All categories

3 years ago

Find the measure of line segment CU. A) 27 B) 16 C) 25 D) 21

Mathematics

2 answers:

Ahat [919]3 years ago

8 0

Do you have the answers to the rest/multiple choice part of the test ??

OleMash [197]3 years ago

5 0

Answer:

D) 21

Step-by-step explanation:

r*t=s*u

14*15= 210

so, the easy way to do it is to just multiply all the answer choices by 10 unitil you get 210.

WHY because r*t = 210 so, 10*x = 210

If you want the break down just set up an equation 10 *( 5x +1 ) = 210

210/10 =21

21-1=20

20/5=4

PLUG IN, 5*4=20

20 +1 = 21

There you go kiddos :)

You might be interested in

I just need the answer

kogti [31]

The answer is 6 with a remainder of 1. (6 R.1)

Hope this helps!!

Please Mark Brainliest!!!

3 0

2 years ago

Consider the work shown for 99÷7. What does the highlighted 1 represent?

kirill [66]

Answer:

The amount left after each group gets 14

Step-by-step explanation:

R1 means there is 1 left after dividing everything

5 0

2 years ago

Read 2 more answers

Find thd <img src="https://tex.z-dn.net/?f=%5Cfrac%7Bdy%7D%7Bdx%7D" id="TexFormula1" title="\frac{dy}{dx}" alt="\frac{dy}{dx}" a

NARA [144]

$x^3y^2+\sin(x\ln y)+e^{xy}=0$

Differentiate both sides, treating $y$ as a function of $x$ . Let's take it one term at a time.

Power, product and chain rules:

$\dfrac{\mathrm d(x^3y^2)}{\mathrm dx}=\dfrac{\mathrm d(x^3)}{\mathrm dx}y^2+x^3\dfrac{\mathrm d(y^2)}{\mathrm dx}$

$=3x^2y^2+x^3(2y)\dfrac{\mathrm dy}{\mathrm dx}$

$=3x^2y^2+6x^3y\dfrac{\mathrm dy}{\mathrm dx}$

Product and chain rules:

$\dfrac{\mathrm d(\sin(x\ln y)}{\mathrm dx}=\cos(x\ln y)\dfrac{\mathrm d(x\ln y)}{\mathrm dx}$

$=\cos(x\ln y)\left(\dfrac{\mathrm d(x)}{\mathrm dx}\ln y+x\dfrac{\mathrm d(\ln y)}{\mathrm dx}\right)$

$=\cos(x\ln y)\left(\ln y+\dfrac1y\dfrac{\mathrm dy}{\mathrm dx}\right)$

$=\cos(x\ln y)\ln y+\dfrac{\cos(x\ln y)}y\dfrac{\mathrm dy}{\mathrm dx}$

Product and chain rules:

$\dfrac{\mathrm d(e^{xy})}{\mathrm dx}=e^{xy}\dfrac{\mathrm d(xy)}{\mathrm dx}$

$=e^{xy}\left(\dfrac{\mathrm d(x)}{\mathrm dx}y+x\dfrac{\mathrm d(y)}{\mathrm dx}\right)$

$=e^{xy}\left(y+x\dfrac{\mathrm dy}{\mathrm dx}\right)$

$=ye^{xy}+xe^{xy}\dfrac{\mathrm dy}{\mathrm dx}$

The derivative of 0 is, of course, 0. So we have, upon differentiating everything,

$3x^2y^2+6x^3y\dfrac{\mathrm dy}{\mathrm dx}+\cos(x\ln y)\ln y+\dfrac{\cos(x\ln y)}y\dfrac{\mathrm dy}{\mathrm dx}+ye^{xy}+xe^{xy}\dfrac{\mathrm dy}{\mathrm dx}=0$

Isolate the derivative, and solve for it:

$\left(6x^3y+\dfrac{\cos(x\ln y)}y+xe^{xy}\right)\dfrac{\mathrm dy}{\mathrm dx}=-\left(3x^2y^2+\cos(x\ln y)\ln y-ye^{xy}\right)$

$\dfrac{\mathrm dy}{\mathrm dx}=-\dfrac{3x^2y^2+\cos(x\ln y)\ln y-ye^{xy}}{6x^3y+\frac{\cos(x\ln y)}y+xe^{xy}}$

(See comment below; all the 6s should be 2s)

We can simplify this a bit by multiplying the numerator and denominator by $y$ to get rid of that fraction in the denominator.

$\dfrac{\mathrm dy}{\mathrm dx}=-\dfrac{3x^2y^3+y\cos(x\ln y)\ln y-y^2e^{xy}}{6x^3y^2+\cos(x\ln y)+xye^{xy}}$

3 0

2 years ago

Hellooo menejkfnfkdjdkdkdmmdmfjf

Vesna [10]

Helloo, i don’t see the image seen above, is that the only picture?

6 0

3 years ago

A ride-share company has a fee that includes a fixed cost and a cost that depends on both the time spent travelling, in minutes,

zlopas [31]

Answer:

the cost of Roy's ride is $23.05

Step-by-step explanation:

According to the Question,

Let, Cost of per minute charge is 'x' & Cost Of Per Kilometre charge is y .

Given, A ride-share company has a fee of the fixed cost of a ride is $2.55 .
And, The Total cost of the Ride depends on both the time spent on travelling(in minutes), and the distance travelled(in kilometres) .

⇒ Judy's ride costs $16.75 . but the actual cost after deducting the fixed charge is 16.75-2.55 = $14.20, took 8 minutes & The distance travelled was 10 km. Thus, the equation for the journey is 8x+10y=14.20 ⇒ Equ. 1

⇒ Pat's ride costs $30.35 . but the actual cost after deducting the fixed charge is 30.35-2.55 = $27.80, took 20 minutes & The distance travelled was 18 km. Thus, the equation for the journey is 20x+18y=27.80 ⇒ Equ. 2

Now, on Solving Equation 1 & 2, We get

x=0.4(Cost of per minute charge) & y=1.1(Cost Of Per Kilometre charge)

Now, Roy's ride took 10 minutes & The distance travelled was 15 km . Thus, the cost of Roy's Ride is 10x+15y ⇔ 10×0.4 + 15×1.1 ⇔ $20.5

Hence, the total cost of Roy's ride is 20.5 + 2.55(fixed cost) = $23.05

8 0

3 years ago