Explain how you could use the distributive property and mental math to find 5 * 198.

For each value of y, determine whether it is a solution to 12 = 2y +8.<br> Is it a solution

grin007 [14]

Answer: 2(0) + 8 does not equal 12, not a solution.

2(2) +8 = 12 yes it is a solution

2(-3) + 8 does not equal 12, not a solution

2(5) + 8 does not equal 12, not a solution.

Step-by-step explanation:

Looks like you need to plug in each y value given and multiplied by 2 and add 8

2(0) + 8 does not equal 12, not a solution.

2(2) +8 = 12 yes it is a solution

2(-3) + 8 does not equal 12, not a solution

2(5) + 8 does not equal 12, not a solution.

3 0

3 years ago

Solve.

Paul [167]

What's the question and are the jobs making food and bringing it to tables

6 0

2 years ago

A point on the rim of a wheel has a linear speed of 33 cm/s. If the radius of the wheel is 50 cm, what is the angular speed of t

Zigmanuir [339]

Answer:

The angular speed of the wheel in radians per second is 0.66.

Step-by-step explanation:

Recall the following statement:

A linear speed (v) is given by,

$v=\omega \times r$ ...... (1)

Here, $\omega$ represents the angular speed of the wheel and <em>r</em> represents the radius of the wheel.

From the given information:

Linear speed (v) = 33 cm/s

Radius of the wheel (r) = 50 cm

Now to find the angular speed in radian per second.

$33 =50 \times \omega$

Divide both sides by 50.

$0.66 rad/sec= \omega$

Hence, the angular speed of the wheel in radians per second is 0.66.

4 0

3 years ago

Find an equation of the sphere with center (2, −10, 3) and radius 5. $$25=(x−2)2+(y−(−10))2+(z−3)2 use an equation to describe i

lana66690 [7]

In the $x,y$ plane, we have $z=0$ everywhere. So in the equation of the sphere, we have

$25=(x-2)^2+(y+10)^2+(-3)^2\implies(x-2)^2+(y+10)^2=16=4^2$

which is a circle centered at (2, -10, 0) of radius 4.

In the $x,z$ plane, we have $y=0$ , which gives

$25=(x-2)^2+10^2+(z-3)^2\implies(x-2)^2+(z-3)^2=-75$

But any squared real quantity is positive, so there is no intersection between the sphere and this plane.

In the $y,z$ plane, $x=0$ , so

$25=(-2)^2+(y+10)^2+(z-3)^2\implies(y+10)^2+(z-3)^2=21=(\sqrt{21})^2$

which is a circle centered at (0, -10, 3) of radius $\sqrt{21}$ .

3 0

3 years ago

Two sides and an angle (ssa) of a triangle are given. determine whether the given measurements produce one triangle, two tria

Setler [38]

These are a huge pain. First set up your initial triangle with A and B as your base angles and C as your vertex angle. Now drop an altitude and call it h. You need to solve for h. Use sin 56 = h/13 to get that h = 10.8. The rule is that if the side length of a is greater than the height but less than the side length of b, you have 2 triangles. h<a<b --> 10.8<12<13. Those are true statements so we have 2 triangles. Side a is the side that swings, this is the one we "move", forming the second triangle. First we have to solve the first triangle using the Law of Sines, then we can solve the second. $\frac{sin56}{12} = \frac{sinB}{13}$ to get that angle B is 64 degrees. Now find C: 180-56-64=60. And now for side c: $\frac{sin56}{12} = \frac{sin60}{c}$ and c=12.5. That's your first triangle. In the second triangle, side a is the swinging side and that length doesn't change. Neither does the angle measure. Angle B has a supplement of 180-64 which is 116. So the new angle B in the second triangle is 116, but the length of b doesn't change, either. I'll show you how you know you're right about that in just a sec. The only angle AND side that both change are C and c. If our new triangle has angles 56 and 116, then C has to be 8 degrees. Using the Law of Sines again, we can solve for c: $\frac{sin8}{c} = \frac{sin56}{12}$ and c = 2.0. We can look at this new triangle and determine the side measures are correct because the longest side will always be across from the largest angle, and the shortest side will always be across from the smallest angle. The new angle B is 116, which is across from the longest side of 13. These are hard. Ugh.

8 0

2 years ago