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3 years ago

Solve for b.

2 answers:

8 0

Expression: a = 9b²c

b² = a/9c

b = √(a/9c)

b = √a/3√c
Or b = 1/3 √(a/c)

In short, Your Answer would be: Option D

Hope this helps!

Ipatiy [6.2K]3 years ago

3 0

Answer:

$b=\pm\frac{1}{3}\sqrt{\frac{a}{c}}$

Step-by-step explanation:

Given expression $a=9b^2c$

We have to find the value in terms of b

Consider the given expression $a=9b^2c$

Divide both side by c, we have

$\frac{a}{c}=9b^2$

Again divide by 9 , we have

$\frac{a}{9c}=b^2$

Taking square root both side, we have,

$\sqrt{\frac{a}{9c}}=\sqrt{b^2}$

Simplify , we have,

$\sqrt{\frac{a}{9c}}=b$

We know $\sqrt{9}=\pm 3$

Thus, $b=\pm\frac{1}{3}\sqrt{\frac{a}{c}}$

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Can someone help me please

andrey2020 [161]

Answer: y = 6

Step-by-step explanation:

A square's area can be done by using s^2, where s is y in this case. Because there are 5 squares, the area of the figure is 5y^2. Because the area is also 180cm, 5y^2=180.

Then divide both sides of the equation by 5 to get y^2 = 36. Then square root both sides of the equation to get y = 6.

Hope it helps <3

7 0

3 years ago

You work at a pioneer historical site. On this site you have handcarts. One cart has a handle that connects to the center of the

Gelneren [198K]

Answer:

a) see below

b) radius = 16.4 in (1 d.p.)

c) 18°. Yes contents will remain. No, handle will not rest on the ground.

d) Yes contents would spill. Max height of handle = 32.8 in (1 d.p.)

Step-by-step explanation:

Part a

A chord is a line segment with endpoints on the circumference of the circle.

The diameter is a chord that passes through the center of a circle.

Therefore, the spokes passing through the center of the wheel are congruent chords.

The spokes on the wheel represent the radii of the circle. Spokes on a wheel are usually evenly spaced, therefore the congruent central angles are the angles formed when two spokes meet at the center of the wheel.

Part b

The tangent of a circle is always perpendicular to the radius.

The tangent to the wheel touches the wheel at point B on the diagram. The radius is at a right angle to this tangent. Therefore, we can model this as a right triangle and use the tan trigonometric ratio to calculate the radius of the wheel (see attached diagram 1).

$\sf \tan(\theta)=\dfrac{O}{A}$

where:

$\theta$ is the angle
O is the side opposite the angle
A is the side adjacent the angle

Given:

$\theta$ = 20°
O = radius (r)
A = 45 in

Substituting the given values into the tan trig ratio:

$\implies \sf \tan(20^{\circ})=\dfrac{r}{45}$

$\implies \sf r=45\tan(20^{\circ})$

$\implies \sf r=16.37866054...$

Therefore, the radius is 16.4 in (1 d.p.).

Part c

The measure of an angle formed by a secant and a tangent from a point outside the circle is half the difference of the measures of the intercepted arcs.

If the measure of the arc AB was changed to 72°, then the other intercepted arc would be 180° - 72° = 108° (since AC is the diameter).

$\implies \sf new\: angle=\dfrac{108^{\circ}-72^{\circ}}{2}=18^{\circ}$

As the handle of the cart needs to be no more than 20° with the ground for the contents not to spill out, the contents will remain in the handcart at an angle of 18°.

The handle will not rest of the ground (see attached diagram 2).

Part d

This can be modeled as a right triangle (see diagram 3), with:

height = (48 - r) in
hypotenuse ≈ 48 in

Use the sin trig ratio to find the angle the handle makes with the horizontal:

$\implies \sf \sin (\theta)=\dfrac{O}{H}$

$\implies \sf \sin (\theta)=\dfrac{48-r}{48}$

$\implies \sf \sin (\theta)=\dfrac{48-45\tan(20^{\circ})}{48}$

$\implies \theta = 41.2^{\circ}\:\sf(1\:d.p.)$

As 41.2° > 20° the contents will spill out the back.

To find the maximum height of the handle from the ground before the contents start spilling out, find the height from center of the wheel (setting the angle to its maximum of 20°):

$\implies \sin(20^{\circ})=\dfrac{h}{48}$

$\implies h=48\sin(20^{\circ})$

Then add it to the radius:

$\implies \sf max\:height=48\sin(20^{\circ})+45\tan(20^{\circ})=32.8\:in\:(1\:d.p.)$

(see diagram 4)

------------------------------------------------------------------------------------------

Circle Theorem vocabulary

Secant: a straight line that intersects a circle at two points.

Arc: the curve between two points on the circumference of a circle

Intercepted arc: the curve between the two points where two chords or line segments (that meet at one point on the other side of the circle) intercept the circumference of a circle.

Tangent: a straight line that touches a circle at only one point.

7 0

2 years ago

Which table does not represent a function?

andrew-mc [135]

the correct option is answer choice D, because there are multiple 2's in the x column with different f(x) values.

4 0

3 years ago

Write a number or expression in each blank space to create true equations.

Rina8888 [55]

Answer:

$7(3+5)=7(3)+7(5)\\15-10=5(3-2)$

Step-by-step explanation:

Both expressions are examples of the distributive property, which basically says "if I have this many groups of some size and that many groups of the same size, I've got this + that groups of that size altogether."

To give an example, if I've got 3 groups of 5 and 2 groups of 5, I've got 3 + 2 = 5 groups of 5 in total. I've attached a visual from Math with Bad Drawings to illustrate this idea.

Mathematically, we'd capture that last example with the equation

$5(3)+5(2)=5(3+2)$ . We can also read that in reverse: 3 + 2 groups of 5 is the same as adding together 3 groups of 5 and 2 groups of 5; both directions get us 8 groups of 5. We can use this fact to rewrite the first expression like this: $7(3+5)=7(3)+7(5)$ .

This idea extends to subtraction too: If we have 3 groups of 4 and we take away 1 group of 4, we'd expect to be left with 3 - 1 = 2 groups of 4, or in symbols: $4(3)-4(1)=4(3-1)=4(2)$ . When we start with two numbers like 15 and 10, our first question should be if we can split them up into groups of the same size. Obviously, you could make 15 groups of 1 and 10 groups of 1, but 15 is also the same as 3 groups of 5 and 10 is the same as 2 groups of 5. Using the distributive property, we could write this as $15-10=3(5)-2(5)=5(3-2)$ , so we can say that $15-10=5(3-2)$ .

4 0

3 years ago

Solve these equations: 3|x+5|=7 | = absolute value

loris [4]

Answer:

x=-8/3

x=-22/3

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers