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2 years ago

Find each sum 1. 56 + (-56) 2. -240 + 370 3. -5.7 + (-4.2)

Mathematics

2 answers:

Zinaida [17]2 years ago

4 0

Answer:

1. 0

2. 130

3. − 9.9

sorry if am wrong

Zolol [24]2 years ago

3 0

Answer:

1. 56 + (-56) = 56 - 56 = 0

2. -240 + 370 = 370-240 = 130

3. -5.7 + (-4.2) = -(5.7+4.2) = -9.9

Let me know if this helps!

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Please answer correctly !!!!!!!!!!!!!!!!!! Will mark brainliest !!!!!!!!!!!!!!!!!

Olin [163]

U is a point in the line segment TV

TU + UV = TV

TU = 12

UV = 3

Hope this will help...

4 0

3 years ago

F(c)= 9/5c+ 32 I just don't understand how to find the inverse of the function

Viefleur [7K]

F = 9/5 C + 32
- 32 -32

subtracting - 32 from both sides

F - 32 = 9/5c
5/9 F -32 = 5/9(9/5)C

Invert 9/5 to 5/9 multiply to both sides

5/9F - 32 = C

The answer is 5/9F-32 = C

5 0

3 years ago

Please help me idk this

PIT_PIT [208]

Answer:

SA=94

Step-by-step explanation:

SA= 2(4*5) + 2(4*3) + 2(5*3)

SA=2(20) + 2(12) + 2(15)

SA=40 + 24 + 30

SA=94

8 0

3 years ago

AC is tangent to circle O at point C. What is the length of OC

Sedaia [141]

Let OC is x, so OC=OB=x

${4}^{2} + x^{2} = {(2 + x)}^{2} \\ 16 + {x}^{2} = 4 + 4x + {x}^{2} \\ 12 = 4x \\ x = 3$

Answer: OC=3 points

6 0

3 years ago

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