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3 years ago

Find the measurements of each angle

Mathematics

1 answer:

svp [43]3 years ago

7 0

From 7-12

60, 65, 115, 55, 120, 60

explanation: according to the picture, ab=bc=ac

cd=ce

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The solution of the system of equations Ax+y=5 and Ax+By=20 is (2,-3). What are the values of A and B

STatiana [176]

Answer:

A = 4
B = -4

Step-by-step explanation:

Put the values of x and y in the equations and solve the resulting system.

A(2) +(-3) = 5

A(2) +B(-3) = 20

The first equation tells the value of A:

2A = 8 . . . . add 3

A = 4 . . . . . .divide by 2

This and the second equation tells the value of B:

(4)(2) -3B = 20

-3B = 12 . . . . . subtract 8

B = -4 . . . . . . . divide by -3

The values of A and B are 4 and -4, respectively.

6 0

3 years ago

Use f’( x ) = lim With h ---> 0 [f( x + h ) - f ( x )]/h to find the derivative at x for the given function. 5-x²

beks73 [17]

<h2>Answer:</h2>

The derivative of the function f(x) is:

$f'(x)=-2x$

<h2>Step-by-step explanation:</h2>

We are given a function f(x) as:

$f(x)=5-x^2$

We have:

$f(x+h)=5-(x+h)^2\\\\i.e.\\\\f(x+h)=5-(x^2+h^2+2xh)$

( Since,

$(a+b)^2=a^2+b^2+2ab$ )

Hence, we get:

$f(x+h)=5-x^2-h^2-2xh$

Also, by using the definition of f'(x) i.e.

$f'(x)= \lim_{h \to 0} \dfrac{f(x+h)-f(x)}{h}$

Hence, on putting the value in the formula:

$f'(x)= \lim_{h \to 0} \dfrac{5-x^2-h^2-2xh-(5-x^2)}{h}\\\\\\f'(x)=\lim_{h \to 0} \dfrac{5-x^2-h^2-2xh-5+x^2}{h}\\\\i.e.\\\\f'(x)=\lim_{h \to 0} \dfrac{-h^2-2xh}{h}\\\\f'(x)=\lim_{h \to 0} \dfrac{-h^2}{h}+\dfrac{-2xh}{h}\\\\f'(x)=\lim_{h \to 0} -h-2x\\\\i.e.\ on\ putting\ the\ limit\ we\ obtain:\\\\f'(x)=-2x$

Hence, the derivative of the function f(x) is:

$f'(x)=-2x$

3 0

3 years ago

Read 2 more answers

-244 is it

fgiga [73]

-244 is an integer, rational and real number

5 0

3 years ago

A plane traveled 240 miles. The trip was with the wind. It took 3 hours. The trip back was into the wind. The trip back took 6 h

sveta [45]

Answer:

Plane speed is 60 miles/hour

Wind speed is 20 miles/hour

Step-by-step explanation:

Lets suppose the speed of the plane in still air is s and speed of the wind is w and since speed in the formula below equals speed of the plane plus the wind's speed so speed=s+w for this journey with the wind

$Speed=\frac{Distance}{Time}$

$Distance=Speed*Time$

For the first journey the trip with the wind:

$Distance=Speed*Time\\240=(s+w)(3)\\240/3=s+w\\s+w=80$

Here we got our first equation for the second equation we use the information from the second trip which was into the wind. Which means the speed here would be speed=s-w

$Distance=Speed*Time\\240=(s-w)*6\\240/6=s-w\\s-w=40$