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3 years ago

14

What are ratios that are equivalent to 6:21

1 answer:

mr_godi [17]3 years ago

7 0

Answer:

$2:7$

Step-by-step explanation:

<u>Step 1: Divide both sides by 3 </u>

$6:21$

$(6/3):(21/3)$

$2:7$

Answer: $2:7$

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Solve the inequality 5y+4 ≤ 22-y and graph the solution set on a number line. Please explain how to do it!

gtnhenbr [62]

Answer:

y≤3

Step-by-step explanation:

5y+4≤22−y

<em>Step 1: Simplify both sides of the inequality.</em>

5y+4≤−y+22

<em>Step 2: Add y to both sides.</em>

5y+4+y≤−y+22+y

6y+4≤22

<em>Step 3: Subtract 4 from both sides.</em>

6y+4−4≤22−4

6y≤18

<em>Step 4: Divide both sides by 6.</em>

6y/6≤18/6

y≤3

4 0

2 years ago

Read 2 more answers

james,priya,and slobhan work in a grocery store.james makes $7.00 per hour . priya makes 20% more than jaimes,and slobhan makes

qaws [65]

Slobhan makes $3.78 per hour

4 0

4 years ago

Please I need help.<br> Find the sum of the interior angles of the figure below.

FrozenT [24]

Ok so to find the sum of interior angles the formula is

s=180(n-2)
s is sum and n is the number of sides

so this is a quadrilateral cause it has 4 sides

because of that n=4

substitute

s=180(4-2)

now you can either distribute or do pemdas but its easier and more efficient to just do pemdas

s=180(2)
s=360 degrees

5 0

3 years ago

Help this is very important

Kay [80]

#2 is C
#3 is B
#4 I’m no that sure
(Ignore the top )

8 0

4 years ago

If w = 5 cos (xy) − sin (xz) and x = 1/t , y = t, z = t^3 ; then find dw/dt

Scrat [10]

In this question, we find the derivatives, using the chain's rule.

Doing this, the derivative is:

$\frac{dw}{dt} = \frac{5}{t}(\sin{1} - \cos{1}) - 2t\cos{t^2}$

Chain Rule:

Suppose we have a function $w(x,y,z), x = x(t), y = y(t), z = z(t)$ , and want to find it's derivative as function of t. It will be given by:

$\frac{dw}{dt} = \frac{dw}{dx}\frac{dx}{dt} + \frac{dw}{dy}\frac{dy}{dt} + \frac{dw}{dz}\frac{dz}{dt}$

Thus, we have to find the desired derivatives, which are:

w of x:

$\frac{dw}{dx} = -5y\sin{(xy)} - z\cos{(xz)}$

Considering $x = \frac{1}{t}, y = t, z = t^3$

$\frac{dw}{dx} = -5t\sin{(1)} - t^3\cos{(t^2)}$

w of y:

$\frac{dw}{dy} = -5x\cos{(xy)}$

Considering $x = \frac{1}{t}, y = t$

$\frac{dw}{dy} = -\frac{5}{t}\cos{1}$

w of z:

$\frac{dw}{dz} = -x\cos{(xz)}$

Considering $x = \frac{1}{t}, z = t^3$

$\frac{dw}{dz} = -\frac{1}{t}\cos{(t^2)}$

Derivatives of x, y and z as functions of t:

$\frac{dx}{dt} = -\frac{1}{t^2}$

$\frac{dy}{dt} = 1$

$\frac{dz}{dt} = 3t^2$

Derivative of w as function of t.

Now, we just replace what we found into the formula. So

$\frac{dw}{dt} = \frac{dw}{dx}\frac{dx}{dt} + \frac{dw}{dy}\frac{dy}{dt} + \frac{dw}{dz}\frac{dz}{dt}$

$\frac{dw}{dt} = (-5t\sin{(1)} - t^3\cos{(t^2)})(-\frac{1}{t^2}) - (\frac{5}{t}\cos{1}) - (\frac{1}{t}\cos{(t^2)})3t^2$

Applying the multiplications:

$\frac{dw}{dt} = \frac{5}{t}\sin{1} + t\cos{t^2} - \frac{5}{t}\cos{1} - 3t\cos{t^2}$

Applying the simplifications:

$\frac{dw}{dt} = \frac{5}{t}(\sin{1} - \cos{1}) - 2t\cos{t^2}$

Which is the derivative.

For more on the chain rule, you can check brainly.com/question/12795383

8 0

3 years ago