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3 years ago

What is mZDAR in circle A?

Mathematics

1 answer:

Andreas93 [3]3 years ago

5 0

Answer:

68°

Step-by-step explanation:

Since, angle subtended on the circumference of the circle is half angle subtended on the center of the circle.

$\therefore m\angle DBR = \frac{1}{2} \times m\angle DAR\\\\\therefore 34\degree = \frac{1}{2} \times m\angle DAR\\\\\therefore 34\degree \times 2= m\angle DAR\\\\\therefore 68\degree = m\angle DAR\\\\\therefore m\angle DAR=68\degree$

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Slav-nsk [51]

Answer:

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Step-by-step explanation:

1)5000

2)6450

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3 years ago

Can someone help me answer

siniylev [52]

Answer: 9 kilos total thats it

Step-by-step explanation: Leaving the store Fatima thinks,“Each bag of groceries seems asheavy as a watermelon!”UseFatima’s idea about theweight of thewatermelon to estimate the total weight of 7 bags.c.The grocer helps carry about 9 kilograms. Fatima carries the rest.Estimate how many kilograms ofgroceries Fatima carries.d.It takes Fatima 12 minutes to drive to the bank after she leaves the store and then 34 more minutesto drive home.How many minutes does Fatima drive after she leaves the store?

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3 years ago

Ramiro buys 50 packs of candy bars for $40 before a tax of 7.5%. He sells each candy var for 175% of what he paid before tax. ho

Zepler [3.9K]

Answer:

Cost price of each candy = $0.80, Selling price of each candy = $1.40

Step-by-step explanation:

Given:

Ramiro buys 50 packs of candy bars for $40 before a tax of 7.5%.

He sells each candy var for 175% of what he paid before tax.

Question asked:

How much does he buy each candy bar for ?

How much does he sell one for ?

Solution:

By unitary method:

Cost of 50 packs of candy bars = $40

Cost price of 1 pack of candy bars = $\frac{40}{50} =\$0.8$

Cost price of each candy = $0.80

Thus, he buy each candy bar for $0.80

Now:-

As he sells each candy var for 175% of what he paid before tax:-

Selling price of each candy = 0.80 $\times$ 175%

= $0.80\times\frac{175}{100} =\frac{140}{100}=\$1.4$

Thus, he sell one candy for $1.40

3 0

3 years ago

Can someone please help me with this

Mama L [17]

Answer:

27.

Equation of 2 tangent lines at the given curve going through point P is:

y = -7x + 1

y = x + 1

28.

Part 1: 400 feet

Part 2: Velocity is +96 feet/second (or 96 feet/second UPWARD) & Speed is 96 feet per second

Part 3: acceleration at any time t is -32 feet/second squared

Part 4: t = 10 seconds

29.

Part 1: Average Rate of Change = -15

Part 2: The instantaneous rate of change at x = 2 is -8 & at x = 3 is -23

Step-by-step explanation:

27.

First of all, the equation of tangent line is given by:

$y-y_1=m(x-x_1)$

Where m is the slope, or the derivative of the function

Now,

If we take a point x, the corresponding y point would be $x^2-3x+5$ , so the point would be $(x,x^2-3x+5)$

Also, the derivative is:

$f(x)=x^2-3x+5\\f'(x)=2x-3$

Hence, we can equate the DERIVATIVE (slope) and the slope expression through the point given (0,1) and the point we found $(x,x^2-3x+5)$

The slope is $\frac{y_2-y_1}{x_2-x_1}$

So we have:

$\frac{x^2-3x+5-1}{x-0}\\=\frac{x^2-3x+4}{x}$

Now, we equate:

$2x-3=\frac{x^2-3x+4}{x}$

We need to solve this for x. Shown below:

$2x-3=\frac{x^2-3x+4}{x}\\x(2x-3)=x^2-3x+4\\2x^2-3x=x^2-3x+4\\x^2=4\\x=-2,2$

So, this is the x values of the point of tangency. We evaluate the derivative at these 2 points, respectively.

$f'(x)=2x-3\\f'(-2)=2(-2)-3=-7\\f'(2)=2(2)-3=1$

Now, we find 2 equations of tangent lines through the point (0,1) and with respective slopes of -7 and 1. Shown below:

$y-y_1=m(x-x_1)\\y-1=-7(x-0)\\y-1=-7x\\y=-7x+1$

and

$y-y_1=m(x-x_1)\\y-1=1(x-0)\\y-1=x\\y=x+1$

So equation of 2 tangent lines at the given curve going through point P is:

y = -7x + 1

y = x + 1

28.

Part 1:

The highest point is basically the maximum value of the position function. To get maximum, we differentiate and set it equal to 0. Let's do this:

$s(t)=160t-16t^2\\s'(t)=160-32t\\s'(t)=0\\160-32t=0\\32t=160\\t=\frac{160}{32}\\t=5$

So, at t = 5, it reaches max height. We plug in t = 5 into position equation to find max height:

$s(t)=160t-16t^2\\s(5)=160(5)-16(5^2)\\=400$

max height = 400 feet

Part 2:

Velocity is speed, but with direction.

We also know the position function differentiated, is the velocity function.

Let's first find time(s) when position is at 256 feet. So we set position function to 256 and find t:

$s(t)=160t-16t^2\\256=160t-16t^2\\16t^2-160t+256=0\\t^2-10t+16=0\\(t-2)(t-8)=0\\t=2,8$

At t = 2, the velocity is:

$s'(t)=v(t)=160-32t\\v(2)=160-32(2)\\v(2)=96$

It is going UPWARD at this point, so the velocity is +96 feet/second or 96 feet/second going UPWARD

The corresponding speed (without +, -, direction) is simply 96 feet/second

Part 3:

We know the acceleration is the differentiation of the velocity function. let's find it:

$v(t)=160-32t\\v'(t)=a(t)=-32$

hence, the acceleration at any time t is -32 feet/second squared

Part 4:

The rock hits the ground when the position is 0 (at ground). So we equate the position function, s(t), to 0 and find time when it hits the ground. Shown below:

$s(t)=160t-16t^2\\0=160t-16t^2\\16t^2-160t=0\\16t(t-10)=0\\t=0,10$