years ago

3 0

Answer:

3.33 m/s

Step-by-step explanation:

speed = $\frac{meters}{seconds} > \frac{20}{6} = 3.33$

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When number can each term of the equation be multiplied by to eliminate the fractions before solving 6-3/4x+1/3=1/2x+5

PolarNik [594]

Answer:

Step-by-step explanation:

You need to find the least common denominator (LCD) to all the denominators of the fractions present in the equation. These denominators are (writing them in their prime factor form to make our calculations easier):

$4=2^2\\3=3\\2=2$

Therefore, we need to include a factor of 3, and two factors of 2 ( $2^2$ ) in our least common denominator, so this LCD will be a perfectly divided by all three given denominators, therefore eliminating all fractions in the equation.

Our LCD is = $3*2^2=13*4=12$

6 0

3 years ago

A theater group made appearances in two cities. The hotel charge before tax in the second city was $1000 lower than in the first

Artyom0805 [142]

Answer:

Let x = the charge in 1st city before taxes

Let y = the charge in 2nd city before taxes

Set up equation before taxes.

y = x - 1500 eq1

Set up equation for total tax paid.

0.065x + 0.06y = 378.75 eq2

Substitute eq1 into eq2.

0.065x + 0.06(x - 1500) = 378.75

0.065x + 0.06x - 90 = 378.75

0.125x - 90 = 378.75

0.125x = 468.75

x = 3750

Substitute this value of x into eq1.

y = 3750 - 1500

y = 2250

The hotel charge in city one is $3750 and the hotel charge in city two is $2250

3 0

3 years ago

HELP!!! ASAP

Mumz [18]

The answer is 12 percent change

3 0

3 years ago

Find equations of the spheres with center(1, −1, 6)that touch the following planes.a) xy-planeb) yz-planec) xz-plane

Misha Larkins [42]

Given :

Center of sphere , C( 1 , -1 , 6 ) .

To Find :

Find equations of the spheres with center (1, −1, 6) that touch the following planes.a) xy-plane b) yz-plane c) xz-plane .

Solution :

Distance of the point from xy-plane is :

d = 6 units .

So , equation of circle with center C and radius 6 units is :

$(x-1)^2+(y-(-1))^2+(z-6)^2=6^2\\\\(x-1)^2+(y+1)^2+(z-6)^2=36$

Distance of point from yz-plane is :

d = 1 unit .

So , equation of circle with center C and radius 1 units is :

$(x-1)^2+(x+1)^2+(z-6)^2=1^2\\\\(x-1)^2+(x+1)^2+(z-6)^2=1$

Distance of point from xz-plane is :

d = 1 unit .

So , equation of circle with center C and radius 1 units is :

$(x-1)^2+(x+1)^2+(z-6)^2=1^2\\\\(x-1)^2+(x+1)^2+(z-6)^2=1$

Hence , this is the required solution .

4 0

3 years ago

Joe is 1.6m tall. His shadow is 2m long when he stands 3m from the base of a floodlight.height1.6 m2m3mWhat is the height of the

alexdok [17]

We are given the height of Joe which is 1.6 meters, the length of his shadow is 2 meters when he stands 3 meters from the base of the floodlight.

First, we have to illustrate the problem. Then we can observe two right triangles formed, one is using Joe and the length of the shadow, the other is the floodlight and the sum of the distance from the base plus the length of the shadow. To determine the height of the floodlight, use ratio and proportion:

1.6 / 2 = x / (2 +3)

where x is the height of the flood light

solve for x, x = 4. Therefore, the height of the floodlight is 4 meters.

6 0

3 years ago