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3 years ago

15

The map shows the location of a mall, library, and school in a city: Coordinate grid shown from negative 16 to positive 16 on x

axis at intervals of 2, and negative 10 to positive 10 on y axis at intervals of 2. A triangle is shown with vertices labeled Library, Mall, and School. Library is the ordered pair negative 15, 6, Mall is the ordered pair 15, 6, and School is the ordered pair 15 and negative 10. Mike traveled from the school to the mall and then from the mall to the library. Henry traveled from the school to the library. How many more miles did Mike travel than Henry? 10 miles 12 miles 34 miles 46 miles

1 answer:

matrenka [14]3 years ago

3 0

16 because 8 is = to 15

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What is the value of x?<br> Enter your answer in the box.

Law Incorporation [45]

Answer:

x=25

Step-by-step explanation:

3x-5=2x+20=4x-30

3(25)-5 = 2(25) + 20 = 4(25) - 30

75-5 = 50 + 20 = 100 - 30

70=70=70

7 0

2 years ago

My smart people help me love yall! ^^

mr_godi [17]

Answer:

<u>x = 36</u>

Step-by-step explanation:

180 - 126 = 54

90 + 54 + x = 180

144 + x = 180

180 - 144 = x

x = 36

Have a nice day!

7 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

3 years ago

Seema covered 3 4/13(three four by thirteen)km by bus,2 5/13(Two five by thirteen)km by metro and 1 2/13(One two by thirteen)km

Ad libitum [116K]

Answer: 89/13 km

Step-by-step explanation:

3 4/13 + 2 5/13 + 1 2/13

= 43/13 + 31/13 + 15/13

= 89/13 km

6 0

3 years ago

Identify the restrictions on the domain of f(x) = quantity x plus 5 over quantity x minus 2. x ≠ 5 x ≠ −5 x ≠ 2 x ≠ −2

horsena [70]

Answer: Third option.

Step-by-step explanation:

By definition, Rational Functions have the following form:

$f(x)=\frac{P(x)}{Q(x)}$

Where $P(x)$ and $Q(x)$ are polynomials.

The Restrictions of the Domain of Rational Functions are those Real numbers that make the denominator equal to zero, because the division by zero is not defined.

In this case, you have the following Rational Function:

$f(x)=\frac{x+5}{x-2}$

The Restrictions of the Domain can be found applying this steps:

- Make the denominator equal to 0:

$x-2=0$

- Solve for "x":

$x=2$

Then, the Domain of this function includes all "x" not equal to 2.

Therefore, the answer is:

$x\neq 2$

8 0

3 years ago