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2 years ago

Anybody ????? I will mark you brainliest

Mathematics

2 answers:

Over [174]2 years ago

6 0

I think the answer is B

lesya [120]2 years ago

4 0

The answer is B ....

You might be interested in

Order these numbers least to greatest. 29/4, 7.177,7.17,7 3/11

vazorg [7]

Answer:

-1/8 < 0.33 < 3/8 < 75% < 1 5/8

Step-by-step explanation:

7 0

2 years ago

Mrs.Rodriquez is purchasing trophies for each child that plays little league she wants to spend no more than $7.00 per thropy

Greeley [361]

Answer:

7 trophies for $49.00

5 trophies for $30.00

-----------------------------------

Explination:

It says that she wants to pay no more than 7 dollars for each trophy she buys, which means 7 and less. So we have to multiply 7 x the number of trophies, And if it gives you the excact number or less its correct.

So we multiply

7 x 7 = 49 (The excact number, correct)

9 x 7 = 63 (The price they are selling it at is higher, incorrect)

12 x 7 = 84 (The price they are selling it at is higher, incorrect)

5 x 7 = 35 (Paying less, Correct)

3 x 7 (The price they are selling it at is higher, incorrect)

---------------------------------------------------------------------------------------

Its really easy if you put your mind into it! Goodluck <3

5 0

3 years ago

A circle is inscribed with 3 lines. Point x is at the center of the circle. A vertical line goes through point x to point z on t

Elza [17]

Answer:

the correct answer is

Line segment X Z

4 0

2 years ago

Which clock shows a time after<br> 3:15 but before 4:00

amid [387]

Answer:

there are no clocks to compare

Step-by-step explanation:

6 0

2 years ago

At 2:00 PM a car's speedometer reads 30 mi/h. At 2:20 PM it reads 50 mi/h. Show that at some time between 2:00 and 2:20 the acce

Bad White [126]

Answer:

Let v(t) be the velocity of the car t hours after 2:00 PM. Then $\frac{v(1/3)-v(0)}{1/3-0}=\frac{50 \:{\frac{mi}{h} }-30\:{\frac{mi}{h} }}{1/3\:h-0\:h} = 60 \:{\frac{mi}{h^2} }$ . By the Mean Value Theorem, there is a number c such that $0 < c$ with $v'(c)=60 \:{\frac{mi}{h^2}}$ . Since v'(t) is the acceleration at time t, the acceleration c hours after 2:00 PM is exactly $60 \:{\frac{mi}{h^2}}$ .

Step-by-step explanation:

The Mean Value Theorem says,

Let be a function that satisfies the following hypotheses:

f is continuous on the closed interval [a, b].
f is differentiable on the open interval (a, b).

Then there is a number c in (a, b) such that

$f'(c)=\frac{f(b)-f(a)}{b-a}$

Note that the Mean Value Theorem doesn’t tell us what c is. It only tells us that there is at least one number c that will satisfy the conclusion of the theorem.

By assumption, the car’s speed is continuous and differentiable everywhere. This means we can apply the Mean Value Theorem.

Let v(t) be the velocity of the car t hours after 2:00 PM. Then $v(0 \:h) = 30 \:{\frac{mi}{h} }$ and $v( \frac{1}{3} \:h) = 50 \:{\frac{mi}{h} }$ (note that 20 minutes is $20/60=1/3$ of an hour), so the average rate of change of v on the interval $[0 \:h, \frac{1}{3} \:h]$ is

$\frac{v(1/3)-v(0)}{1/3-0}=\frac{50 \:{\frac{mi}{h} }-30\:{\frac{mi}{h} }}{1/3\:h-0\:h} = 60 \:{\frac{mi}{h^2} }$

We know that acceleration is the derivative of speed. So, by the Mean Value Theorem, there is a time c in $(0 \:h, \frac{1}{3} \:h)$ at which $v'(c)=60 \:{\frac{mi}{h^2}}$ .

c is a time time between 2:00 and 2:20 at which the acceleration is $60 \:{\frac{mi}{h^2}}$ .

4 0

3 years ago